power formula hypothesis testing

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25.2 - power functions, example 25-2 section .

Let's take a look at another example that involves calculating the power of a hypothesis test.

Let $X$ denote the IQ of a randomly selected adult American. Assume, a bit unrealistically, that $X$ is normally distributed with unknown mean $\mu$ and standard deviation 16. Take a random sample of $n=16$ students, so that, after setting the probability of committing a Type I error at $\alpha=0.05$, we can test the null hypothesis $H_0:\mu=100$ against the alternative hypothesis that $H_A:\mu>100$.

What is the power of the hypothesis test if the true population mean were $\mu=108$?

Setting $\alpha$, the probability of committing a Type I error, to 0.05, implies that we should reject the null hypothesis when the test statistic $Z\ge 1.645$, or equivalently, when the observed sample mean is 106.58 or greater:

because we transform the test statistic $Z$ to the sample mean by way of:

$Z=\dfrac{\bar{X}-\mu}{\frac{\sigma}{\sqrt{n}}}\qquad \Rightarrow \bar{X}=\mu+Z\dfrac{\sigma}{\sqrt{n}} \qquad \bar{X}=100+1.645\left(\dfrac{16}{\sqrt{16}}\right)=106.58$

Now, that implies that the power, that is, the probability of rejecting the null hypothesis, when $\mu=108$ is 0.6406 as calculated here (recalling that $Phi(z)$ is standard notation for the cumulative distribution function of the standard normal random variable):

$ \text{Power}=P(\bar{X}\ge 106.58\text{ when } \mu=108) = P\left(Z\ge \dfrac{106.58-108}{\frac{16}{\sqrt{16}}}\right) \\ = P(Z\ge -0.36)=1-P(Z<-0.36)=1-\Phi(-0.36)=1-0.3594=0.6406 $

and illustrated here:

In summary, we have determined that we have (only) a 64.06% chance of rejecting the null hypothesis $H_0:\mu=100$ in favor of the alternative hypothesis $H_A:\mu>100$ if the true unknown population mean is in reality $\mu=108$.

What is the power of the hypothesis test if the true population mean were $\mu=112$?

Because we are setting $\alpha$, the probability of committing a Type I error, to 0.05, we again reject the null hypothesis when the test statistic $Z\ge 1.645$, or equivalently, when the observed sample mean is 106.58 or greater. That means that the probability of rejecting the null hypothesis, when $\mu=112$ is 0.9131 as calculated here:

$ \text{Power}=P(\bar{X}\ge 106.58\text{ when }\mu=112)=P\left(Z\ge \frac{106.58-112}{\frac{16}{\sqrt{16}}}\right) \\ = P(Z\ge -1.36)=1-P(Z<-1.36)=1-\Phi(-1.36)=1-0.0869=0.9131 $

In summary, we have determined that we now have a 91.31% chance of rejecting the null hypothesis $H_0:\mu=100$ in favor of the alternative hypothesis $H_A:\mu>100$ if the true unknown population mean is in reality $\mu=112$. Hmm.... it should make sense that the probability of rejecting the null hypothesis is larger for values of the mean, such as 112, that are far away from the assumed mean under the null hypothesis.

What is the power of the hypothesis test if the true population mean were $\mu=116$?

Again, because we are setting $\alpha$, the probability of committing a Type I error, to 0.05, we reject the null hypothesis when the test statistic $Z\ge 1.645$, or equivalently, when the observed sample mean is 106.58 or greater. That means that the probability of rejecting the null hypothesis, when $\mu=116$ is 0.9909 as calculated here:

$\text{Power}=P(\bar{X}\ge 106.58\text{ when }\mu=116) =P\left(Z\ge \dfrac{106.58-116}{\frac{16}{\sqrt{16}}}\right) = P(Z\ge -2.36)=1-P(Z<-2.36)= 1-\Phi(-2.36)=1-0.0091=0.9909 $

In summary, we have determined that, in this case, we have a 99.09% chance of rejecting the null hypothesis $H_0:\mu=100$ in favor of the alternative hypothesis $H_A:\mu>100$ if the true unknown population mean is in reality $\mu=116$. The probability of rejecting the null hypothesis is the largest yet of those we calculated, because the mean, 116, is the farthest away from the assumed mean under the null hypothesis.

Are you growing weary of this? Let's summarize a few things we've learned from engaging in this exercise:

First and foremost, my instructor can be tedious at times..... errrr, I mean, first and foremost, the power of a hypothesis test depends on the value of the parameter being investigated. In the above, example, the power of the hypothesis test depends on the value of the mean $\mu$.
As the actual mean $\mu$ moves further away from the value of the mean $\mu=100$ under the null hypothesis, the power of the hypothesis test increases.

It's that first point that leads us to what is called the power function of the hypothesis test . If you go back and take a look, you'll see that in each case our calculation of the power involved a step that looks like this:

$\text{Power } =1 - \Phi (z) $ where $z = \frac{106.58 - \mu}{16 / \sqrt{16}} $

That is, if we use the standard notation $K(\mu)$ to denote the power function, as it depends on $\mu$, we have:

$K(\mu) = 1- \Phi \left( \frac{106.58 - \mu}{16 / \sqrt{16}} \right) $

So, the reality is your instructor could have been a whole lot more tedious by calculating the power for every possible value of $\mu$ under the alternative hypothesis! What we can do instead is create a plot of the power function, with the mean $\mu$ on the horizontal axis and the power $K(\mu)$ on the vertical axis. Doing so, we get a plot in this case that looks like this:

Now, what can we learn from this plot? Well:

We can see that $\alpha$ (the probability of a Type I error), $\beta$ (the probability of a Type II error), and $K(\mu)$ are all represented on a power function plot, as illustrated here:

We can see that the probability of a Type I error is $\alpha=K(100)=0.05$, that is, the probability of rejecting the null hypothesis when the null hypothesis is true is 0.05.

We can see the power of a test $K(\mu)$, as well as the probability of a Type II error $\beta(\mu)$, for each possible value of $\mu$.

We can see that $\beta(\mu)=1-K(\mu)$ and vice versa, that is, $K(\mu)=1-\beta(\mu)$.

And we can see graphically that, indeed, as the actual mean $\mu$ moves further away from the null mean $\mu=100$, the power of the hypothesis test increases.

Now, what would do you suppose would happen to the power of our hypothesis test if we were to change our willingness to commit a Type I error? Would the power for a given value of $\mu$ increase, decrease, or remain unchanged? Suppose, for example, that we wanted to set $\alpha=0.01$ instead of $\alpha=0.05$? Let's return to our example to explore this question.

Example 25-2 (continued) Section

Let $X$ denote the IQ of a randomly selected adult American. Assume, a bit unrealistically, that $X$ is normally distributed with unknown mean $\mu$ and standard deviation 16. Take a random sample of $n=16$ students, so that, after setting the probability of committing a Type I error at $\alpha=0.01$, we can test the null hypothesis $H_0:\mu=100$ against the alternative hypothesis that $H_A:\mu>100$.

Setting $\alpha$, the probability of committing a Type I error, to 0.01, implies that we should reject the null hypothesis when the test statistic $Z\ge 2.326$, or equivalently, when the observed sample mean is 109.304 or greater:

$\bar{x} = \mu + z \left( \frac{\sigma}{\sqrt{n}} \right) =100 + 2.326\left( \frac{16}{\sqrt{16}} \right)=109.304 $

That means that the probability of rejecting the null hypothesis, when $\mu=108$ is 0.3722 as calculated here:

So, the power when $\mu=108$ and $\alpha=0.01$ is smaller (0.3722) than the power when $\mu=108$ and $\alpha=0.05$ (0.6406)! Perhaps we can see this graphically:

By the way, we could again alternatively look at the glass as being half-empty. In that case, the probability of a Type II error when $\mu=108$ and $\alpha=0.01$ is $1-0.3722=0.6278$. In this case, the probability of a Type II error is greater than the probability of a Type II error when $\mu=108$ and $\alpha=0.05$.

All of this can be seen graphically by plotting the two power functions, one where $\alpha=0.01$ and the other where $\alpha=0.05$, simultaneously. Doing so, we get a plot that looks like this:

This last example illustrates that, providing the sample size $n$ remains unchanged, a decrease in $\alpha$ causes an increase in $\beta$ , and at least theoretically, if not practically, a decrease in $\beta$ causes an increase in $\alpha$. It turns out that the only way that $\alpha$ and $\beta$ can be decreased simultaneously is by increasing the sample size $n$.

Power and Sample Size Determination

Lisa Sullivan, PhD

Professor of Biosatistics

Boston Univeristy School of Public Health

Introduction

A critically important aspect of any study is determining the appropriate sample size to answer the research question. This module will focus on formulas that can be used to estimate the sample size needed to produce a confidence interval estimate with a specified margin of error (precision) or to ensure that a test of hypothesis has a high probability of detecting a meaningful difference in the parameter.

Studies should be designed to include a sufficient number of participants to adequately address the research question. Studies that have either an inadequate number of participants or an excessively large number of participants are both wasteful in terms of participant and investigator time, resources to conduct the assessments, analytic efforts and so on. These situations can also be viewed as unethical as participants may have been put at risk as part of a study that was unable to answer an important question. Studies that are much larger than they need to be to answer the research questions are also wasteful.

The formulas presented here generate estimates of the necessary sample size(s) required based on statistical criteria. However, in many studies, the sample size is determined by financial or logistical constraints. For example, suppose a study is proposed to evaluate a new screening test for Down Syndrome. Suppose that the screening test is based on analysis of a blood sample taken from women early in pregnancy. In order to evaluate the properties of the screening test (e.g., the sensitivity and specificity), each pregnant woman will be asked to provide a blood sample and in addition to undergo an amniocentesis. The amniocentesis is included as the gold standard and the plan is to compare the results of the screening test to the results of the amniocentesis. Suppose that the collection and processing of the blood sample costs $250 per participant and that the amniocentesis costs $900 per participant. These financial constraints alone might substantially limit the number of women that can be enrolled. Just as it is important to consider both statistical and clinical significance when interpreting results of a statistical analysis, it is also important to weigh both statistical and logistical issues in determining the sample size for a study.

Learning Objectives

After completing this module, the student will be able to:

Provide examples demonstrating how the margin of error, effect size and variability of the outcome affect sample size computations.
Compute the sample size required to estimate population parameters with precision.
Interpret statistical power in tests of hypothesis.
Compute the sample size required to ensure high power when hypothesis testing.

Issues in Estimating Sample Size for Confidence Intervals Estimates

The module on confidence intervals provided methods for estimating confidence intervals for various parameters (e.g., μ , p, ( μ 1 - μ 2 ), μ d , (p 1 -p 2 )). Confidence intervals for every parameter take the following general form:

Point Estimate + Margin of Error

In the module on confidence intervals we derived the formula for the confidence interval for μ as

In practice we use the sample standard deviation to estimate the population standard deviation. Note that there is an alternative formula for estimating the mean of a continuous outcome in a single population, and it is used when the sample size is small (n<30). It involves a value from the t distribution, as opposed to one from the standard normal distribution, to reflect the desired level of confidence. When performing sample size computations, we use the large sample formula shown here. [Note: The resultant sample size might be small, and in the analysis stage, the appropriate confidence interval formula must be used.]

The point estimate for the population mean is the sample mean and the margin of error is

In planning studies, we want to determine the sample size needed to ensure that the margin of error is sufficiently small to be informative. For example, suppose we want to estimate the mean weight of female college students. We conduct a study and generate a 95% confidence interval as follows 125 + 40 pounds, or 85 to 165 pounds. The margin of error is so wide that the confidence interval is uninformative. To be informative, an investigator might want the margin of error to be no more than 5 or 10 pounds (meaning that the 95% confidence interval would have a width (lower limit to upper limit) of 10 or 20 pounds). In order to determine the sample size needed, the investigator must specify the desired margin of error . It is important to note that this is not a statistical issue, but a clinical or a practical one. For example, suppose we want to estimate the mean birth weight of infants born to mothers who smoke cigarettes during pregnancy. Birth weights in infants clearly have a much more restricted range than weights of female college students. Therefore, we would probably want to generate a confidence interval for the mean birth weight that has a margin of error not exceeding 1 or 2 pounds.

The margin of error in the one sample confidence interval for μ can be written as follows:

Our goal is to determine the sample size, n, that ensures that the margin of error, " E ," does not exceed a specified value. We can take the formula above and, with some algebra, solve for n :

First, multipy both sides of the equation by the square root of n . Then cancel out the square root of n from the numerator and denominator on the right side of the equation (since any number divided by itself is equal to 1). This leaves:

Now divide both sides by "E" and cancel out "E" from the numerator and denominator on the left side. This leaves:

Finally, square both sides of the equation to get:

This formula generates the sample size, n , required to ensure that the margin of error, E , does not exceed a specified value. To solve for n , we must input " Z ," " σ ," and " E ."

Z is the value from the table of probabilities of the standard normal distribution for the desired confidence level (e.g., Z = 1.96 for 95% confidence)
E is the margin of error that the investigator specifies as important from a clinical or practical standpoint.
σ is the standard deviation of the outcome of interest.

Sometimes it is difficult to estimate σ . When we use the sample size formula above (or one of the other formulas that we will present in the sections that follow), we are planning a study to estimate the unknown mean of a particular outcome variable in a population. It is unlikely that we would know the standard deviation of that variable. In sample size computations, investigators often use a value for the standard deviation from a previous study or a study done in a different, but comparable, population. The sample size computation is not an application of statistical inference and therefore it is reasonable to use an appropriate estimate for the standard deviation. The estimate can be derived from a different study that was reported in the literature; some investigators perform a small pilot study to estimate the standard deviation. A pilot study usually involves a small number of participants (e.g., n=10) who are selected by convenience, as opposed to by random sampling. Data from the participants in the pilot study can be used to compute a sample standard deviation, which serves as a good estimate for σ in the sample size formula. Regardless of how the estimate of the variability of the outcome is derived, it should always be conservative (i.e., as large as is reasonable), so that the resultant sample size is not too small.

Sample Size for One Sample, Continuous Outcome

In studies where the plan is to estimate the mean of a continuous outcome variable in a single population, the formula for determining sample size is given below:

where Z is the value from the standard normal distribution reflecting the confidence level that will be used (e.g., Z = 1.96 for 95%), σ is the standard deviation of the outcome variable and E is the desired margin of error. The formula above generates the minimum number of subjects required to ensure that the margin of error in the confidence interval for μ does not exceed E .

An investigator wants to estimate the mean systolic blood pressure in children with congenital heart disease who are between the ages of 3 and 5. How many children should be enrolled in the study? The investigator plans on using a 95% confidence interval (so Z=1.96) and wants a margin of error of 5 units. The standard deviation of systolic blood pressure is unknown, but the investigators conduct a literature search and find that the standard deviation of systolic blood pressures in children with other cardiac defects is between 15 and 20. To estimate the sample size, we consider the larger standard deviation in order to obtain the most conservative (largest) sample size.

In order to ensure that the 95% confidence interval estimate of the mean systolic blood pressure in children between the ages of 3 and 5 with congenital heart disease is within 5 units of the true mean, a sample of size 62 is needed. [ Note : We always round up; the sample size formulas always generate the minimum number of subjects needed to ensure the specified precision.] Had we assumed a standard deviation of 15, the sample size would have been n=35. Because the estimates of the standard deviation were derived from studies of children with other cardiac defects, it would be advisable to use the larger standard deviation and plan for a study with 62 children. Selecting the smaller sample size could potentially produce a confidence interval estimate with a larger margin of error.

An investigator wants to estimate the mean birth weight of infants born full term (approximately 40 weeks gestation) to mothers who are 19 years of age and under. The mean birth weight of infants born full-term to mothers 20 years of age and older is 3,510 grams with a standard deviation of 385 grams. How many women 19 years of age and under must be enrolled in the study to ensure that a 95% confidence interval estimate of the mean birth weight of their infants has a margin of error not exceeding 100 grams? Try to work through the calculation before you look at the answer.

Sample Size for One Sample, Dichotomous Outcome

In studies where the plan is to estimate the proportion of successes in a dichotomous outcome variable (yes/no) in a single population, the formula for determining sample size is:

where Z is the value from the standard normal distribution reflecting the confidence level that will be used (e.g., Z = 1.96 for 95%) and E is the desired margin of error. p is the proportion of successes in the population. Here we are planning a study to generate a 95% confidence interval for the unknown population proportion, p . The equation to determine the sample size for determining p seems to require knowledge of p, but this is obviously this is a circular argument, because if we knew the proportion of successes in the population, then a study would not be necessary! What we really need is an approximate value of p or an anticipated value. The range of p is 0 to 1, and therefore the range of p(1-p) is 0 to 1. The value of p that maximizes p(1-p) is p=0.5. Consequently, if there is no information available to approximate p, then p=0.5 can be used to generate the most conservative, or largest, sample size.

Example 2:

An investigator wants to estimate the proportion of freshmen at his University who currently smoke cigarettes (i.e., the prevalence of smoking). How many freshmen should be involved in the study to ensure that a 95% confidence interval estimate of the proportion of freshmen who smoke is within 5% of the true proportion?

Because we have no information on the proportion of freshmen who smoke, we use 0.5 to estimate the sample size as follows:

In order to ensure that the 95% confidence interval estimate of the proportion of freshmen who smoke is within 5% of the true proportion, a sample of size 385 is needed.

Suppose that a similar study was conducted 2 years ago and found that the prevalence of smoking was 27% among freshmen. If the investigator believes that this is a reasonable estimate of prevalence 2 years later, it can be used to plan the next study. Using this estimate of p, what sample size is needed (assuming that again a 95% confidence interval will be used and we want the same level of precision)?

An investigator wants to estimate the prevalence of breast cancer among women who are between 40 and 45 years of age living in Boston. How many women must be involved in the study to ensure that the estimate is precise? National data suggest that 1 in 235 women are diagnosed with breast cancer by age 40. This translates to a proportion of 0.0043 (0.43%) or a prevalence of 43 per 10,000 women. Suppose the investigator wants the estimate to be within 10 per 10,000 women with 95% confidence. The sample size is computed as follows:

A sample of size n=16,448 will ensure that a 95% confidence interval estimate of the prevalence of breast cancer is within 0.10 (or to within 10 women per 10,000) of its true value. This is a situation where investigators might decide that a sample of this size is not feasible. Suppose that the investigators thought a sample of size 5,000 would be reasonable from a practical point of view. How precisely can we estimate the prevalence with a sample of size n=5,000? Recall that the confidence interval formula to estimate prevalence is:

Assuming that the prevalence of breast cancer in the sample will be close to that based on national data, we would expect the margin of error to be approximately equal to the following:

Thus, with n=5,000 women, a 95% confidence interval would be expected to have a margin of error of 0.0018 (or 18 per 10,000). The investigators must decide if this would be sufficiently precise to answer the research question. Note that the above is based on the assumption that the prevalence of breast cancer in Boston is similar to that reported nationally. This may or may not be a reasonable assumption. In fact, it is the objective of the current study to estimate the prevalence in Boston. The research team, with input from clinical investigators and biostatisticians, must carefully evaluate the implications of selecting a sample of size n = 5,000, n = 16,448 or any size in between.

Sample Sizes for Two Independent Samples, Continuous Outcome

In studies where the plan is to estimate the difference in means between two independent populations, the formula for determining the sample sizes required in each comparison group is given below:

where n i is the sample size required in each group (i=1,2), Z is the value from the standard normal distribution reflecting the confidence level that will be used and E is the desired margin of error. σ again reflects the standard deviation of the outcome variable. Recall from the module on confidence intervals that, when we generated a confidence interval estimate for the difference in means, we used Sp, the pooled estimate of the common standard deviation, as a measure of variability in the outcome (based on pooling the data), where Sp is computed as follows:

If data are available on variability of the outcome in each comparison group, then Sp can be computed and used in the sample size formula. However, it is more often the case that data on the variability of the outcome are available from only one group, often the untreated (e.g., placebo control) or unexposed group. When planning a clinical trial to investigate a new drug or procedure, data are often available from other trials that involved a placebo or an active control group (i.e., a standard medication or treatment given for the condition under study). The standard deviation of the outcome variable measured in patients assigned to the placebo, control or unexposed group can be used to plan a future trial, as illustrated below.

Note that the formula for the sample size generates sample size estimates for samples of equal size. If a study is planned where different numbers of patients will be assigned or different numbers of patients will comprise the comparison groups, then alternative formulas can be used.

An investigator wants to plan a clinical trial to evaluate the efficacy of a new drug designed to increase HDL cholesterol (the "good" cholesterol). The plan is to enroll participants and to randomly assign them to receive either the new drug or a placebo. HDL cholesterol will be measured in each participant after 12 weeks on the assigned treatment. Based on prior experience with similar trials, the investigator expects that 10% of all participants will be lost to follow up or will drop out of the study over 12 weeks. A 95% confidence interval will be estimated to quantify the difference in mean HDL levels between patients taking the new drug as compared to placebo. The investigator would like the margin of error to be no more than 3 units. How many patients should be recruited into the study?

The sample sizes are computed as follows:

A major issue is determining the variability in the outcome of interest (σ), here the standard deviation of HDL cholesterol. To plan this study, we can use data from the Framingham Heart Study. In participants who attended the seventh examination of the Offspring Study and were not on treatment for high cholesterol, the standard deviation of HDL cholesterol is 17.1. We will use this value and the other inputs to compute the sample sizes as follows:

Samples of size n 1 =250 and n 2 =250 will ensure that the 95% confidence interval for the difference in mean HDL levels will have a margin of error of no more than 3 units. Again, these sample sizes refer to the numbers of participants with complete data. The investigators hypothesized a 10% attrition (or drop-out) rate (in both groups). In order to ensure that the total sample size of 500 is available at 12 weeks, the investigator needs to recruit more participants to allow for attrition.

N (number to enroll) * (% retained) = desired sample size

Therefore N (number to enroll) = desired sample size/(% retained)

N = 500/0.90 = 556

If they anticipate a 10% attrition rate, the investigators should enroll 556 participants. This will ensure N=500 with complete data at the end of the trial.

An investigator wants to compare two diet programs in children who are obese. One diet is a low fat diet, and the other is a low carbohydrate diet. The plan is to enroll children and weigh them at the start of the study. Each child will then be randomly assigned to either the low fat or the low carbohydrate diet. Each child will follow the assigned diet for 8 weeks, at which time they will again be weighed. The number of pounds lost will be computed for each child. Based on data reported from diet trials in adults, the investigator expects that 20% of all children will not complete the study. A 95% confidence interval will be estimated to quantify the difference in weight lost between the two diets and the investigator would like the margin of error to be no more than 3 pounds. How many children should be recruited into the study?

Again the issue is determining the variability in the outcome of interest (σ), here the standard deviation in pounds lost over 8 weeks. To plan this study, investigators use data from a published study in adults. Suppose one such study compared the same diets in adults and involved 100 participants in each diet group. The study reported a standard deviation in weight lost over 8 weeks on a low fat diet of 8.4 pounds and a standard deviation in weight lost over 8 weeks on a low carbohydrate diet of 7.7 pounds. These data can be used to estimate the common standard deviation in weight lost as follows:

We now use this value and the other inputs to compute the sample sizes:

Samples of size n 1 =56 and n 2 =56 will ensure that the 95% confidence interval for the difference in weight lost between diets will have a margin of error of no more than 3 pounds. Again, these sample sizes refer to the numbers of children with complete data. The investigators anticipate a 20% attrition rate. In order to ensure that the total sample size of 112 is available at 8 weeks, the investigator needs to recruit more participants to allow for attrition.

N = 112/0.80 = 140

Sample Size for Matched Samples, Continuous Outcome

In studies where the plan is to estimate the mean difference of a continuous outcome based on matched data, the formula for determining sample size is given below:

where Z is the value from the standard normal distribution reflecting the confidence level that will be used (e.g., Z = 1.96 for 95%), E is the desired margin of error, and σ d is the standard deviation of the difference scores. It is extremely important that the standard deviation of the difference scores (e.g., the difference based on measurements over time or the difference between matched pairs) is used here to appropriately estimate the sample size.

Sample Sizes for Two Independent Samples, Dichotomous Outcome

In studies where the plan is to estimate the difference in proportions between two independent populations (i.e., to estimate the risk difference), the formula for determining the sample sizes required in each comparison group is:

where n i is the sample size required in each group (i=1,2), Z is the value from the standard normal distribution reflecting the confidence level that will be used (e.g., Z = 1.96 for 95%), and E is the desired margin of error. p 1 and p 2 are the proportions of successes in each comparison group. Again, here we are planning a study to generate a 95% confidence interval for the difference in unknown proportions, and the formula to estimate the sample sizes needed requires p 1 and p 2 . In order to estimate the sample size, we need approximate values of p 1 and p 2 . The values of p 1 and p 2 that maximize the sample size are p 1 =p 2 =0.5. Thus, if there is no information available to approximate p 1 and p 2 , then 0.5 can be used to generate the most conservative, or largest, sample sizes.

Similar to the situation for two independent samples and a continuous outcome at the top of this page, it may be the case that data are available on the proportion of successes in one group, usually the untreated (e.g., placebo control) or unexposed group. If so, the known proportion can be used for both p 1 and p 2 in the formula shown above. The formula shown above generates sample size estimates for samples of equal size. If a study is planned where different numbers of patients will be assigned or different numbers of patients will comprise the comparison groups, then alternative formulas can be used. Interested readers can see Fleiss for more details. 4

An investigator wants to estimate the impact of smoking during pregnancy on premature delivery. Normal pregnancies last approximately 40 weeks and premature deliveries are those that occur before 37 weeks. The 2005 National Vital Statistics report indicates that approximately 12% of infants are born prematurely in the United States. 5 The investigator plans to collect data through medical record review and to generate a 95% confidence interval for the difference in proportions of infants born prematurely to women who smoked during pregnancy as compared to those who did not. How many women should be enrolled in the study to ensure that the 95% confidence interval for the difference in proportions has a margin of error of no more than 4%?

The sample sizes (i.e., numbers of women who smoked and did not smoke during pregnancy) can be computed using the formula shown above. National data suggest that 12% of infants are born prematurely. We will use that estimate for both groups in the sample size computation.

Samples of size n 1 =508 women who smoked during pregnancy and n 2 =508 women who did not smoke during pregnancy will ensure that the 95% confidence interval for the difference in proportions who deliver prematurely will have a margin of error of no more than 4%.

Is attrition an issue here?

Issues in Estimating Sample Size for Hypothesis Testing

In the module on hypothesis testing for means and proportions, we introduced techniques for means, proportions, differences in means, and differences in proportions. While each test involved details that were specific to the outcome of interest (e.g., continuous or dichotomous) and to the number of comparison groups (one, two, more than two), there were common elements to each test. For example, in each test of hypothesis, there are two errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. In the first step of any test of hypothesis, we select a level of significance, α , and α = P(Type I error) = P(Reject H 0 | H 0 is true). Because we purposely select a small value for α , we control the probability of committing a Type I error. The second type of error is called a Type II error and it is defined as the probability we do not reject H 0 when it is false. The probability of a Type II error is denoted β , and β =P(Type II error) = P(Do not Reject H 0 | H 0 is false). In hypothesis testing, we usually focus on power, which is defined as the probability that we reject H 0 when it is false, i.e., power = 1- β = P(Reject H 0 | H 0 is false). Power is the probability that a test correctly rejects a false null hypothesis. A good test is one with low probability of committing a Type I error (i.e., small α ) and high power (i.e., small β, high power).

Here we present formulas to determine the sample size required to ensure that a test has high power. The sample size computations depend on the level of significance, aα, the desired power of the test (equivalent to 1-β), the variability of the outcome, and the effect size. The effect size is the difference in the parameter of interest that represents a clinically meaningful difference. Similar to the margin of error in confidence interval applications, the effect size is determined based on clinical or practical criteria and not statistical criteria.

The concept of statistical power can be difficult to grasp. Before presenting the formulas to determine the sample sizes required to ensure high power in a test, we will first discuss power from a conceptual point of view.

Suppose we want to test the following hypotheses at aα=0.05: H 0 : μ = 90 versus H 1 : μ ≠ 90. To test the hypotheses, suppose we select a sample of size n=100. For this example, assume that the standard deviation of the outcome is σ=20. We compute the sample mean and then must decide whether the sample mean provides evidence to support the alternative hypothesis or not. This is done by computing a test statistic and comparing the test statistic to an appropriate critical value. If the null hypothesis is true (μ=90), then we are likely to select a sample whose mean is close in value to 90. However, it is also possible to select a sample whose mean is much larger or much smaller than 90. Recall from the Central Limit Theorem (see page 11 in the module on Probability), that for large n (here n=100 is sufficiently large), the distribution of the sample means is approximately normal with a mean of

If the null hypothesis is true, it is possible to observe any sample mean shown in the figure below; all are possible under H 0 : μ = 90.

Normal distribution of X when the mean of X is 90. A bell-shaped curve with a value of X-90 at the center.

Rejection Region for Test H 0 : μ = 90 versus H 1 : μ ≠ 90 at α =0.05

Standard normal distribution showing a mean of 90. The rejection areas are in the two tails at the extremes above and below the mean. If the alpha level is 0.05, then each tail accounts for an arean of 0.025.

The areas in the two tails of the curve represent the probability of a Type I Error, α= 0.05. This concept was discussed in the module on Hypothesis Testing.

Now, suppose that the alternative hypothesis, H 1 , is true (i.e., μ ≠ 90) and that the true mean is actually 94. The figure below shows the distributions of the sample mean under the null and alternative hypotheses.The values of the sample mean are shown along the horizontal axis.

Two overlapping normal distributions, one depicting the null hypothesis with a mean of 90 and the other showing the alternative hypothesis with a mean of 94. A more complete explanation of the figure is provided in the text below the figure.

If the true mean is 94, then the alternative hypothesis is true. In our test, we selected α = 0.05 and reject H 0 if the observed sample mean exceeds 93.92 (focusing on the upper tail of the rejection region for now). The critical value (93.92) is indicated by the vertical line. The probability of a Type II error is denoted β, and β = P(Do not Reject H 0 | H 0 is false), i.e., the probability of not rejecting the null hypothesis if the null hypothesis were true. β is shown in the figure above as the area under the rightmost curve (H 1 ) to the left of the vertical line (where we do not reject H 0 ). Power is defined as 1- β = P(Reject H 0 | H 0 is false) and is shown in the figure as the area under the rightmost curve (H 1 ) to the right of the vertical line (where we reject H 0 ).

Note that β and power are related to α, the variability of the outcome and the effect size. From the figure above we can see what happens to β and power if we increase α. Suppose, for example, we increase α to α=0.10.The upper critical value would be 92.56 instead of 93.92. The vertical line would shift to the left, increasing α, decreasing β and increasing power. While a better test is one with higher power, it is not advisable to increase α as a means to increase power. Nonetheless, there is a direct relationship between α and power (as α increases, so does power).

β and power are also related to the variability of the outcome and to the effect size. The effect size is the difference in the parameter of interest (e.g., μ) that represents a clinically meaningful difference. The figure above graphically displays α, β, and power when the difference in the mean under the null as compared to the alternative hypothesis is 4 units (i.e., 90 versus 94). The figure below shows the same components for the situation where the mean under the alternative hypothesis is 98.

Overlapping bell-shaped distributions - one with a mean of 90 and the other with a mean of 98

Notice that there is much higher power when there is a larger difference between the mean under H 0 as compared to H 1 (i.e., 90 versus 98). A statistical test is much more likely to reject the null hypothesis in favor of the alternative if the true mean is 98 than if the true mean is 94. Notice also in this case that there is little overlap in the distributions under the null and alternative hypotheses. If a sample mean of 97 or higher is observed it is very unlikely that it came from a distribution whose mean is 90. In the previous figure for H 0 : μ = 90 and H 1 : μ = 94, if we observed a sample mean of 93, for example, it would not be as clear as to whether it came from a distribution whose mean is 90 or one whose mean is 94.

Ensuring That a Test Has High Power

In designing studies most people consider power of 80% or 90% (just as we generally use 95% as the confidence level for confidence interval estimates). The inputs for the sample size formulas include the desired power, the level of significance and the effect size. The effect size is selected to represent a clinically meaningful or practically important difference in the parameter of interest, as we will illustrate.

The formulas we present below produce the minimum sample size to ensure that the test of hypothesis will have a specified probability of rejecting the null hypothesis when it is false (i.e., a specified power). In planning studies, investigators again must account for attrition or loss to follow-up. The formulas shown below produce the number of participants needed with complete data, and we will illustrate how attrition is addressed in planning studies.

In studies where the plan is to perform a test of hypothesis comparing the mean of a continuous outcome variable in a single population to a known mean, the hypotheses of interest are:

H 0 : μ = μ 0 and H 1 : μ ≠ μ 0 where μ 0 is the known mean (e.g., a historical control). The formula for determining sample size to ensure that the test has a specified power is given below:

where α is the selected level of significance and Z 1-α /2 is the value from the standard normal distribution holding 1- α/2 below it. For example, if α=0.05, then 1- α/2 = 0.975 and Z=1.960. 1- β is the selected power, and Z 1-β is the value from the standard normal distribution holding 1- β below it. Sample size estimates for hypothesis testing are often based on achieving 80% or 90% power. The Z 1-β values for these popular scenarios are given below:

For 80% power Z 0.80 = 0.84
For 90% power Z 0.90 =1.282

ES is the effect size , defined as follows:

where μ 0 is the mean under H 0 , μ 1 is the mean under H 1 and σ is the standard deviation of the outcome of interest. The numerator of the effect size, the absolute value of the difference in means | μ 1 - μ 0 |, represents what is considered a clinically meaningful or practically important difference in means. Similar to the issue we faced when planning studies to estimate confidence intervals, it can sometimes be difficult to estimate the standard deviation. In sample size computations, investigators often use a value for the standard deviation from a previous study or a study performed in a different but comparable population. Regardless of how the estimate of the variability of the outcome is derived, it should always be conservative (i.e., as large as is reasonable), so that the resultant sample size will not be too small.

Example 7:

An investigator hypothesizes that in people free of diabetes, fasting blood glucose, a risk factor for coronary heart disease, is higher in those who drink at least 2 cups of coffee per day. A cross-sectional study is planned to assess the mean fasting blood glucose levels in people who drink at least two cups of coffee per day. The mean fasting blood glucose level in people free of diabetes is reported as 95.0 mg/dL with a standard deviation of 9.8 mg/dL. 7 If the mean blood glucose level in people who drink at least 2 cups of coffee per day is 100 mg/dL, this would be important clinically. How many patients should be enrolled in the study to ensure that the power of the test is 80% to detect this difference? A two sided test will be used with a 5% level of significance.

The effect size is computed as:

The effect size represents the meaningful difference in the population mean - here 95 versus 100, or 0.51 standard deviation units different. We now substitute the effect size and the appropriate Z values for the selected α and power to compute the sample size.

Therefore, a sample of size n=31 will ensure that a two-sided test with α =0.05 has 80% power to detect a 5 mg/dL difference in mean fasting blood glucose levels.

In the planned study, participants will be asked to fast overnight and to provide a blood sample for analysis of glucose levels. Based on prior experience, the investigators hypothesize that 10% of the participants will fail to fast or will refuse to follow the study protocol. Therefore, a total of 35 participants will be enrolled in the study to ensure that 31 are available for analysis (see below).

N (number to enroll) * (% following protocol) = desired sample size

N = 31/0.90 = 35.

Sample Size for One Sample, Dichotomous Outcome

In studies where the plan is to perform a test of hypothesis comparing the proportion of successes in a dichotomous outcome variable in a single population to a known proportion, the hypotheses of interest are:

where p 0 is the known proportion (e.g., a historical control). The formula for determining the sample size to ensure that the test has a specified power is given below:

where α is the selected level of significance and Z 1-α /2 is the value from the standard normal distribution holding 1- α/2 below it. 1- β is the selected power and Z 1-β is the value from the standard normal distribution holding 1- β below it , and ES is the effect size, defined as follows:

where p 0 is the proportion under H 0 and p 1 is the proportion under H 1 . The numerator of the effect size, the absolute value of the difference in proportions |p 1 -p 0 |, again represents what is considered a clinically meaningful or practically important difference in proportions.

Example 8:

A recent report from the Framingham Heart Study indicated that 26% of people free of cardiovascular disease had elevated LDL cholesterol levels, defined as LDL > 159 mg/dL. 9 An investigator hypothesizes that a higher proportion of patients with a history of cardiovascular disease will have elevated LDL cholesterol. How many patients should be studied to ensure that the power of the test is 90% to detect a 5% difference in the proportion with elevated LDL cholesterol? A two sided test will be used with a 5% level of significance.

We first compute the effect size:

We now substitute the effect size and the appropriate Z values for the selected α and power to compute the sample size.

A sample of size n=869 will ensure that a two-sided test with α =0.05 has 90% power to detect a 5% difference in the proportion of patients with a history of cardiovascular disease who have an elevated LDL cholesterol level.

A medical device manufacturer produces implantable stents. During the manufacturing process, approximately 10% of the stents are deemed to be defective. The manufacturer wants to test whether the proportion of defective stents is more than 10%. If the process produces more than 15% defective stents, then corrective action must be taken. Therefore, the manufacturer wants the test to have 90% power to detect a difference in proportions of this magnitude. How many stents must be evaluated? For you computations, use a two-sided test with a 5% level of significance. (Do the computation yourself, before looking at the answer.)

In studies where the plan is to perform a test of hypothesis comparing the means of a continuous outcome variable in two independent populations, the hypotheses of interest are:

where μ 1 and μ 2 are the means in the two comparison populations. The formula for determining the sample sizes to ensure that the test has a specified power is:

where n i is the sample size required in each group (i=1,2), α is the selected level of significance and Z 1-α /2 is the value from the standard normal distribution holding 1- α /2 below it, and 1- β is the selected power and Z 1-β is the value from the standard normal distribution holding 1- β below it. ES is the effect size, defined as:

where | μ 1 - μ 2 | is the absolute value of the difference in means between the two groups expected under the alternative hypothesis, H 1 . σ is the standard deviation of the outcome of interest. Recall from the module on Hypothesis Testing that, when we performed tests of hypothesis comparing the means of two independent groups, we used Sp, the pooled estimate of the common standard deviation, as a measure of variability in the outcome.

Sp is computed as follows:

If data are available on variability of the outcome in each comparison group, then Sp can be computed and used to generate the sample sizes. However, it is more often the case that data on the variability of the outcome are available from only one group, usually the untreated (e.g., placebo control) or unexposed group. When planning a clinical trial to investigate a new drug or procedure, data are often available from other trials that may have involved a placebo or an active control group (i.e., a standard medication or treatment given for the condition under study). The standard deviation of the outcome variable measured in patients assigned to the placebo, control or unexposed group can be used to plan a future trial, as illustrated.

Note also that the formula shown above generates sample size estimates for samples of equal size. If a study is planned where different numbers of patients will be assigned or different numbers of patients will comprise the comparison groups, then alternative formulas can be used (see Howell 3 for more details).

An investigator is planning a clinical trial to evaluate the efficacy of a new drug designed to reduce systolic blood pressure. The plan is to enroll participants and to randomly assign them to receive either the new drug or a placebo. Systolic blood pressures will be measured in each participant after 12 weeks on the assigned treatment. Based on prior experience with similar trials, the investigator expects that 10% of all participants will be lost to follow up or will drop out of the study. If the new drug shows a 5 unit reduction in mean systolic blood pressure, this would represent a clinically meaningful reduction. How many patients should be enrolled in the trial to ensure that the power of the test is 80% to detect this difference? A two sided test will be used with a 5% level of significance.

In order to compute the effect size, an estimate of the variability in systolic blood pressures is needed. Analysis of data from the Framingham Heart Study showed that the standard deviation of systolic blood pressure was 19.0. This value can be used to plan the trial.

The effect size is:

Samples of size n 1 =232 and n 2 = 232 will ensure that the test of hypothesis will have 80% power to detect a 5 unit difference in mean systolic blood pressures in patients receiving the new drug as compared to patients receiving the placebo. However, the investigators hypothesized a 10% attrition rate (in both groups), and to ensure a total sample size of 232 they need to allow for attrition.

N = 232/0.90 = 258.

The investigator must enroll 258 participants to be randomly assigned to receive either the new drug or placebo.

An investigator is planning a study to assess the association between alcohol consumption and grade point average among college seniors. The plan is to categorize students as heavy drinkers or not using 5 or more drinks on a typical drinking day as the criterion for heavy drinking. Mean grade point averages will be compared between students classified as heavy drinkers versus not using a two independent samples test of means. The standard deviation in grade point averages is assumed to be 0.42 and a meaningful difference in grade point averages (relative to drinking status) is 0.25 units. How many college seniors should be enrolled in the study to ensure that the power of the test is 80% to detect a 0.25 unit difference in mean grade point averages? Use a two-sided test with a 5% level of significance.

Answer

In studies where the plan is to perform a test of hypothesis on the mean difference in a continuous outcome variable based on matched data, the hypotheses of interest are:

where μ d is the mean difference in the population. The formula for determining the sample size to ensure that the test has a specified power is given below:

where α is the selected level of significance and Z 1-α/2 is the value from the standard normal distribution holding 1- α/2 below it, 1- β is the selected power and Z 1-β is the value from the standard normal distribution holding 1- β below it and ES is the effect size, defined as follows:

where μ d is the mean difference expected under the alternative hypothesis, H 1 , and σ d is the standard deviation of the difference in the outcome (e.g., the difference based on measurements over time or the difference between matched pairs).

Example 10:

An investigator wants to evaluate the efficacy of an acupuncture treatment for reducing pain in patients with chronic migraine headaches. The plan is to enroll patients who suffer from migraine headaches. Each will be asked to rate the severity of the pain they experience with their next migraine before any treatment is administered. Pain will be recorded on a scale of 1-100 with higher scores indicative of more severe pain. Each patient will then undergo the acupuncture treatment. On their next migraine (post-treatment), each patient will again be asked to rate the severity of the pain. The difference in pain will be computed for each patient. A two sided test of hypothesis will be conducted, at α =0.05, to assess whether there is a statistically significant difference in pain scores before and after treatment. How many patients should be involved in the study to ensure that the test has 80% power to detect a difference of 10 units on the pain scale? Assume that the standard deviation in the difference scores is approximately 20 units.

First compute the effect size:

Then substitute the effect size and the appropriate Z values for the selected α and power to compute the sample size.

A sample of size n=32 patients with migraine will ensure that a two-sided test with α =0.05 has 80% power to detect a mean difference of 10 points in pain before and after treatment, assuming that all 32 patients complete the treatment.

Sample Sizes for Two Independent Samples, Dichotomous Outcomes

In studies where the plan is to perform a test of hypothesis comparing the proportions of successes in two independent populations, the hypotheses of interest are:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2

where p 1 and p 2 are the proportions in the two comparison populations. The formula for determining the sample sizes to ensure that the test has a specified power is given below:

where n i is the sample size required in each group (i=1,2), α is the selected level of significance and Z 1-α/2 is the value from the standard normal distribution holding 1- α/2 below it, and 1- β is the selected power and Z 1-β is the value from the standard normal distribution holding 1- β below it. ES is the effect size, defined as follows:

where |p 1 - p 2 | is the absolute value of the difference in proportions between the two groups expected under the alternative hypothesis, H 1 , and p is the overall proportion, based on pooling the data from the two comparison groups (p can be computed by taking the mean of the proportions in the two comparison groups, assuming that the groups will be of approximately equal size).

Example 11:

An investigator hypothesizes that there is a higher incidence of flu among students who use their athletic facility regularly than their counterparts who do not. The study will be conducted in the spring. Each student will be asked if they used the athletic facility regularly over the past 6 months and whether or not they had the flu. A test of hypothesis will be conducted to compare the proportion of students who used the athletic facility regularly and got flu with the proportion of students who did not and got flu. During a typical year, approximately 35% of the students experience flu. The investigators feel that a 30% increase in flu among those who used the athletic facility regularly would be clinically meaningful. How many students should be enrolled in the study to ensure that the power of the test is 80% to detect this difference in the proportions? A two sided test will be used with a 5% level of significance.

We first compute the effect size by substituting the proportions of students in each group who are expected to develop flu, p 1 =0.46 (i.e., 0.35*1.30=0.46) and p 2 =0.35 and the overall proportion, p=0.41 (i.e., (0.46+0.35)/2):

We now substitute the effect size and the appropriate Z values for the selected α and power to compute the sample size.

Samples of size n 1 =324 and n 2 =324 will ensure that the test of hypothesis will have 80% power to detect a 30% difference in the proportions of students who develop flu between those who do and do not use the athletic facilities regularly.

Donor Feces? Really? Clostridium difficile (also referred to as "C. difficile" or "C. diff.") is a bacterial species that can be found in the colon of humans, although its numbers are kept in check by other normal flora in the colon. Antibiotic therapy sometimes diminishes the normal flora in the colon to the point that C. difficile flourishes and causes infection with symptoms ranging from diarrhea to life-threatening inflammation of the colon. Illness from C. difficile most commonly affects older adults in hospitals or in long term care facilities and typically occurs after use of antibiotic medications. In recent years, C. difficile infections have become more frequent, more severe and more difficult to treat. Ironically, C. difficile is first treated by discontinuing antibiotics, if they are still being prescribed. If that is unsuccessful, the infection has been treated by switching to another antibiotic. However, treatment with another antibiotic frequently does not cure the C. difficile infection. There have been sporadic reports of successful treatment by infusing feces from healthy donors into the duodenum of patients suffering from C. difficile. (Yuk!) This re-establishes the normal microbiota in the colon, and counteracts the overgrowth of C. diff. The efficacy of this approach was tested in a randomized clinical trial reported in the New England Journal of Medicine (Jan. 2013). The investigators planned to randomly assign patients with recurrent C. difficile infection to either antibiotic therapy or to duodenal infusion of donor feces. In order to estimate the sample size that would be needed, the investigators assumed that the feces infusion would be successful 90% of the time, and antibiotic therapy would be successful in 60% of cases. How many subjects will be needed in each group to ensure that the power of the study is 80% with a level of significance α = 0.05?

Determining the appropriate design of a study is more important than the statistical analysis; a poorly designed study can never be salvaged, whereas a poorly analyzed study can be re-analyzed. A critical component in study design is the determination of the appropriate sample size. The sample size must be large enough to adequately answer the research question, yet not too large so as to involve too many patients when fewer would have sufficed. The determination of the appropriate sample size involves statistical criteria as well as clinical or practical considerations. Sample size determination involves teamwork; biostatisticians must work closely with clinical investigators to determine the sample size that will address the research question of interest with adequate precision or power to produce results that are clinically meaningful.

The following table summarizes the sample size formulas for each scenario described here. The formulas are organized by the proposed analysis, a confidence interval estimate or a test of hypothesis.

Buschman NA, Foster G, Vickers P. Adolescent girls and their babies: achieving optimal birth weight. Gestational weight gain and pregnancy outcome in terms of gestation at delivery and infant birth weight: a comparison between adolescents under 16 and adult women. Child: Care, Health and Development. 2001; 27(2):163-171.
Feuer EJ, Wun LM. DEVCAN: Probability of Developing or Dying of Cancer. Version 4.0 .Bethesda, MD: National Cancer Institute, 1999.
Howell DC. Statistical Methods for Psychology. Boston, MA: Duxbury Press, 1982.
Fleiss JL. Statistical Methods for Rates and Proportions. New York, NY: John Wiley and Sons, Inc.,1981.
National Center for Health Statistics. Health, United States, 2005 with Chartbook on Trends in the Health of Americans. Hyattsville, MD : US Government Printing Office; 2005.
Plaskon LA, Penson DF, Vaughan TL, Stanford JL. Cigarette smoking and risk of prostate cancer in middle-aged men. Cancer Epidemiology Biomarkers & Prevention. 2003; 12: 604-609.
Rutter MK, Meigs JB, Sullivan LM, D'Agostino RB, Wilson PW. C-reactive protein, the metabolic syndrome and prediction of cardiovascular events in the Framingham Offspring Study. Circulation. 2004;110: 380-385.
Ramachandran V, Sullivan LM, Wilson PW, Sempos CT, Sundstrom J, Kannel WB, Levy D, D'Agostino RB. Relative importance of borderline and elevated levels of coronary heart disease risk factors. Annals of Internal Medicine. 2005; 142: 393-402.
Wechsler H, Lee JE, Kuo M, Lee H. College Binge Drinking in the 1990s:A Continuing Problem Results of the Harvard School of Public Health 1999 College Health, 2000; 48: 199-210.

Answers to Selected Problems

Answer to birth weight question - page 3.

In order to ensure that the 95% confidence interval estimate of the mean birthweight is within 100 grams of the true mean, a sample of size 57 is needed. In planning the study, the investigator must consider the fact that some women may deliver prematurely. If women are enrolled into the study during pregnancy, then more than 57 women will need to be enrolled so that after excluding those who deliver prematurely, 57 with outcome information will be available for analysis. For example, if 5% of the women are expected to delivery prematurely (i.e., 95% will deliver full term), then 60 women must be enrolled to ensure that 57 deliver full term. The number of women that must be enrolled, N, is computed as follows:

N (number to enroll) * (% retained) = desired sample size

N (0.95) = 57

N = 57/0.95 = 60.

Answer Freshmen Smoking - Page 4

In order to ensure that the 95% confidence interval estimate of the proportion of freshmen who smoke is within 5% of the true proportion, a sample of size 303 is needed. Notice that this sample size is substantially smaller than the one estimated above. Having some information on the magnitude of the proportion in the population will always produce a sample size that is less than or equal to the one based on a population proportion of 0.5. However, the estimate must be realistic.

Answer to Medical Device Problem - Page 7

Then substitute the effect size and the appropriate z values for the selected alpha and power to comute the sample size.

A sample size of 364 stents will ensure that a two-sided test with α=0.05 has 90% power to detect a 0.05, or 5%, difference in jthe proportion of defective stents produced.

Answer to Alcohol and GPA - Page 8

First compute the effect size.

Now substitute the effect size and the appropriate z values for alpha and power to compute the sample size.

Sample sizes of n i =44 heavy drinkers and 44 who drink few fewer than five drinks per typical drinking day will ensure that the test of hypothesis has 80% power to detect a 0.25 unit difference in mean grade point averages.

Answer to Donor Feces - Page 8

We first compute the effect size by substituting the proportions of patients expected to be cured with each treatment, p 1 =0.6 and p 2 =0.9, and the overall proportion, p=0.75:

We now substitute the effect size and the appropriate Z values for the selected a and power to compute the sample size.

Samples of size n 1 =33 and n 2 =33 will ensure that the test of hypothesis will have 80% power to detect this difference in the proportions of patients who are cured of C. diff. by feces infusion versus antibiotic therapy.

In fact, the investigators enrolled 38 into each group to allow for attrition. Nevertheless, the study was stopped after an interim analysis. Of 16 patients in the infusion group, 13 (81%) had resolution of C. difficile–associated diarrhea after the first infusion. The 3 remaining patients received a second infusion with feces from a different donor, with resolution in 2 patients. Resolution of C. difficile infection occurred in only 4 of 13 patients (31%) receiving the antibiotic vancomycin.

Teach yourself statistics

How to Find the Power of a Statistical Test

When a researcher designs a study to test a hypothesis, he/she should compute the power of the test (i.e., the likelihood of avoiding a Type II error).

How to Compute the Power of a Hypothesis Test

To compute the power of a hypothesis test, use the following three-step procedure.

Define the region of acceptance . Previously, we showed how to compute the region of acceptance for a hypothesis test.
Specify the critical parameter value. The critical parameter value is an alternative to the value specified in the null hypothesis. The difference between the critical parameter value and the value from the null hypothesis is called the effect size . That is, the effect size is equal to the critical parameter value minus the value from the null hypothesis.
Compute power. Assume that the true population parameter is equal to the critical parameter value, rather than the value specified in the null hypothesis. Based on that assumption, compute the probability that the sample estimate of the population parameter will fall outside the region of acceptance. That probability is the power of the test.

The following examples illustrate how this works. The first example involves a mean score; and the second example, a proportion.

Sample Size Calculator

The steps required to compute the power of a hypothesis test can be time-consuming and complex. Stat Trek's Sample Size Calculator does this work for you - quickly and accurately. The calculator is easy to use, and it is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.

Example 1: Power of the Hypothesis Test of a Mean Score

Two inventors have developed a new, energy-efficient lawn mower engine. One inventor says that the engine will run continuously for 5 hours (300 minutes) on a single ounce of regular gasoline. Suppose a random sample of 50 engines is tested. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. The inventor tests the null hypothesis that the mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300 minutes, using a 0.05 level of significance.

The other inventor says that the new engine will run continuously for only 290 minutes on a ounce of gasoline. Find the power of the test to reject the null hypothesis, if the second inventor is correct.

Solution: The steps required to compute power are presented below.

Define the region of acceptance . In a previous lesson, we showed that the region of acceptance for this problem consists of the values between 294.46 and 305.54 (see previous lesson ).
Specify the critical parameter value . The null hypothesis tests the hypothesis that the run time of the engine is 300 minutes. We are interested in determining the probability that the hypothesis test will reject the null hypothesis, if the true run time is actually 290 minutes. Therefore, the critical parameter value is 290. (Another way to express the critical parameter value is through effect size. The effect size is equal to the critical parameter value minus the hypothesized value. Thus, effect size is equal to 290 - 300 or -10.)

Therefore, we need to compute the probability that the sampled run time will be less than 294.46 or greater than 305.54. To do this, we make the following assumptions:

The sampling distribution of the mean is normally distributed. (Because the sample size is relatively large, this assumption can be justified by the central limit theorem .)
The mean of the sampling distribution is the critical parameter value, 290.
The standard error of the sampling distribution is 2.83. The standard error of the sampling distribution was computed in a previous lesson (see previous lesson ).

Given these assumptions, we first assess the probability that the sample run time will be less than 294.46. This is easy to do, using the Normal Calculator . We enter the following values into the calculator: normal random variable = 294.46; mean = 290; and standard deviation = 2.83. Given these inputs, we find that the cumulative probability is 0.942. This means the probability that the sample mean will be less than 294.46 is 0.942.

Next, we assess the probability that the sample mean is greater than 305.54. Again, we use the Normal Calculator . We enter the following values into the calculator: normal random variable = 305.54; mean = 290; and standard deviation = 2.83. Given these inputs, we find that the probability that the sample mean is less than 305.54 (i.e., the cumulative probability) is 1.0. Thus, the probability that the sample mean is greater than 305.54 is 1 - 1.0 or 0.0.

Example 2: Power of the Hypothesis Test of a Proportion

A major corporation offers a large bonus to all of its employees if at least 80 percent of the corporation's 1,000,000 customers are very satisfied. The company conducts a survey of 100 randomly sampled customers to determine whether or not to pay the bonus. The null hypothesis states that the proportion of very satisfied customers is at least 0.80. If the null hypothesis cannot be rejected, given a significance level of 0.05, the company pays the bonus.

Suppose the true proportion of satisfied customers is 0.75. Find the power of the test to reject the null hypothesis.

Define the region of acceptance . In a previous lesson, we showed that the region of acceptance for this problem consists of the values between 0.734 and 1.00. (see previous lesson ).
Specify the critical parameter value . The null hypothesis tests the hypothesis that the proportion of very satisfied customers is 0.80. We are interested in determining the probability that the hypothesis test will reject the null hypothesis, if the true satisfaction level is 0.75. Therefore, the critical parameter value is 0.75. (Another way to express the critical parameter value is through effect size. The effect size is equal to the critical parameter value minus the hypothesized value. Thus, effect size is equal to [0.75 - 0.80] or - 0.05.)

Therefore, we need to compute the probability that the sample proportion will be less than 0.734. To do this, we take the following steps:

Assume that the sampling distribution of the mean is normally distributed. (Because the sample size is relatively large, this assumption can be justified by the central limit theorem .)
Assume that the mean of the sampling distribution is the critical parameter value, 0.75. (This assumption is justified because, for the purpose of calculating power, we assume that the true population proportion is equal to the critical parameter value. And the mean of all possible sample proportions is equal to the population proportion. Hence, the mean of the sampling distribution is equal to the critical parameter value.)

σ P = sqrt[ P * ( 1 - P ) / n ]

σ P = sqrt[ ( 0.75 * 0.25 ) / 100 ] = 0.0433

Power Analysis

Power analysis is an important aspect of experimental design. It allows us to determine the sample size required to detect an effect of a given size with a given degree of confidence. Conversely, it allows us to determine the probability of detecting an effect of a given size with a given level of confidence, under sample size constraints. If the probability is unacceptably low, we would be wise to alter or abandon the experiment.

The following four quantities have an intimate relationship:

sample size
effect size
significance level = P(Type I error) = probability of finding an effect that is not there
power = 1 - P(Type II error) = probability of finding an effect that is there

Given any three, we can determine the fourth.

Power Analysis in R

The pwr package develped by Stéphane Champely, impliments power analysis as outlined by Cohen (!988) . Some of the more important functions are listed below.

For each of these functions, you enter three of the four quantities (effect size, sample size, significance level, power) and the fourth is calculated.

The significance level defaults to 0.05. Therefore, to calculate the significance level, given an effect size, sample size, and power, use the option "sig.level=NULL".

Specifying an effect size can be a daunting task. ES formulas and Cohen's suggestions (based on social science research) are provided below. Cohen's suggestions should only be seen as very rough guidelines. Your own subject matter experience should be brought to bear.

(To explore confidence intervals and drawing conclusions from samples try this interactive course on the foundations of inference.)

For t-tests, use the following functions:

pwr.t.test(n = , d = , sig.level = , power = , type = c("two.sample", "one.sample", "paired"))

where n is the sample size, d is the effect size, and type indicates a two-sample t-test, one-sample t-test or paired t-test. If you have unequal sample sizes, use

pwr.t2n.test(n1 = , n2= , d = , sig.level =, power = )

where n1 and n2 are the sample sizes.

For t-tests, the effect size is assessed as

Cohen suggests that d values of 0.2, 0.5, and 0.8 represent small, medium, and large effect sizes respectively.

You can specify alternative="two.sided", "less", or "greater" to indicate a two-tailed, or one-tailed test. A two tailed test is the default.

For a one-way analysis of variance use

pwr.anova.test(k = , n = , f = , sig.level = , power = )

where k is the number of groups and n is the common sample size in each group.

For a one-way ANOVA effect size is measured by f where

Correlations

For correlation coefficients use

pwr.r.test(n = , r = , sig.level = , power = )

where n is the sample size and r is the correlation. We use the population correlation coefficient as the effect size measure. Cohen suggests that r values of 0.1, 0.3, and 0.5 represent small, medium, and large effect sizes respectively.

Linear Models

For linear models (e.g., multiple regression) use

pwr.f2.test(u =, v = , f2 = , sig.level = , power = )

where u and v are the numerator and denominator degrees of freedom. We use f2 as the effect size measure.

The first formula is appropriate when we are evaluating the impact of a set of predictors on an outcome. The second formula is appropriate when we are evaluating the impact of one set of predictors above and beyond a second set of predictors (or covariates). Cohen suggests f2 values of 0.02, 0.15, and 0.35 represent small, medium, and large effect sizes.

Tests of Proportions

When comparing two proportions use

pwr.2p.test(h = , n = , sig.level =, power = )

where h is the effect size and n is the common sample size in each group.

Cohen suggests that h values of 0.2, 0.5, and 0.8 represent small, medium, and large effect sizes respectively.

For unequal n's use

pwr.2p2n.test(h = , n1 = , n2 = , sig.level = , power = )

To test a single proportion use

pwr.p.test(h = , n = , sig.level = power = )

For both two sample and one sample proportion tests, you can specify alternative="two.sided", "less", or "greater" to indicate a two-tailed, or one-tailed test. A two tailed test is the default.

Chi-square Tests

For chi-square tests use

pwr.chisq.test(w =, N = , df = , sig.level =, power = )

where w is the effect size, N is the total sample size, and df is the degrees of freedom. The effect size w is defined as

Cohen suggests that w values of 0.1, 0.3, and 0.5 represent small, medium, and large effect sizes respectively.

Some Examples

Creating power or sample size plots.

The functions in the pwr package can be used to generate power and sample size graphs.

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Knowledge Base

Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a or H 1 ).
Collect data in a way designed to test the hypothesis.
Perform an appropriate statistical test .
Decide whether to reject or fail to reject your null hypothesis.
Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

Step 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

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For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

an estimate of the difference in average height between the two groups.
a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

Normal distribution
Descriptive statistics
Measures of central tendency
Correlation coefficient

Methodology

Cluster sampling
Stratified sampling
Types of interviews
Cohort study
Thematic analysis

Research bias

Implicit bias
Cognitive bias
Survivorship bias
Availability heuristic
Nonresponse bias
Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

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Power Calculation

Suppose we wish to determine the statistical power of a test of the null hypothesis that the population mean is 100 if in fact the population mean is 105 with a population standard deviation of 10, and we use a sample size of 25 cases and set alpha error to .05 for a one-tailed test.

Statistical power is equal to (1 – beta error), so to find statistical power we can solve for Z β .

We can rearrange the terms in Formula 1 to solve for Z β :

Using the BEAN acronym, we wish to solve for B because power is (1 – beta error). We need to specify the other three terms: E, A, and N.

E ffect size d = ( μ 1 – μ 0 ) / σ = (105 – 100) / 10 = 0.50.

A lpha error rate is set at .05, which corresponds to a Z score of 1.645 for a one-tailed test.

N is set at 25.

Thus, Z β = 1.645 – 5 (0.50) = 1.645 – 2.500 = -0.855

Z β corresponds to the Z score for the critical value on the pink (alternate hypothesis) sampling distribution. We will reject the null hypothesis if we observe a score greater than this critical value. We consult a Z table or the WISE p - z converter applet to find that the probability of observing a Z score greater than ‑0.855 is .804. This is the value given by the WISE power applet.

Thus, in this scenario our statistical power is about 80%. If we collect data and conduct a test of statistical significance, there is an 80% chance that the test will attain statistical significance, and a 20% chance that the test will fail to detect statistical significance.

Sample Size Calculation

We can rearrange the terms in Formula 1 to solve for n .

Suppose we wish to have a sample large enough to have power of 90% to detect a difference between means of 4.0 where the standard deviation is 10.0, using one-tailed alpha error rate of .01. Recalling BEAN, we need to specify B, E, and A to solve for N.

B : Z β corresponds to the Z score on the pink distribution where 90% of the distribution falls above that score. We can use the WISE p - z converter or a Z table to find Z β = -1.282.
E : The effect size d = ( μ 1 – μ 0 ) / σ = (4.0 / 10.0) = 0.40.
A : The alpha error rate of .01 corresponds to Z α = 2.326.

Applying the formula for n :

Thus, we need a sample size of about 82 to attain the desired level of power in this scenario.

It is important to note carefully that the sign on Z β is often negative. Subtracting a negative value is equivalent to adding a positive value.

Questions, comments, difficulties? See our technical support page or contact us: [email protected] .

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Power of a hypothesis test
How to Get the Power of Test in Hypothesis Testing with Binomial
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Hypothesis Testing Formula
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Power of a Hypothesis Test

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Statistical Inference
AP Statistics: Power of a Hypothesis Test
Normal Distribution
Hypothesis Testing Statistics
Class 2: Logistic Model: OR formula, hypothesis testing/interval estimation; EVW Model
Hypothesis Testing

COMMENTS

25.2
In the above, example, the power of the hypothesis test depends on the value of the mean $\mu$. As the actual mean $\mu$ moves further away from the value of the mean $\mu=100$ under the null hypothesis, the power of the hypothesis test increases. It's that first point that leads us to what is called the power function of the hypothesis ...
Power and Sample Size Determination
The inputs for the sample size formulas include the desired power, the level of significance and the effect size. ... Samples of size n 1 =232 and n 2 = 232 will ensure that the test of hypothesis will have 80% power to detect a 5 unit difference in mean systolic blood pressures in patients receiving the new drug as compared to patients ...
Power of Hypothesis Test
Effect Size. To compute the power of the test, one offers an alternative view about the "true" value of the population parameter, assuming that the null hypothesis is false. The effect size is the difference between the true value and the value specified in the null hypothesis. Effect size = True value - Hypothesized value.
Finding the Power of a Hypothesis Test
To calculate power, you basically work two problems back-to-back. First, find a percentile assuming that H 0 is true. Then, turn it around and find the probability that you'd get that value assuming H 0 is false (and instead H a is true). Assume that H 0 is true, and. Find the percentile value corresponding to.
Power of a test
Power of a test. In statistics, the power of a binary hypothesis test is the probability that the test correctly rejects the null hypothesis ( ) when a specific alternative hypothesis ( ) is true. It is commonly denoted by , and represents the chances of a true positive detection conditional on the actual existence of an effect to detect.
How to Find the Power of a Statistical Test
Compute power. The power of the test is the probability of rejecting the null hypothesis, assuming that the true population proportion is equal to the critical parameter value. Since the region of acceptance is 0.734 to 1.00, the null hypothesis will be rejected when the sample proportion is less than 0.734.
PDF 6: THE POWER FUNCTION
6: THE POWER FUNCTION-b The power function of a hypothesis test is the pro ability of rejecting H. This will be a function of t 0 he true value of the parameter. For example, if the, t parameter is the mean µ of a normal distribution hen we write K 1(µ) for the power function, which 0 e m is the probability of rejecting H, given that the tru ...
Quick-R: Power Analysis
The second formula is appropriate when we are evaluating the impact of one set of predictors above and beyond a second set of predictors (or covariates). ... pwr.2p2n.test(h = , n1 = , n2 = , sig.level = , power = ) To test a single proportion use. pwr.p.test(h = , n = , sig.level = power = ) For both two sample and one sample proportion tests ...
Power in Tests of Significance
Power is the probability of making a correct decision (to reject the null hypothesis) when the null hypothesis is false. Power is the probability that a test of significance will pick up on an effect that is present. Power is the probability that a test of significance will detect a deviation from the null hypothesis, should such a deviation exist.
What is Power in Statistics?
High statistical power occurs when a hypothesis test is likely to find an effect that exists in the population. A low power test is unlikely to detect that effect. For example, if statistical power is 80%, a hypothesis test has an 80% chance of detecting an effect that actually exists. Now imagine you're performing a study that has only 10%.
Statistical Power: What it is, How to Calculate it
Power analysis is a method for finding statistical power: the probability of finding an effect, assuming that the effect is actually there. To put it another way, power is the probability of rejecting a null hypothesis when it's false. Note that power is different from a Type II error, which happens when you fail to reject a false null ...
PDF Power and Sample Size Determination
Solution. as z = 1:96 cuts o the bottom =2 = 0:025 of the area. The power when = 3:30 is essentially just the area to the left of a under the normal density centered at 3.30 as the area in the right part of the rejection region is essentially 0. Normal(3:30; 0:15= n) density.
Introduction to power in significance tests
Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: https://www.khanacademy.org/math/ap-statistics/xfb5d8e68:infere...
Hypothesis Testing
There are 5 main steps in hypothesis testing: State your research hypothesis as a null hypothesis and alternate hypothesis (H o) and (H a or H 1 ). Collect data in a way designed to test the hypothesis. Perform an appropriate statistical test. Decide whether to reject or fail to reject your null hypothesis. Present the findings in your results ...
How to Calculate Sample Size Needed for Power
Statistical power and sample size analysis provides both numeric and graphical results, as shown below. The text output indicates that we need 15 samples per group (total of 30) to have a 90% chance of detecting a difference of 5 units. The dot on the Power Curve corresponds to the information in the text output.
Power Of A Test
The power of a test formula is the probability of rejecting the null hypothesis with the assumption that the true population fraction is equal to the critical parameter value. The region of acceptance is 0.734 to 1.00. Therefore, the null hypothesis will be rejected when the sample proportion is less than 0.734.
Power
If we collect data and conduct a test of statistical significance, there is an 80% chance that the test will attain statistical significance, and a 20% chance that the test will fail to detect statistical significance. Sample Size Calculation. We can rearrange the terms in Formula 1 to solve for n.
A Gentle Introduction to Statistical Power and Power Analysis in Python
The statistical power of a hypothesis test is the probability of detecting an effect, if there is a true effect present to detect. ... Just some suggestions from a reader perspective, maybe discuss what is the formula to calculate sample size given the other 3 factors in power analysis. It would be much better for someone to understand it ...
Formula for calculating power in hypothesis testing
Formula for calculating power in hypothesis testing. We calculate power in hypothesis testing as 1-beta (probability that null hypothesis is false given that alternative hypothesis is true). As this probability increases, power also increases. By similar relationship can't we say that power could also be calculated as equal to alpha ...

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Example 25-2 (continued) Section

Introduction

Learning Objectives

Issues in Estimating Sample Size for Confidence Intervals Estimates

Sample Size for One Sample, Continuous Outcome

Sample Size for One Sample, Dichotomous Outcome

Sample Sizes for Two Independent Samples, Continuous Outcome

Sample Size for Matched Samples, Continuous Outcome

Sample Sizes for Two Independent Samples, Dichotomous Outcome

Issues in Estimating Sample Size for Hypothesis Testing

Ensuring That a Test Has High Power

Sample Size for One Sample, Dichotomous Outcome

Sample Sizes for Two Independent Samples, Dichotomous Outcomes

Answers to Selected Problems

Answer Freshmen Smoking - Page 4

Answer to Medical Device Problem - Page 7

Answer to Alcohol and GPA - Page 8

Answer to Donor Feces - Page 8

How to Find the Power of a Statistical Test

How to Compute the Power of a Hypothesis Test

Sample Size Calculator

Example 1: Power of the Hypothesis Test of a Mean Score

Example 2: Power of the Hypothesis Test of a Proportion

Power Analysis

Power Analysis in R

Correlations

Linear Models

Tests of Proportions

Chi-square Tests

Some Examples

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Hypothesis Testing | A Step-by-Step Guide with Easy Examples

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Power Calculation

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