For the following procedures, the assumption is that both \(np \geq 10\) and \(n(1-p) \geq 10\). When we're constructing confidence intervals \(p\) is typically unknown, in which case we use \(\widehat{p}\) as an estimate of \(p\).

Note that \(n \widehat p\) is the number of successes in the sample and \(n(1- \widehat p)\) is the number of failures in the sample. 

This means that our sample needs to have at least 10 "successes" and at least 10 "failures" in order to construct a confidence interval using the normal approximation method. 

Below is the general form of a confidence interval.

The sample statistic here is the sample proportion, \(\widehat p\). When using the normal approximation method the multiplier is taken from the standard normal distribution (i.e., z distribution).  And, the standard error is computed using \(\widehat p\) as an estimate of \(p\): \(\sqrt{\frac{\hat{p} (1-\hat{p})}{n}}\). This leaves us with the following formula to construct a confidence interval for a population proportion:

The value of the \(z^*\) multiplier depends on the level of confidence. The multiplier for the confidence interval for a population proportion can be found using the standard normal distribution [i.e., z distribution, N(0,1)]. The most commonly used level of confidence is 95%. As shown on the probability distribution plot below, the multiplier associated with a 95% confidence interval is 1.960, often rounded to 2 (recall the Empirical Rule and 95% Rule).

Below is a table of frequently used \(z^*\) multipliers.

The value of the multiplier increases as the confidence level increases. This leads to wider intervals for higher confidence levels. We are more confident of catching the population value when we use a wider interval.

Before we can construct a confidence interval for a proportion we must first determine if we should use the exact method or the normal approximation method. Recall that if \(np \geq 10\) and \(n(1-p) \geq 10\) then the sampling distribution can be approximated by a normal distribution. Since we don't have the population proportion (\(p\)), we using \(\widehat p\) as an estimate. Note that \(n\widehat p\) is the number of successes in the sample and \(n(1-\widehat p)\) is the number of failures in the sample.

If this assumption has not been met, then the sampling distribution is constructed using a binomial distribution which Minitab refers to as the "exact method." 

To check this assumption we can construct a frequency table. You first learned how to construct a frequency table in Lesson 2.1.1.2.1 of these online notes. Here is another example:

From the frequency table above we can see that there were at least 10 "successes" and at least 10 "failures" in the sample. In this example a success is defined as answering "yes" to the question "do you own a dog?" A failure is defined as answering "no." Because both \(n \widehat p \geq 10\) and \(n(1- \widehat p) \geq 10\), the normal approximation method may be used. In Minitab, the exact method is the default method. If there are at least 10 successes and at least 10 failures, then you need to change the method to the normal approximation method.

If the number of successes or the number of failures in the sample is less than 10, then the exact method should be used instead of the normal approximation method. In Minitab, this means that in step 8 above the default setting of Exact method should not be changed.

If you do not have a Minitab worksheet filled with data concerning individuals, but instead have summarized data (e.g., the number of successes and the number of failures), you would not load the data set, but in step 3 you would select Summarized data. For Number of events, enter the number of successes (i.e., \(n \widehat p\)) and for Number of trials enter the total sample size (i.e., \(n\)). 

When we begin a study to estimate a population parameter we typically have an idea as how confident we want to be in our results and within what degree of accuracy. This means we get started with a set level of confidence and margin of error. We can use these pieces to determine a minimum sample size needed to produce these results by using algebra to solve for \(n\):

If we have no preconceived idea of the value of the population proportion, then we use \(\tilde{p}=0.50\) because it is most conservative and it will give use the largest sample size calculation.

Here we will be using the five step hypothesis testing procedure to compare the proportion in one random sample to a specified population proportion using the normal approximation method.

In order to use the normal approximation method, the assumption is that both \(n p_0 \geq 10\) and \(n (1-p_0) \geq 10\). Recall that \(p_0\) is the population proportion in the null hypothesis.

Where \(p_0\) is the hypothesized population proportion that you are comparing your sample to.

When using the normal approximation method we will be using a z test statistic. The z test statistic tells us how far our sample proportion is from the hypothesized population proportion in standard error units. Note that this formula follows the basic structure of a test statistic that you learned in the last lesson:

\(test\;statistic=\dfrac{sample\;statistic-null\;parameter}{standard\;error}\)

Given that the null hypothesis is true, the p value is the probability that a randomly selected sample of n would have a sample proportion as different, or more different, than the one in our sample, in the direction of the alternative hypothesis. We can find the p value by mapping the test statistic from step 2 onto the z distribution. 

Note that p-values are also symbolized by \(p\). Do not confuse this with the population proportion which shares the same symbol.

We can look up the \(p\)-value using Minitab by constructing the sampling distribution.  Because we are using the normal approximation here, we have a \(z\) test statistic that we can map onto the \(z\) distribution. Recall, the z distribution is a normal distribution with a mean of 0 and standard deviation of 1. If we are conducting a one-tailed (i.e., right- or left-tailed) test, we look up the area of the sampling distribution that is beyond our test statistic. If we are conducting a two-tailed (i.e., non-directional) test there is one additional step: we need to multiple the area by two to take into account the possibility of being in the right or left tail. 

We can decide between the null and alternative hypotheses by examining our p-value. If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis. Unless stated otherwise, assume that \(\alpha=.05\).

When we reject the null hypothesis our results are said to be statistically significant.

Based on our decision in step 4, we will write a sentence or two concerning our decision in relation to the original research question.

A hypothesis test for one proportion can be conducted in Minitab. This can be done using raw data or summarized data.

• If you have a data file with every individual's observation, then you have raw data.
• If you do not have each individual observation, but rather have the sample size and number of successes in the sample, then you have summarized data.

The next two pages will show you how to use Minitab to conduct this analysis using either raw data or summarized data.

Note that the default method for constructing the sampling distribution in Minitab is to use the exact method.  If \(np_0 \geq 10\) and \(n(1-p_0) \geq 10\) then you will need to change this to the normal approximation method.  This must be done manually. Minitab will use the method that you select, it will not check assumptions for you!

Earlier in this lesson we considered confidence intervals for proportions and the multiplier in our intervals was a value from the standard normal (i.e., \(z\)) distribution. But, what if our variable of interest is a quantitative variable and we want to estimate a population mean? 

We apply similar techniques when constructing a confidence interval for a mean, but now we are interested in estimating the population mean (\(\mu\)) by using the sample statistic (\(\overline{x}\)) and the multiplier is a \(t\) value. Similar to the \(z\) values that you used as the multiplier for constructing confidence intervals for population proportions, here you will use \(t\) values as the multipliers. Because \(t\) values vary depending on the number of degrees of freedom (df), you will need to use statistical software to look up the appropriate \(t\) value for each confidence interval that you construct. The degrees of freedom will be based on the sample size. Since we are working with one sample here, \(df=n-1\).

Let’s review some of symbols and equations that we learned in previous lessons:

Recall the general form for a confidence interval:

When constructing a confidence interval for a population mean the point estimate is the sample mean, \(\overline{x}\). The multiplier is taken from a \(t\) distribution. And, the standard error is equal to \(\frac{s}{\sqrt{n}}\).

On the following pages we will walk through examples of constructing confidence intervals for population means by hand. Then, you will learn how to compute confidence intervals using Minitab.

Here you will learn how to use Minitab to construct a confidence interval for a mean. The procedure is similar to the one that you learned earlier in this lesson for constructing a confidence interval for a proportion. The following example walks through this procedure when data are in a Minitab work. At the bottom of this page you will find instructions for using Minitab with summarized data.

If you do not have a Minitab worksheet filled with data concerning individuals, but instead have summarized data (e.g., the values of \(s\), \(\overline{x}\), and \(n\)), you would skip step 1 above and in step 3 you would select Summarized data. 

Calculating the sample size necessary for estimating a population mean with a given margin of error and level of confidence is similar to that for estimating a population proportion. However, since the \(t\) distribution is not as “neat” as the standard normal distribution, the process can be iterative. (Recall, the shape of the \(t\) distribution is different for each degree of freedom). This means that we would solve, reset, solve, reset, etc. until we reached a conclusion. Yet, we can avoid this iterative process if we employ an approximate method based on \(t\) distribution approaching the standard normal distribution as the sample size increases. This approximate method invokes the following formula:

The sample standard deviation may be estimated on the basis of prior research studies.

Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large (\(\ge\ 30\)) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are:

where \( \mu_{0} \) is the hypothesized population mean.

For the test of one group mean we will be using a \(t\) test statistic:

Note that structure of this formula is similar to the general formula for a test statistic:

\(\dfrac{sample\;statistic-null\;value}{standard\;error}\)

When testing hypotheses about a mean or mean difference, a \(t\) distribution is used to find the \(p\)-value. These \(t\) distributions are indexed by a quantity called degrees of freedom, calculated as \(df = n – 1\) for the situation involving a test of one mean or test of mean difference. The \(p\)-value can be found using Minitab.

If \(p \leq \alpha\) reject the null hypothesis.

If \(p>\alpha\) fail to reject the null hypothesis.

Based on your decision in Step 4, write a conclusion in terms of the original research question.

The new few pages will walk you through examples before giving you the opportunity to do two on your own.

A hypothesis test for one group mean can be conducted in Minitab using raw data or summarized data. 

• If you have a data file with every individual's observation, then you have raw data. 
• If you do not have each individual's observation, but rather have the sample mean, sample standard deviation, and sample size, then you have summarized data. 

The next two pages will show you how to use Minitab to conduct a one-sample mean t-test using either raw data or summarized data. There is also one example of using Minitab to conduct a one-sample mean z test which is only performed if the population is known to be normally distributed and the population standard deviation (\(\sigma\)) is available. 

A one sample mean \(z\) test is used when the population is known to be normally distributed and when the population standard deviation (\(\sigma\)) is known. This most frequently occurs in the social sciences when standardized measures are used such as IQ, SAT, ACT, or GRE scores, for which the population parameters are known. 

The formula for computing a \(z\) test statistic for one sample mean is identical to that of computing a \(t\) test statistic for one sample mean, except now the population standard deviation is known and can be used in computing the standard error.

The other primary difference between the one sample mean \(t\) test and the one sample mean \(z\) test is the latter uses the standard normal distribution (i.e., \(z\) distribution) in determining the \(p\)-value. Below are the directions for conducting a one sample mean \(z\) test in Minitab. 

We could summarize these results using the five step hypothesis testing procedure:

The population is known to be normally distributed and the population standard deviation is known to be 15. With these two conditions met we can conduct a one sample mean z test

\(H_0\colon \mu = 100\)
\(H_a\colon \mu > 100\)

From the Minitab output, \(z = 3.33\)

From the Minitab output, \(p = 0.000\)

\(p \le \alpha\), reject the null hypothesis

There is evidence that the mean IQ score of all students at this school is greater than 100. 

An educational research study is designed so that participants complete a measure of demonstrated knowledge twice. The researcher wants to estimate the change in scores from the first to second administrations (i.e., pre- and post-test). Data are paired by participant. The researcher subtracted pre-test scores from the post test scores and found a mean increase of 6.560 with a standard deviation of 3.867 for \(n=100\). She wants to construct a 95% confidence interval for the mean difference.

First, we'll find the appropriate multiplier.

\(df=n-1=100-1=99\)

For a 95% confidence interval: \(t_{df=99}=1.984\)

\(6.560 \pm 1.984 \left(\frac{3.867}{\sqrt{100}}\right)=6.560 \pm 0.767=[5.793, 7.327]\)

We are 95% confident that the difference between post- and pre- test scores is between 5.793 and 7.327.

Data from Zimmerman, W. A. (2015). Impact of Instructional Materials Eliciting Low and High Cognitive Load on Self-Efficacy and Demonstrated Knowledge (Unpublished doctoral dissertation). The Pennsylvania State University, University Park, PA.