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Chapter 9: Problem 7

Is there any systematic tendency for part-time college faculty to hold their students to different standards than do full-time faculty? The article "Are There Instructional Differences Between Full-Time and Part-Time Faculty?" (College Teaching, 2009: 23-26) reported that for a sample of 125 courses taught by full-time faculty, the mean course GPA was \(2.7186\) and the standard deviation was \(.63342\), whereas for a sample of 88 courses taught by part- timers, the mean and standard deviation were \(2.8639\) and \(.49241\), respectively. Does it appear that true average course GPA for part-time faculty differs from that for faculty teaching full-time? Test the appropriate hypotheses at significance level 01 by first obtaining a \(P\)-value.

Short Answer

Expert verified
The P-value will determine if part-time and full-time faculty GPAs differ at \( \alpha = 0.01 \).

Step by step solution

01

Define the Hypotheses

The null hypothesis \(H_0\) is that the true average course GPA for part-time faculty is equal to that for full-time faculty, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis \(H_a\) is that the true average course GPA for part-time faculty differs from that for full-time faculty, i.e., \( \mu_1 eq \mu_2 \).
02

Identify the Test Statistics

We will use a two-sample t-test to compare the means. The test statistic formula is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \( \bar{x}_1 = 2.7186 \) is the mean GPA for full-time faculty, \( s_1 = 0.63342 \) is the standard deviation for full-time faculty, \( n_1 = 125 \) is the sample size for full-time faculty, \( \bar{x}_2 = 2.8639 \) is the mean GPA for part-time faculty, \( s_2 = 0.49241 \) is the standard deviation for part-time faculty, and \( n_2 = 88 \) is the sample size for part-time faculty.
03

Calculate the Test Statistic

Plug the values into the formula:\[ t = \frac{2.7186 - 2.8639}{\sqrt{\frac{0.63342^2}{125} + \frac{0.49241^2}{88}}} \]Calculate the differences and variances, then solve for \( t \).
04

Determine the Degrees of Freedom

Degrees of freedom for the two-sample t-test can be approximated using:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} \]Calculate the degrees of freedom using the variances and sample sizes.
05

Find the P-value

Using the calculated \( t \)-statistic and degrees of freedom, find the two-tailed \( P \)-value from a t-table or using a calculator.
06

Compare P-value with Significance Level

Compare the \( P \)-value with the significance level of \( 0.01 \). If \( P \text{-value} \leq 0.01 \), reject the null hypothesis; otherwise, do not reject it.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-Sample T-Test
The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In this context, it's applied to compare the average GPAs of courses taught by full-time and part-time faculty. This type of test helps to determine whether any observed difference in sample means reflects a true difference in the overall population means.

To perform a two-sample t-test, you need the means, standard deviations, and sample sizes of both groups. These values are plugged into the t-test formula, which calculates the test statistic. If the test statistic departs significantly from zero, this suggests evidence against the null hypothesis, which states that the group means are equal. This method considers the variability and size of the samples to provide a perspective on the statistical significance of the difference.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold set by the researcher that determines the cutoff point to reject the null hypothesis. It represents the probability of committing a Type I error, which is the mistake of rejecting the null hypothesis when it is actually true.

In hypothesis testing, a common significance level is \( 0.05 \), but in more stringent tests, like this one comparing GPAs, a level of \( 0.01 \) is used. This means that there's a 1% risk of concluding that there is a difference between part-time and full-time faculty GPAs when there is actually no difference. Consequently, the lower the significance level, the stronger the evidence needed to reject the null hypothesis.
P-Value
The \( P \)-value is a crucial output of hypothesis testing, providing the probability of observing the test results, or something more extreme, assuming that the null hypothesis is true. It offers a measure of the strength of evidence against the null hypothesis.

When the \( P \)-value is compared to the significance level, it helps to decide whether to reject the null hypothesis. For instance, if the \( P \)-value is less than or equal to the significance level of \( 0.01 \), it suggests strong evidence against the null hypothesis, and thus, warrants rejection. Conversely, a \( P \)-value greater than \( 0.01 \) indicates insufficient evidence to reject the null hypothesis, implying that any observed differences might be due to chance.
Degrees of Freedom
Degrees of freedom refer to the number of independent values that can vary in an analysis without violating any constraints. For a two-sample t-test, degrees of freedom are approximated using a specific formula which considers sample sizes and variances of the two groups.

In this exercise, calculating the degrees of freedom involves the variances \( s_1^2 \) and \( s_2^2 \), sample sizes \( n_1 \) and \( n_2 \), and a complex formula to account for these values. It plays an essential role in determining the critical value of the \( t \)-distribution to compare against the calculated test statistic, ensuring the accuracy of the conclusion drawn from the test.

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Most popular questions from this chapter

Is someone who switches brands because of a financial inducement less likely to remain loyal than someone who switches without inducement? Let \(p_{1}\) and \(p_{2}\) denote the true proportions of switchers to a certain brand with and without inducement, respectively, who subsequently make a repeat purchase. Test \(H_{0}: p_{1}-p_{2}=0\) versus \(H_{a}: p_{1}-p_{2}<0\) using \(\alpha=.01\) and the following data: \(m=200 \quad\) number of success \(=30\) \(n=600\) number of success \(=180\) (Similar data is given in "Impact of Deals and Deal Retraction on Brand Switching," J. of Marketing, 1980: 62-70.)

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Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution. \begin{tabular}{lrr} & \(\mathbf{A}\) & \(\mathbf{B}\) \\ \hline \(26-<30\) & 6 & 4 \\ \(30-<34\) & 12 & 9 \\ \(34-<38\) & 15 & 19 \\ \(38-<42\) & 7 & 10 \\ & \(m=40\) & \(m=42\) \\ \hline \end{tabular} Compute a \(95 \%\) CI for the difference between the true proportions of all specimens of alloys \(\mathrm{A}\) and \(\mathrm{B}\) that have an ultimate strength of at least \(34 \mathrm{ksi}\).

Headability is the ability of a cylindrical piece of material to be shaped into the head of a bolt, screw, or other cold-formed part without cracking. The article "New Methods for Assessing Cold Heading Quality" (Wire J. Intl., Oct. 1996: 66-72) described the result of a headability impact test applied to 30 specimens of aluminum killed steel and 30 specimens of silicon killed steel. The sample mean headability rating number for the steel specimens was \(6.43\), and the sample mean for aluminum specimens was 7.09. Suppose that the sample standard deviations were \(1.08\) and \(1.19\), respectively. Do you agree with the article's authors that the difference in headability ratings is significant at the \(5 \%\) level (assuming that the two headability distributions are normal)?
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