91Ó°ÊÓ

A professor who teaches a large introductory statistics class (197 students) with eight discussion sections would like to test if student performance differs by discussion section, where each discussion section has a different teaching assistant. The summary table below shows the average final exam score for each discussion section as well as the standard deviation of scores and the number of students in each section. $$ \begin{array}{rrrrrrrrrr} \hline & \text { Sec 1 } & \text { Sec 2 } & \text { Sec 3 } & \text { Sec 4 } & \text { Sec 5 } & \text { Sec 6 } & \text { Sec 7 } & \text { Sec 8 } \\ \hline n_{i} & 33 & 19 & 10 & 29 & 33 & 10 & 32 & 31 \\ \bar{x}_{i} & 92.94 & 91.11 & 91.80 & 92.45 & 89.30 & 88.30 & 90.12 & 93.35 \\\ s_{i} & 4.21 & 5.58 & 3.43 & 5.92 & 9.32 & 7.27 & 6.93 & 4.57 \\ \hline \end{array} $$ The ANOVA output below can be used to test for differences between the average scores from the different discussion sections. $$ \begin{array}{lrrrrr} \hline & \text { Df } & \text { Sum Sq } & \text { Mean Sq } & \text { F value } & \operatorname{Pr}(>\mathrm{F}) \\ \hline \text { section } & 7 & 525.01 & 75.00 & 1.87 & 0.0767 \\ \text { Residuals } & 189 & 7584.11 & 40.13 & & \\ \hline \end{array} $$ Conduct a hypothesis test to determine if these data provide convincing evidence that the average score varies across some (or all) groups. Check conditions and describe any assumptions you must make to proceed with the test.

Step by step solution

01

State the Hypotheses

The hypothesis test for ANOVA involves checking if there is a difference between group means. The null hypothesis \(H_0\) states that all group means are equal: \( \mu_1 = \mu_2 = \cdots = \mu_8 \). The alternative hypothesis \(H_A\) is that at least one group mean is different: \( \exists\ i,j \text{ such that } \mu_i eq \mu_j \).

02

Check ANOVA Assumptions

For ANOVA, we assume independence of observations, normality of distributions within groups, and homogeneity of variances. Given the large sample size in most groups, the Central Limit Theorem suggests normality is a reasonable assumption. We assume independence since students are in separate sections. We will proceed assuming variances are approximately equal.

03

Analyze ANOVA Table

Look at the ANOVA table provided. The F statistic is 1.87 and the p-value is 0.0767. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed F statistic assuming the null hypothesis is true.

04

Make a Decision

Set a significance level, typically \( \alpha = 0.05 \). Compare the p-value to \( \alpha \). Here, \( \text{Pr}(>F) = 0.0767 \) is greater than \( 0.05 \), so we fail to reject the null hypothesis.

05

State the Conclusion

Since we failed to reject the null hypothesis, we do not have convincing evidence at the \( \alpha = 0.05 \) level that the average exam scores differ between discussion sections.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis

In hypothesis testing, the null hypothesis serves as a starting point for any statistical analysis. It is a statement that assumes there is no effect or no difference in the context of the test. For the professor's statistics class example, the null hypothesis (H_0) claims that the average exam scores are identical across all discussion sections.So, if you were to write it out fully in terms of the example, it looks like this: H_0: \( \mu_1 = \mu_2 = \cdots = \mu_8 \). This expression indicates that all the group means (average scores for each discussion section) are equal.Rejecting the null hypothesis means there is enough statistical evidence to suggest that not all group means are equal. However, in hypothesis testing, we begin with the assumption that the null hypothesis is true until proven otherwise. In this exercise, the null hypothesis serves as a benchmark for comparison against the alternative hypothesis.

Alternative Hypothesis

The alternative hypothesis offers a clear contrast to the null hypothesis. It suggests that there is a significant difference in the average scores among at least some of the discussion sections. In shorthand for our scenario, the alternative hypothesis (H_A) can be stated as: \( \exists\ i,j \text{ such that } \mu_i eq \mu_j \). This means that, for at least one pair of discussion sections, the average scores are not the same.The alternative hypothesis is important because it underpins the goal of our ANOVA test: to find evidence that there is variation among the group means. By doing so, it opens up the possibility that different teaching assistants might have influenced student performances differently. In the context of the professor's study, supporting the alternative hypothesis implies a potential teaching discrepancy needing further investigation.

P-Value

The p-value plays a crucial role in hypothesis testing as it measures the evidence against the null hypothesis. Specifically, it is the probability of observing a test statistic at least as extreme as the one calculated, under the assumption that the null hypothesis is true.In this exercise, the calculated p-value is 0.0767. To make a decision, we need a chosen significance level, commonly expressed as \( \alpha \). Popular choices for \( \alpha \) include 0.05 or 0.01. When the p-value is smaller than \( \alpha \), we reject the null hypothesis because it suggests the data are inconsistent with the null hypothesis.In this scenario, since the p-value of 0.0767 exceeds 0.05, we fail to reject the null hypothesis. Essentially, this means there isn't sufficiently strong evidence in the data to conclude that the average exam scores differ by discussion section.

Homogeneity of Variances

Homogeneity of variances, also known as homoscedasticity, is an essential assumption in ANOVA tests. It implies that the variance within each group should be similar across all groups. This assumption ensures that the F statistic, used in ANOVA testing, follows the appropriate distribution under the null hypothesis. Why is this important? Homogeneity of variances assures us that any observed differences among the group means are not masked or confounded by unequal variability within the groups. In our exercise, the summary table provides clues about homogeneity. We see different standard deviations for each section, but to apply the ANOVA appropriately, we assume these differences are not substantial enough to affect the F test's reliability significantly. Meeting this assumption bolsters our confidence in the test results. However, if unequal variances were suspected, other methods like Welch’s ANOVA might be considered for more accurate conclusions.