Question

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In a study of a group of women science majors who remained in their profession and a group who left their profession within a few months of graduation, the researchers collected the data shown here on a self-esteem questionnaire. At \(\alpha=0.05,\) can it be concluded that there is a difference in the selfesteem scores of the two groups? Use the \(P\) -value method. $$ \begin{array}{ll} \text { Leavers } & \text { Stayers } \\ \hline \bar{X}_{1}=3.05 & \bar{X}_{2}=2.96 \\ \sigma_{1}=0.75 & \sigma_{2}=0.75 \\ n_{1}=103 & n_{2}=225 \end{array} $$

Short answer

There is no significant difference in self-esteem scores between the two groups.

Step-by-step solution

State the Hypotheses

Identify the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). The claim is that there is a difference in self-esteem scores between the two groups. Thus, we state:\[H_0: \mu_1 = \mu_2\]\[H_1: \mu_1 eq \mu_2\]Here, \(\mu_1\) and \(\mu_2\) represent the mean self-esteem scores of Leavers and Stayers, respectively.

Determine the Critical Value

Using a significance level \(\alpha = 0.05\) for a two-tailed test, find the critical values from the standard normal distribution table. The critical values for a two-tailed test at \(\alpha = 0.05\) are approximately \(z = \pm 1.96\).

Compute the Test Value

Calculate the test value using the formula for the test statistic for the difference between two population means:\[ z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Plugging in the given values:\[ z = \frac{3.05 - 2.96}{\sqrt{\frac{0.75^2}{103} + \frac{0.75^2}{225}}} \]Calculate the standard error:\[ \sqrt{\frac{0.75^2}{103} + \frac{0.75^2}{225}} = \sqrt{0.00547 + 0.0025} = \sqrt{0.00797} \approx 0.0893 \]Then:\[ z = \frac{0.09}{0.0893} \approx 1.007 \]

Make the Decision

Compare the computed test value \(z = 1.007\) with the critical values \(\pm 1.96\). Since \(1.007\) is within the range \(-1.96\) to \(1.96\), we fail to reject the null hypothesis \(H_0\).

Summarize the Results

Since we fail to reject the null hypothesis, we conclude that there is not enough statistical evidence at the \(\alpha = 0.05\) significance level to claim that there is a difference in self-esteem scores between the Leavers and Stayers groups.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis, denoted by \( H_0 \), represents a statement of no effect or no difference. It is the hypothesis that researchers typically aim to test against. In this exercise, the null hypothesis asserts that the mean self-esteem scores of women who left their science professions (Leavers) and those who stayed (Stayers) are equal. Mathematically, this is represented as \( H_0: \mu_1 = \mu_2 \). Here, \( \mu_1 \) is the mean score for Leavers, and \( \mu_2 \) is for Stayers.The null hypothesis assumes that any observed difference in sample means is due to random sampling variation rather than a real effect. Consequently, it acts as a benchmark for determining whether observed data departs significantly from expectations.

Alternative Hypothesis

The alternative hypothesis, denoted by \( H_1 \), challenges the null hypothesis. It is the statement that there is an effect or a difference. For the self-esteem scores in this study, the alternative hypothesis posits that there is indeed a difference between the two groups' mean scores. Formally, it is written as \( H_1: \mu_1 eq \mu_2 \).Unlike the null hypothesis, \( H_1 \) suggests that any observed variation in sample means reflects a true difference in the population means. Researchers use statistical tests to provide evidence for rejecting \( H_0 \) in favor of \( H_1 \). The choice of \( H_1 \) being \( \mu_1 eq \mu_2 \) signifies a two-tailed test, suggesting interest in any difference, whether positive or negative.

Critical Value

The critical value is a key threshold in hypothesis testing that helps determine whether to reject the null hypothesis. It represents the point(s) beyond which the test statistic is considered unlikely if the null hypothesis is true. In this study, with a significance level \( \alpha = 0.05 \), the critical values for a two-tailed test can be found using the standard normal distribution table, resulting in critical values of \( z = \pm 1.96 \).A two-tailed test is used because \( H_1 \) is non-directional, indicating simply a difference, rather than specifying direction. Thus, each tail of the distribution accounts for 2.5% of the total significance level. If the test statistic surpasses the critical values, it suggests sufficient evidence to reject the null hypothesis.

Test Statistic

The test statistic provides a standardized way to assess the level of agreement between the observed data and the null hypothesis. For this study, the test statistic for comparing two population means is computed using the following formula: \[ z = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Substitute the given values to calculate: \[ z = \frac{3.05 - 2.96}{\sqrt{\frac{0.75^2}{103} + \frac{0.75^2}{225}}} = \frac{0.09}{0.0893} \approx 1.007 \]This test statistic of 1.007 falls within the range defined by the critical values \(-1.96\) and \(1.96\). As a result, it indicates that the observed mean difference is not enough to conclude a significant difference at the 0.05 significance level.

P-value Method

The \( P \)-value method offers an alternative assessment to the critical value approach. The \( P \)-value represents the probability of observing a test statistic as extreme as, or more extreme than, the one calculated if the null hypothesis is true. A small \( P \)-value suggests evidence against \( H_0 \).Given a calculated \( z \)-value of 1.007, the \( P \)-value can be found using statistical software or tables. This \( P \)-value relates to the probability in the tails of the normal distribution beyond \( z = \pm 1.007 \).If the \( P \)-value is less than the chosen \( \alpha = 0.05 \), it indicates strong evidence against the null hypothesis, warranting rejection. However, in this exercise, the computed \( P \)-value would be larger than 0.05, confirming the decision to fail to reject the null hypothesis, indicating insufficient evidence of a true mean difference in self-esteem scores.