Questions: Please submit a rough draft of your hypothesis test, including all aspects (see the template from class notes, or the "Inference Test" portion of the outline for the final paper). I will try to give feedback on this submission within 24 hours so that you can correct any errors or oversights before the final version. female mean =4.53 Variance =0.2481 s D=0.4981 male mean =4.7333 variance =2.0619 s D=1.436 t=-0.423 a=0.05 a=0.05 since t=0.423 is less than t

Solution Steps

To perform a hypothesis test comparing the means of two groups (female and male), we can use a t-test. The steps include:

1. State the Hypotheses:

   • Null Hypothesis (\(H_0\)): There is no difference in means (\(\mu_{\text{female}} = \mu_{\text{male}}\)).
   • Alternative Hypothesis (\(H_a\)): There is a difference in means (\(\mu_{\text{female}} \neq \mu_{\text{male}}\)).

2. Calculate the Test Statistic: Use the given means, variances, and sample sizes to compute the t-statistic.

3. Determine the Critical Value: Use the significance level (\(\alpha = 0.05\)) to find the critical t-value from the t-distribution table.

4. Make a Decision: Compare the calculated t-statistic with the critical t-value to accept or reject the null hypothesis.

• Null Hypothesis (\(H_0\)): \(\mu_{\text{female}} = \mu_{\text{male}}\)
• Alternative Hypothesis (\(H_a\)): \(\mu_{\text{female}} \neq \mu_{\text{male}}\)

The test statistic is calculated as: \[ t = \frac{\bar{x}_{\text{female}} - \bar{x}_{\text{male}}}{\sqrt{\frac{s_{\text{female}}^2}{n_{\text{female}}} + \frac{s_{\text{male}}^2}{n_{\text{male}}}}} \] Substituting the given values: \[ t = \frac{4.53 - 4.7333}{\sqrt{\frac{0.2481}{30} + \frac{2.0619}{30}}} = -0.7326 \]

Using a significance level of \(\alpha = 0.05\) and degrees of freedom \(df = 29\), the critical t-value is: \[ t_{\text{critical}} = 2.0452 \]

Compare the absolute value of the test statistic with the critical value: \[ |t| = 0.7326 < t_{\text{critical}} = 2.0452 \] Since \(|t|\) is less than \(t_{\text{critical}}\), we fail to reject the null hypothesis.

Final Answer