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

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
Transcript text: 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. \[ \begin{array}{l} \text { female mean }=4.53 \\ \text { Variance }=0.2481 \\ s D=0.4981 \\ \text { male mean }=4.7333 \\ \text { variance }=2.0619 \\ s D=1.436 \\ t=-0.423 \\ a=0.05 \end{array} \] \[ \begin{array}{l} a=0.05 \\ \text { since } \mid t)=0.423 \text { is less than } t \end{array} \]
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Solution

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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.

Step 1: State the Hypotheses
  • Null Hypothesis (\(H_0\)): \(\mu_{\text{female}} = \mu_{\text{male}}\)
  • Alternative Hypothesis (\(H_a\)): \(\mu_{\text{female}} \neq \mu_{\text{male}}\)
Step 2: Calculate the Test Statistic

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: \[ t = \frac{4.53 - 4.7333}{\sqrt{\frac{0.2481}{30} + \frac{2.0619}{30}}} = -0.7326 \]

Step 3: Determine the Critical Value

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

Step 4: Make a Decision

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

We fail to reject the null hypothesis. There is not enough evidence to suggest a significant difference in means between females and males at the \(\alpha = 0.05\) level.

\[ \boxed{\text{Fail to reject } H_0} \]

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