Questions: Do you expect that this study will yield stronger, weaker, or the same evidence against the null hypothesis? Explain.

Solution Steps

The Standard Error \( (SE) \) is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{628.5}{8} + \frac{726.0}{8}} = 13.012 \]

The test statistic \( (t) \) is computed as follows:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{60.25 - 50.5}{13.012} = 0.7493 \]

The degrees of freedom \( (df) \) are calculated using the formula:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{28666.7227}{2058.2327} = 13.9278 \]

The p-value \( (P) \) is determined using the formula:

\[ P = 2(1 - T(|t|)) = 2(1 - T(0.7493)) = 0.4661 \]

The critical value for a two-tailed test at a significance level of \( \alpha = 0.05 \) with \( df = 13.9278 \) is approximately:

\[ \text{Critical value} = 2.1458 \]

Since the p-value \( (0.4661) \) is greater than the significance level \( (0.05) \), we conclude that the evidence against the null hypothesis is weak.

Final Answer