Questions: 11. In the situation described in Question 10, suppose you used the .05 level of significance when you analyzed the data. Your analysis indicated that you should reject the null hypothesis (T/F)?

Solution Steps

The Standard Error \( (SE) \) is calculated as follows:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.133}{5} + \frac{0.157}{5}} = 0.2408 \]

The test statistic \( (t) \) is computed using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{2.76 - 2.62}{0.2408} = 0.5813 \]

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

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{0.0034}{0.0004} = 7.9456 \]

The P-value is calculated as:

\[ P = 2(1 - T(|t|)) = 2(1 - T(0.5813)) = 0.5771 \]

The critical value for a two-tailed test at the \( \alpha = 0.05 \) significance level with \( df = 7.9456 \) is:

\[ \text{Critical Value} = 2.3088 \]

Since the P-value \( (0.5771) \) is greater than the significance level \( (0.05) \), we do not reject the null hypothesis.

Final Answer