Questions: Does the average Presbyterian donate more than the average Catholic in church on Sundays? The 49 randomly observed members of the Presbyterian church donated an average of 20 with a standard deviation of 5. The 53 randomly observed members of the Catholic church donated an average of 17 with a standard deviation of 9. What can be concluded at the α=0.05 level of significance?

Solution Steps

The standard error \( SE \) is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{5^2}{49} + \frac{9^2}{53}} = \sqrt{\frac{25}{49} + \frac{81}{53}} \approx 1.575 \]

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

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{20.0 - 17.0}{1.575} \approx 1.897 \]

The degrees of freedom \( df \) for Welch's t-test is calculated using the formula:

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

Substituting the values:

\[ df = \frac{\left(\frac{25}{49} + \frac{81}{53}\right)^2}{\frac{\left(\frac{25}{49}\right)^2}{48} + \frac{\left(\frac{81}{53}\right)^2}{52}} \approx 99.999 \]

The p-value is calculated based on the test statistic. For \( t \approx 1.897 \) and \( df \approx 99.999 \):

\[ P = 1 - T(t) \approx 0.029 \]

Since the p-value \( P \approx 0.029 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis. Thus, we conclude that there is sufficient evidence to suggest that the average Presbyterian donates more than the average Catholic.

Final Answer