Step 1: Restate the question as hypotheses about the populations.
Null Hypothesis (H0): There is no difference in hearing ability between people who eat a large meal and those who eat a small meal.
Alternative Hypothesis (H1): People who eat a large meal have better hearing ability compared to those who eat a small meal.
Step 2: Determine the characteristics of the comparison distribution.
a. M = Mean (average) of the differences between the two groups.
b. Figure the estimated population variance for each sample.
For the Big Meal Group:
\[ S_{1}^{2} = \frac{\sum (X_{1} - \bar{X}_{1})^2}{(N_{1} - 1)} \]
\[ S_{1}^{2} = \frac{(22-24)^2 + (25-24)^2 + (25-24)^2}{(3-1)} \]
\[ S_{1}^{2} = \frac{4 + 1 + 1}{2} \]
\[ S_{1}^{2} = \frac{6}{2} \]
\[ S_{1}^{2} = 3 \]
For the Small Meal Group:
\[ S_{2}^{2} = \frac{\sum (X_{2} - \bar{X}_{2})^2}{(N_{2} - 1)} \]
\[ S_{2}^{2} = \frac{(19-21)^2 + (23-21)^2 + (21-21)^2}{(3-1)} \]
\[ S_{2}^{2} = \frac{4 + 4 + 0}{2} \]
\[ S_{2}^{2} = \frac{8}{2} \]
\[ S_{2}^{2} = 4 \]
c. Figure the pooled estimate of the population variance.
\[ S_{Pooled}^{2} = \frac{(N_{1}-1)S_{1}^{2} + (N_{2}-1)S_{2}^{2}}{(N_{1} + N_{2} - 2)} \]
\[ S_{Pooled}^{2} = \frac{(3-1)(3) + (3-1)(4)}{(3 + 3 - 2)} \]
\[ S_{Pooled}^{2} = \frac{6 + 8}{4} \]
\[ S_{Pooled}^{2} = \frac{14}{4} \]
\[ S_{Pooled}^{2} = 3.5 \]
d. Figure the variance of each distribution of means.
\[ S_{M1}^{2} = \frac{S_{Pooled}^{2}}{N_{1}} \]
\[ S_{M1}^{2} = \frac{3.5}{3} \]
\[ S_{M1}^{2} = 1.1667 \]
\[ S_{M2}^{2} = \frac{S_{Pooled}^{2}}{N_{2}} \]
\[ S_{M2}^{2} = \frac{3.5}{3} \]
\[ S_{M2}^{2} = 1.1667 \]
e. Figure the variance of the distribution of differences between means.
\[ S_{Difference}^{2} = S_{M1}^{2} + S_{M2}^{2} \]
\[ S_{Difference}^{2} = 1.1667 + 1.1667 \]
\[ S_{Difference}^{2} = 2.3334 \]
f. Figure the standard deviation of the distribution between means.
\[ SDifference = \sqrt{S_{Difference}^{2}} \]
\[ SDifference = \sqrt{2.3334} \]
\[ SDifference ≈ 1.5276 \]
Step 3: Determine the cutoff score at which the null can be rejected (use Table A-2).
For a two-tailed test with a significance level of 0.05, the critical t-value would be obtained from the t-distribution table with degrees of freedom (df) equal to \( N_1 + N_2 - 2 = 6 - 2 = 4 \). From the table, \( t_{\text{critical}} \) is approximately 2.776.
Step 4: Determine the sample t score.
\[ t = \frac{\bar{X}_1 - \bar{X}_2}{SDifference} \]
\[ t = \frac{24 - 21}{1.5276} \]
\[ t ≈ \frac{3}{1.5276} \]
\[ t ≈ 1.9611 \]
Step 5: Decision
The sample t-score (1.9611) does not exceed the critical t-value (2.776) for a significance level of 0.05. Therefore, we fail to reject the null hypothesis.
Conclusion:
Based on the data and the statistical analysis, there is insufficient evidence to conclude that people who eat a large meal have better hearing ability compared to those who eat a small meal at a significance level of 0.05.
