Questions: Is memory ability before a meal worse than after a meal? Ten people were given memory tests before their meal and then again after their meal. The data is shown below. A higher score indicates a better memory ability. Score on the Memory Test Before a Meal: 82, 71, 62, 73, 75, 61, 81, 64, 72, 78 After a Meal: 85, 77, 56, 77, 86, 67, 77, 55, 77, 73 Assume a Normal distribution. What can be concluded at the the α=0.10 level of significance? For this study, we should use t-test for the difference between two dependent population means a. The null and alternative hypotheses would be: H0: μd = 0 H1: μd ≠ 0 b. The test statistic = (please show your answer to 3 decimal places.) c. The p-value = (Please show your answer to 4 decimal places.)

Solution Steps

We set up the null and alternative hypotheses as follows: \[ H_{0}: \mu_d = 0 \quad \text{(There is no difference in memory ability before and after a meal)} \] \[ H_{1}: \mu_d \neq 0 \quad \text{(There is a difference in memory ability before and after a meal)} \]

The test statistic \( t \) is calculated using the formula: \[ t = \frac{\bar{d}}{SE} \] where \( \bar{d} = -1.1 \) is the mean difference and \( SE = 2.0787 \) is the standard error. Thus, we have: \[ t = \frac{-1.1}{2.0787} = -0.5292 \]

For a two-tailed test at \( \alpha = 0.1 \), the critical value is: \[ t_{\alpha/2, df} = t_{(0.05, 9)} = 1.8331 \] The p-value is calculated as: \[ P = 2 \times (1 - T(|t|)) = 2 \times (1 - T(0.5292)) = 0.6095 \]

Since the p-value \( 0.6095 \) is greater than the significance level \( \alpha = 0.1 \), we fail to reject the null hypothesis. This indicates that there is no significant difference in memory ability before and after a meal.

Final Answer