We are given two sets of scores, Time 1 ($X_1$) and Time 2 ($X_2$), for 12 subjects. We need to calculate the difference scores $D = X_2 - X_1$, perform a hypothesis test at the $\alpha = 0.05$ significance level to determine if there is a statistically significant improvement (increase) in scores from Time 1 to Time 2, and estimate the average increase.

1. Problem Description

We are given two sets of scores, Time 1 ($X_1$) and Time 2 ($X_2$), for 12 subjects. We need to calculate the difference scores $D = X_2 - X_1$, perform a hypothesis test at the $\alpha = 0.05$ significance level to determine if there is a statistically significant improvement (increase) in scores from Time 1 to Time 2, and estimate the average increase.

2. Solution Steps

First, calculate the difference scores $D$ for each subject:

$D = X_2 - X_1$

$D_1 = 63 - 61 = 2$

$D_2 = 79 - 75 = 4$

$D_3 = 97 - 91 = 6$

$D_4 = 93 - 83 = 10$

$D_5 = 81 - 74 = 7$

$D_6 = 87 - 82 = 5$

$D_7 = 98 - 98 = 0$

$D_8 = 79 - 82 = -3$

$D_9 = 82 - 69 = 13$

$D_{10} = 77 - 76 = 1$

$D_{11} = 93 - 91 = 2$

$D_{12} = 80 - 70 = 10$

Next, calculate the mean difference ($\bar{D}$) and the standard deviation of the differences ($s_D$).

$\bar{D} = \frac{\sum{D_i}}{n} = \frac{2+4+6+10+7+5+0+(-3)+13+1+2+10}{12} = \frac{57}{12} = 4.75$

$s_D = \sqrt{\frac{\sum{(D_i - \bar{D})^2}}{n-1}}$

To calculate $s_D$, we first need to calculate $(D_i - \bar{D})^2$ for each $D_i$:

$(2-4.75)^2 = (-2.75)^2 = 7.5625$

$(4-4.75)^2 = (-0.75)^2 = 0.5625$

$(6-4.75)^2 = (1.25)^2 = 1.5625$

$(10-4.75)^2 = (5.25)^2 = 27.5625$

$(7-4.75)^2 = (2.25)^2 = 5.0625$

$(5-4.75)^2 = (0.25)^2 = 0.0625$

$(0-4.75)^2 = (-4.75)^2 = 22.5625$

$(-3-4.75)^2 = (-7.75)^2 = 60.0625$

$(13-4.75)^2 = (8.25)^2 = 68.0625$

$(1-4.75)^2 = (-3.75)^2 = 14.0625$

$(2-4.75)^2 = (-2.75)^2 = 7.5625$

$(10-4.75)^2 = (5.25)^2 = 27.5625$

$\sum{(D_i - \bar{D})^2} = 7.5625 + 0.5625 + 1.5625 + 27.5625 + 5.0625 + 0.0625 + 22.5625 + 60.0625 + 68.0625 + 14.0625 + 7.5625 + 27.5625 = 242.25$

$s_D = \sqrt{\frac{242.25}{12-1}} = \sqrt{\frac{242.25}{11}} = \sqrt{22.0227} \approx 4.6928$

Now, perform a paired t-test:

$t = \frac{\bar{D}}{s_D / \sqrt{n}} = \frac{4.75}{4.6928 / \sqrt{12}} = \frac{4.75}{4.6928 / 3.4641} = \frac{4.75}{1.3547} \approx 3.506$

Degrees of freedom $df = n - 1 = 12 - 1 = 11$

For a one-tailed t-test with $\alpha = 0.05$ and $df = 11$, the critical t-value is $t_{crit} = 1.796$. Since our calculated t-value ($3.506$) is greater than the critical t-value ($1.796$), we reject the null hypothesis. This means that there is a statistically significant increase in scores from Time 1 to Time

2.

The average increase in scores is $\bar{D} = 4.75$.

3. Final Answer

Yes, scores increased significantly.

The average increase in scores is 4.

7

5.