1. **Problem Statement:** We want to determine if 7 hours of sleep deprivation significantly increases reaction time using the given before and after data for 10 subjects. 2. **Test Selection:** Since the data are paired (before and after measurements on the same subjects), the appropriate test is the **paired t-test**. 3. **Formulas and Important Rules:** - Calculate the differences $d_i = \text{After}_i - \text{Before}_i$ for each subject. - Compute the mean difference $\bar{d} = \frac{1}{n} \sum_{i=1}^n d_i$. - Compute the standard deviation of differences $s_d = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (d_i - \bar{d})^2}$. - The test statistic is $$ t = \frac{\bar{d}}{s_d / \sqrt{n}} $$ - Degrees of freedom $df = n-1$. - Compare $t$ to the critical value (CV) from the t-distribution at $\alpha=0.05$. 4. **Calculate Differences:** Before: 205, 220, 189, 234, 198, 215, 228, 200, 212, 190 After: 218, 225, 195, 248, 210, 228, 240, 208, 220, 205 Differences $d_i$ = 13, 5, 6, 14, 12, 13, 12, 8, 8, 15 5. **Calculate Mean Difference:** $$ \bar{d} = \frac{13 + 5 + 6 + 14 + 12 + 13 + 12 + 8 + 8 + 15}{10} = \frac{106}{10} = 10.6 $$ 6. **Calculate Standard Deviation of Differences:** Calculate each squared deviation: $(13-10.6)^2=5.76$, $(5-10.6)^2=31.36$, $(6-10.6)^2=21.16$, $(14-10.6)^2=11.56$, $(12-10.6)^2=1.96$, $(13-10.6)^2=5.76$, $(12-10.6)^2=1.96$, $(8-10.6)^2=6.76$, $(8-10.6)^2=6.76$, $(15-10.6)^2=19.36$ Sum = 112.4 $$ s_d = \sqrt{\frac{112.4}{9}} = \sqrt{12.49} = 3.535 $$ 7. **Calculate Test Statistic:** $$ t = \frac{10.6}{3.535 / \sqrt{10}} = \frac{10.6}{3.535 / 3.162} = \frac{10.6}{1.117} = 9.485 $$ 8. **Critical Value:** Given $CV = 1.833$ at $\alpha=0.05$ with $df=9$. 9. **Conclusion:** Since $t = 9.485 > CV = 1.833$, we reject the null hypothesis. **Interpretation:** There is significant evidence that 7 hours of sleep deprivation increases reaction time.