### Impact of Exercise Program on Employee Absences**Context:**Lester Hollar, Vice President for human resources at a large manufacturing company, has observed an increase in employee absences. Believing this might be linked to general health, a fitness program was initiated, allowing employees to exercise during lunch. Hollar evaluated the program with a random sample of 8 employees, comparing their absences 6 months before and after the program's start. Here are the results:**Employee Absence Data:**| Employee | Before | After ||----------|--------|-------|| 1 | 7 | 3 || 2 | 5 | 6 || 3 | 5 | 3 || 4 | 4 | 4 || 5 | 5 | 7 || 6 | 6 | 4 || 7 | 4 | 3 || 8 | 5 | 2 |**Data Analysis:**At the 0.05 significance level, can it be concluded that the number of absences has declined?**Steps:**a. **Hypotheses:**- We need to determine the correct null and alternate hypotheses. The choices include: - \( H_0: \mu_{\text{Before}} - \mu_{\text{After}} \leq 0 \) vs. \( H_1: \mu_d > 0 \) - \( H_0: \mu_{\text{After}} - \mu_{\text{Before}} \leq 0 \) vs. \( H_1: \mu_d > 0 \) - Other options as per the data and hypothesis test setup.b. **Test Statistic:**- Compute the test statistic. (Instructions specify rounding to 3 decimal places.)c. **p-Value:**- Compute the p-value. (Instructions specify rounding to 4 decimal places.)**Note:** Calculations for the test statistic and p-value are based on comparing the means of the 'Before' and 'After' data, typically using a paired t-test, considering the conditions and requirements for statistical hypothesis testing.

The data are given, level of significance = α = 0.050 Here we use paired-t test a) The null and alternative hypotheses are, H0 : μBefore - μAfter = μd≤0H1 : μBefore - μAfter = μd>0 Calculations: The sample standard deviation can be calculated as, s=∑di-d2n-1 =19.8758-1 =1.6850 b) The test statistics can be calculated as, t=x1-x2s/n =5.625-41.6850/8 =2.7277 Therefore, the test statistics value is 2.728 c) Degrees of freedom = n-1 = 8-1 = 7 P-value = 0.0147 Criteria: if p-value < 0.05 then we reject H0 at 5% level of significance. Conclusion: Here we may reject H0 at 5% level of significance because p-value is < 0.05. Interpretation: There is a significant difference between before exercise program and after exercise program.