SOLUTION: A Basic Industries Corporation started a program which encourages its employees to exercise during the lunch hour to maintain good health. The reason for this intuitive is to reduc

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Question 1178193: A Basic Industries Corporation started a program which encourages its employees to exercise during the lunch hour to maintain good health. The reason for this intuitive is to reduce the number of absentees due to health issues. To evaluate the program, Dr. Mohammad randomly selected 8 participants and examined number of missing days before and after the program started for each participant. Use the following statistics to answer the following two questions
Di=14 (Di-D)=59.4804 n=8
Can Dr. Mohammad conclude that the number of missing days has declined?
In terms of the random variable of number of missing days, , the null and the alternative hypotheses are
H0:μD=0 & H1:μD≠0
H0:μD<0 & H1:μD≥0
H0:μD<0 & H1:μD≥0
H0:μD≤0 & H1:μD>0

Answer by CPhill(1987)   (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Understanding the Problem**
* We are testing if the number of missing days has declined after the program.
* We have paired data (before and after) for 8 participants.
* We are using a paired t-test.
* Di represents the difference between the number of missing days before and after the program.
* We are given:
* ΣDi = 14
* Σ(Di - D̄)² = 59.4804
* n = 8
**2. Calculating the Mean Difference (D̄)**
* D̄ = ΣDi / n = 14 / 8 = 1.75
**3. Calculating the Sample Standard Deviation of the Differences (sD)**
* sD = √[Σ(Di - D̄)² / (n - 1)] = √(59.4804 / 7) ≈ √8.4972 ≈ 2.915
**4. Setting up the Hypotheses**
* We want to see if the number of missing days has *declined*. This means the difference (before - after) should be positive.
* Therefore, the null and alternative hypotheses are:
* H0: μD ≤ 0 (The mean difference is less than or equal to zero, meaning no decline or an increase)
* H1: μD > 0 (The mean difference is greater than zero, meaning a decline)
**5. Calculating the t-statistic**
* t = (D̄ - μD) / (sD / √n)
* Under the null hypothesis, μD = 0.
* t = 1.75 / (2.915 / √8) ≈ 1.75 / (2.915 / 2.828) ≈ 1.75 / 1.0307 ≈ 1.698
**6. Determining the Degrees of Freedom**
* df = n - 1 = 8 - 1 = 7
**7. Finding the P-value**
* We need to find the p-value for a one-tailed t-test with df = 7 and t = 1.698.
* Using a t-table or calculator, we find that the p-value is approximately 0.067.
**8. Making a Decision**
* Let's assume a significance level (alpha) of 0.05.
* Since the p-value (0.067) > alpha (0.05), we fail to reject the null hypothesis.
**9. Conclusion**
* Dr. Mohammad cannot conclude that the number of missing days has declined at the 0.05 significance level.
**10. Correct Hypotheses**
* The correct null and alternative hypotheses are:
* H0: μD ≤ 0
* H1: μD > 0

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