A company is evaluating the impact of a wellness program offered on-site as a me

Question

A company is evaluating the impact of a wellness program offered on-site as a means of reducing employee sick days. A total of 8 employces agree to participate in the evaluation which lasts 12 weeks. Their sick days in the 12 months prior to the start of the wellness program and again over the 12 months after the completion of the program are recorded and are shown below. Is there a significant reduction in the number of sick days taken after completing the wellness program? Use the Sign Test at a 5% level of significance. 1. Complete the table below Employee Sick Days Taken in Sick Days Taken Difference 12 Months Prior to in 12 Months (ReductionSign Following Prior-After) 12 10 Alpha Critical value- Smaller of the number of positive or negative signs- Based on comparing the smaller of the number of positive or negative signs to the Critical value which of the following is (are) true? A. There is statistically significant evidence at alpha-0.05 to show that the median difference (reduction) is positive. B. There is not statistically significant evidence at alpha 0.05 to show that the median difference (reduction) is positive HLTH 501 C. There are not enough data points to reach a conclusion. D. The tied value makes the analysis suspect. A small study (n-10) is designed to assess if there is an association between smoking in pregnancy and low birth weight. Low birth weight babies are those born less than 5.5 pounds. The following data represent the birth weights, in pounds, of babies born to mothers who reported smoking in pregnancy and those who did not. 2. 5.0 4.2 4.8 3.3 3.9 Mother smoked in pregnancy Mother did not smoke during pregnancy 5.1 4.9 5.3 54 Is there a significant difference in birth weights between mother who smoked during gnancy and those who did not? Apply the Mann Whitney U Test at a 5% level of

Explanation / Answer

Positive sign count = 6
Negative sign count = 1
Total count = 7
Z-score Calculation

z = (X - pn) / npq

z = (6 - 3.5) / 1.75

z = 1.889822

critical Z value is 1.645 (one tailed test)

The z-value is 1.889822. The p-value is .029391. The result is significant at p < 0.05.

There is statistically significant evidenceat alpha= 0.05 to show that median difference in reduction is positive.

2)

Sample 1
Sum of ranks: 18
Mean of ranks: 3.6
Expected sum of ranks: 27.5
Expected mean of ranks: 5.5
U-value: 22
Expected U-value: 12.5

Sample 2
Sum of ranks: 37
Mean of ranks: 7.4
Expected sum of ranks: 27.5
Expected mean of ranks: 5.5
U-value: 3
Expected U-value: 12.5

Sample 1 & 2 Combined
Sum of ranks: 55
Mean of ranks: 5.5
Standard Deviation: 4.7871

The U-value is 3. The critical value of U at p < .05 is 2. Therefore, the result is not significant at p < .05.

Prior to program After Program DIF (T1-T2) SIGN 8 7 1 + 6 6 0 n/a 4 5 -1 - 12 11 1 + 10 7 3 + 8 4 4 + 6 3 3 + 2 1 1 +

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