A local company is concerned about the number of days missed by its employees due to illness. A random sample of 10 employees is selected. The number of days absent in one year is listed below. An incentive program is offered in an attempt to decrease the number of days absent. The number of days absent in one year after the incentive program is listed below.

A B C ד D E G H J 14 11 10 6 6 3 7 2 11 2 Employee Days Absent Before the Incentive Days Absent After the Incentive 4 3 6 6 9

1. Create and interpret a 95% confidence interval as an estimate for the true mean difference

2. Conduct an appropriate significance test:

  • Condition check
  • State the claim
  • Write the hypotheses
  • State the test statistic and the p-value.
  • Make a decision about Ho and state your decision in the context of the claim
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2 Answers

 

In the table, Before and After refer to the introduction of the incentive program. Some statistics have been calculated from the data. The target data is the difference between the before and after values.

The sample variance, s² = (∑(B-A)²-(∑(B-A)²)/n)/(n-1) where n=10, the sample size.

From this the sample SD is the square root of the variance.

From t tables with 9 DOF, we read a value of 2.262 (2-tail because we are considering a range below and above the mean difference). This tell us how far from the mean is the standard deviation for a 95% CI. But we have to adjust s to s/√n=5.2546/√10=1.662 approx.

Answer to part 1:

The CI is 2.5±2.262×1.662=2.5±3.76, giving us the interval:

-1.26≤mean difference≤6.26 days. This can be written [-1.26,6.26].

So, the mean difference between the average number of days lost before and after introduction of the incentive program lies between -1.26 and 6.26 days.

Answer to part 2:

CONDITION CHECK

See comment.

CLAIM

The incentive program reduces the number of days lost.

HYPOTHESES

Without the incentive program, no difference would be expected so H0: mean difference=0.

The incentive program was designed to reduce the absenteeism so H1: A<B, A-B<0, or B-A>0, mean difference is B-A, so mean difference>0. This implies a one-tailed t test.

p-VALUE

t=2.5/1.662=1.504, corresponding to a p-value of 0.0834. (9 DOF, 1-tail test 0.05 is implied.) Compared to significance level of 0.05. The p-value is not more extreme than the required significance level.

CONCLUSION

See comment.

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by Top Rated User (1.1m points)

The system would not allow me to provide a complete answer. And adding to comments also failed. Conclusion is that we can’t reject H0. Insufficient proof because level was not significant.

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