Question

A manufacturer wants to compare the number of defects on the day shift with the number on the evening shift. A sample of production from recent shifts showed the following defects:

Day Shift 5 8 7 6 9 7

Evening Shift 8 10 7 11 9 12 14 9

The objective is to determine whether the mean number of defects on the night shift is greater than the mean number on the day shift at the 95% confidence level.

1. State the null and alternate hypotheses.

2. What is the level of significance?

3. What is the test statistic?

4. What is the decision rule?

5. Use the Excel Data Analysis pack to analyze the problem. Include the output with your answer. (Note: You may calculate by hand if you prefer).

6. What is your conclusion? Explain.

7. Does the decision change at the 99% confidence level?

Answer 1

Using excel:

Go to Data,select Data Analysis, choose t-Test: Two-Sample Assuming Unequal Variances

H0: µ1 =µ2 ,mean number of defects on the night shift is equal to mean number on the day shift

H1: µ1<µ2, mean number of defects on the night shift is greater than the mean number on the day shift

t-Test: Two-Sample Assuming Unequal Variances
	Day Shift	Evening Shift
Mean	7	10
Variance	2	5.142857143
Observations	6	8
Hypothesized Mean Difference	0
df	12
t Stat	-3.036364964
P(T<=t) one-tail	0.005171757
t Critical one-tail	1.782287556
P(T<=t) two-tail	0.010343513
t Critical two-tail	2.17881283

Level of significance = 1-0.95 = 0.05

Test statistic = -3.036

Decision rule: Since p-value(0.0052) is less than 0.05(level of significance), we reject null hypothesis a conclude that µ1<µ2, mean number of defects on the night shift is greater than the mean number on the day shift.

At 99% confidence interval,

Level of significance = 1-0.99 = 0.01

Since p-value(0.0052) is less than 0.01(level of significance), we reject null hypothesis a conclude that µ1<µ2, mean number of defects on the night shift is greater than the mean number on the day shift.

So, decision remains same at 99% confidence interval.