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 6 9 8 7 10 8

Evening Shift 9 11 8 12 10 13 15 10

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.

a) State the null and alternate hypotheses.

b) What is the level of significance?

c) What is the test statistic?

d) What is the decision rule?

e) 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).

f) What is your conclusion? Explain.

g) Does the decision change at the 99% confidence level?

Answer 1

using excel>ata>data analysis>two sample t

we have

t-Test: Two-Sample Assuming Equal Variances
	day shift	Evening shift
Mean	8	11
Variance	2	5.142857
Observations	6	8
Pooled Variance	3.833333
Hypothesized Mean Difference	0
df	12
t Stat	-2.8372
P(T<=t) one-tail	0.007487
t Critical one-tail	1.782288
P(T<=t) two-tail	0.014974
t Critical two-tail	2.178813

a) the null and alternate hypotheses is

H0:u(day shift)= u(evening shift)

H0:u(day shift)> u(evening shift)

b) the level of significance is 0.05

c) the test statistic t =-2.8372

d) we reject H0 if | t |>1.7822

e)

t-Test: Two-Sample Assuming Equal Variances
	day shift	Evening shift
Mean	8	11
Variance	2	5.142857
Observations	6	8
Pooled Variance	3.833333
Hypothesized Mean Difference	0
df	12
t Stat	-2.8372
P(T<=t) one-tail	0.007487
t Critical one-tail	1.782288
P(T<=t) two-tail	0.014974
t Critical two-tail	2.178813

f) since | t | 2= 2.8372>1.7822 so we reject Ho . and conclude that  the mean number of defects on the night shift is greater than the mean number on the day shift

g) since p value of t is 0.0075<0.10 , so our decison will not change