Question

The null and alternative hypotheses are:

H0:μd≤0H0:μd≤0

H1:μd>0H1:μd>0

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

	Day
	1	2	3	4
Day shift	10	12	15	18
Afternoon shift	8	9	13	16

At the 0.01 significance level, can we conclude there are more defects produced on the Afternoon shift?

a. State the decision rule. (Round the final answer to 3 decimal places.)

H₀ is  (Click to select)  rejected  not rejected  if t >  .

b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)

Test statistic

c. What is your decision regarding the null hypothesis?

H₀ should  (Click to select)  not be rejected  be rejected  .

d. Determine the p-value. (Round the final answer to 4 decimal places.)

The p-value is  .

Answer 1

H0:μd≤0

H1:μd>0

a)

H₀ is    rejected if t > 4.541

b)

TS = 9

c)

Null hypothesis is rejected

d)

p-value = 0.0014