Question

The null and alternate hypotheses are:

H₀ : μ_d ≤ 0

H₁ : μ_d > 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	11	12	14	18
Afternoon shift	8	9	13	16

At the 0.025 significance level, can we conclude there are more defects produced on the day shift? Hint: For the calculations, assume the day shift as the first sample.

State the decision rule. (Round your answer to 2 decimal places.)

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

What is the p-value?

Between 0.005 and 0.01

Between 0.001 and 0.005

Between 0.01 and 0.025

What is your decision regarding H₀?

Reject H₀

Do not reject H₀

rev: 08_30_2017_QC_CS-97216, 11_25_2017_QC_CS-109804

Answer 1

H0 : μd ≤ 0. there is no significant difference in the mean number of defective units produced on the day shift and the afternoon shift.
H1 : μd > 0. there are more defects produced on the day shift.

part1	part2	di
11	8	3
12	9	3
14	13	1
18	16	2

mean_diff=   2.25   [Excel function used -> AVERAGE]
sd_diff=   0.957   [Excel function used -> STDEV]
n=   4   [Excel function used -> COUNT]
      
test statistic, t= (mean_diff)/(sd_diff/sqrt(n))      
t=   2.25/(0.9574/sqrt(4))  
t=   4.7002  
t = 4.700

alpha=   0.025              
t(a,n-1) = t(0.025,4-1) = abs(T.INV(0.025,4-1)) = 3.182
decision rule: reject Ho if t > 3.18

p-value = P(T>|t|  
p-value = P(T>(4.70022978901191)  
p-value = T.DIST.rT(4.70022978901191,4-1)  
p-value = 0.009109228
p-value is Between 0.005 and 0.01

decision regarding H0: Reject Ho.

Conclusion: μd > 0. there are more defects produced on the day shift.