Question

Exercise 11-17 (LO11-3)

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	10	14	19
Afternoon shift	10	9	14	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.

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

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

3. What is the p-value?

• Between 0.05 and 0.10

• Between 0.001 and 0.005

• Between 0.005 and 0.01

4. What is your decision regarding H₀?

• Do not reject H₀

• Reject H₀

Answer 1

Sample #1	Sample #2	difference , Di =sample1-sample2	(Di - Dbar)²
11	10	1	0.06
10	9	1	0.06
14	14	0	1.56
19	16	3	3.06

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	54	49	5	4.750

mean of difference ,    D̅ =ΣDi / n =   1.250
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    1.258

Ho :   µd≤ 0
Ha :   µd >   0

a)

Degree of freedom, DF=   n - 1 =    3  
t-critical value , t* =        3.182   [excel function: =t.inv(α,df) ]
reject Ho, if test stat >3.182

b)

std error , SE = Sd / √n =    1.2583   / √   4   =   0.6292      
                          
t-statistic = (D̅ - µd)/SE = (   1.2500   -   0   ) /    0.6292   =   1.987

c)

p value =Between 0.05 and 0.10

d)

p-value>α , Do not reject null hypothesis