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	12	12	16	19
Afternoon shift	12	10	16	18

At the 0.100 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 0.10 and 0.15

• 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

a. Decision rule

Sample size : n= 4

Degrees of freedom :df =n-1 =4-1=3

Significance level : = 0.10

Right tailed test :

For right tailed test Critical value : t =t0.10

t0.10 for 3 degrees of freedom = 1.638

Reject the null hypothesis if Value of the test statistic > Critical value 1.638

b. Value of the test statistic :

Difference : d : Defect prodcued in day shift - defects produced in  afternoon shift

Sample Mean difference

Sample standard deviation of difference

Day	Day shift	Afternoon shift	d	(d-)	(d-)2
1	12	12	0	-0.7500	0.5625
2	12	10	2	1.2500	1.5625
3	16	16	0	-0.7500	0.5625
4	19	18	1	0.2500	0.0625
			d=3		(d--)2=2.75
			=0.75

value of the test statistic = 1.567

a. What is the p-value?

For 3 degrees of freedom P(t>1.5667) =0.1076

p-value = 0.1076 ; Therefore; p-value is between 0.10 and 0.15

Answer

Between 0.10 and 0.15

a. What is your decision regarding H0?

Do not reject H0

As P-Value i.e. is greater than Level of significance i.e (P-value:0.1076 > 0.1:Level of significance); Do not Reject Null Hypothesis