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	10	12	15	18
Afternoon shift	8	9	12	17

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 0.005 and 0.01

• Between 0.001 and 0.005

• Between 0.01 and 0.025

4. What is your decision regarding H₀?

• Reject H₀

• Do not reject H₀

Answer 1

The table given below ,

Day	Day shift(X)	Afternoon shift(Y)	di=X-Y	di^2
1	10	8	2	4
2	12	9	3	9
3	15	12	3	9
4	18	17	1	1
		Sum	9	23

From table ,

The null and alternative hypothesis is ,

The test is right-tailed test.

Now , the df=degrees of freedom=n-1=4-1=3

The critical value is , ; From t-table

1. The decision rule is , Reject Ho if t-stat>3.18

2. The test statistic is ,

3. The p-value is

between 0.005 to 0.01

Decision : Here , t-stat=4.700>3.18

Therefore , reject Ho.