Question

The null and alternate hypotheses are: H0 : μd ≤ 0 H1 : 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 12 12 16 19 Afternoon shift 10 10 12 15 At the 0.010 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.

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

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

Answer 1

The table given below ,

Day	Day shift (X)	Afternoon shift(Y)	di=X-Y	di^2
1	12	10	2	4
2	12	10	2	4
3	16	12	4	16
4	19	15	4	16
		Sum	12	40

From the table ,

We have , Hypothesis : Vs

a) The critical value is ,

; From t-table

Decision rule : Reject Ho if

b) The test statistic is ,

Decision : here ,

Therefore , reject Ho

Conlcusion : Hence , there is sufficient evidence to support the claim there are more defects produced on the day shift.