Question

The null and alternative hypotheses are: H H :Hd < 0 :Hd > 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 Day shift Afternoon shift 1 10 8 2 12 9 3 15 12 4 11 15 At the 0.05 significance level, can we conclude there are more defects produced on the Afternoon shift? a. State the decision rule. (Round the final answer to 3 decimal places.) Ho is (Click to select) 9 if t>O . b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.) Test statistic c. What is your decision regarding the null hypothesis? Ho should (Click to select) 9. d. Determine the p-value. (Round the final answer to 4 decimal places.) The p-value is O .

Answer 1

Ans a ) here total number of observation are 4

t at 0.05 with 3 df 2.353

Ho is rejected if t>2.353

b ) using minitab>stat>basic stat>paired t

we have

Paired T-Test and CI: day, afternoon

Paired T for day - afternoon

N Mean StDev SE Mean
day 4 12.00 2.16 1.08
afternoon 4 11.00 3.16 1.58
Difference 4 1.00 3.37 1.68

95% upper bound for mean difference: 4.96
T-Test of mean difference = 0 (vs < 0): T-Value = 0.5941 P-Value = 0.7029

b ) the value of test statistic =0.594

c ) Ho should not rejected

d ) the p value is 0.7029