The null and alternate hypotheses are:
H_{0} : μ_{d} ≤ 0
H_{1} : 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 | 12 | 13 | 19 |
Afternoon shift | 9 | 10 | 14 | 15 |
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.
State the decision rule. (Round your answer to 2 decimal places.)
Compute the value of the test statistic. (Round your answer to 3 decimal places.)
What is the p-value?
Between 0.05 and 0.10
Between 0.001 and 0.005
Between 0.005 and 0.01
What is your decision regarding H_{0}?
Do not reject H_{0}
Reject H_{0}
Sol:
Rcode:
Dayshift <- c(11,12,13,19)
Afternoon <- c(9,10,14,15)
t.test(Dayshift,Afternoon,paired = TRUE,alternative =
"greater")
Output:
Paired t-test
data: Dayshift and Afternoon
t = 1.6977, df = 3, p-value = 0.09406
alternative hypothesis: true difference in means is greater than
0
95 percent confidence interval:
-0.6757915 Inf
sample estimates:
mean of the differences
1.75
State the decision rule. (Round your answer to 2 decimal places.)
if p<0.025,Reject Ho
if p>0.025,Fail to reject Ho.
Compute the value of the test statistic. (
t=1.698
What is the p-value?
p value is 0.09406
Between 0.05 and 0.10
What is your decision regarding H_{0}?
alpha=0.025
p=0.09406
p>alpha
Do not reject Ho.
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