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 | 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.
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 0.005 and 0.01
Between 0.001 and 0.005
Between 0.01 and 0.025
What is your decision regarding H_{0}?
Reject H_{0}
Do not reject H_{0}
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.
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