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.
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 H0?
Do not reject H0
The table given below ,
|Day||Day shift(X)||Afternoon shift(Y)||di=X-Y||di^2|
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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