Question 400989: The production manager at Bellevue Steel, a manufacturer of wheelchairs, wants to perform a second analysis of the number of defective wheelchairs produced on the day shift and afternoon shift. A sample of the production from 6 day shifts and 8 afternoon shifts revealed the following number of defects.
At the 0.10 significance level, is there a difference in the mean number of defects per shift?
Day 10 9 6 6 4 9
Afternoon 8 10 7 5 8 10 5 6
1. State the null and alternate hypotheses.
2. What is the decision rule?
3. What is the value of the test statistic?
4. What is your decision regarding the null hypothesis?
5. What is the p-value?
6. Interpret the result.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The production manager at Bellevue Steel, a manufacturer of wheelchairs, wants to perform a second analysis of the number of defective wheelchairs produced on the day shift and afternoon shift. A sample of the production from 6 day shifts and 8 afternoon shifts revealed the following number of defects.
At the 0.10 significance level, is there a difference in the mean number of defects per shift?
Day 10 9 6 6 4 9
Afternoon 8 10 7 5 8 10 5 6
1. State the null and alternate hypotheses.
Ho: u1-u2 = 0
Ha: u1-u2 is not equal to 0
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Run a two-sample T-test.
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x-bar1 = 7.6 and x-bar2 = 7.375
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2. What is the decision rule?
Reject Ho if the ts is < -1.89 or > +1.89
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3. What is the value of the test statistic?
t = 0.1697
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4. What is your decision regarding the null hypothesis?
Fail to reject Ho.
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5. What is the p-value?
P-value = 2*P(t > 0.1697 when df=7)
p = 0.87
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6. Interpret the result.
The test shows strong evidence that the means are
statistically the same at the 10% significance level.
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Cheers,
Stan H.
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