Question 263144
The production manager at Bellevue Steel, a manufacturer of wheelchairs, wants to compare the number of defective wheelchairs produced on the day shift with the number on the afternoon shift. 
A sample of the production from 6 day shifts and 8 afternoon shifts revealed the following number of defects. Assume this is a two sample “Unknown Variances, Assumed Equal”. The level of significance is 0.05. The standard deviation for Day is 1.4142 and the standard deviation for Afternoon is 2.2678. 
Day:::::: 5  8 7  6 9 7 
Afternoon 8 10 7 11 9 12 14 9 
1. Set up the null and alternative hypothesis. 
Ho: u(day)-u(aft) = 0
Ha: u(day)-u(aft) is not equal to 0
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2. Calculate the value of the test statistic.
I ran a 2-Sample Ttest and got:
t = -3.0363... 

3. What is the critical value
If alpha = 5% cv = +-invT(0.025,df = 12) = +-2.1788
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4. What is the decision rule (DR)?
Reject Ho if the test stat is > 2.1788 or <-2.1788
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5. Reject Ho if (complete the DR in next box) 
6. What is your decision regarding Ho? 
Accept or Reject Ho?
Reject Ho.
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Cheers,
Stan H.