SOLUTION: A company that sells mail order systems has been planning inventory and staffing based on an assumption that the mean of their weekly sales is 180. The weekly sales are normally di

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Question 135343: A company that sells mail order systems has been planning inventory and staffing based on an assumption that the mean of their weekly sales is 180. The weekly sales are normally distributed. The company selects 10 weeks at random from the past year and obtains the data shown below.
Weekly sales:
179 172 170 179 194


189 176 198 193 178
Set up the hypothesis test to test weather the sample mean is different from 185
At the 0.05 significance, what can you conclude about the company's assumption?
Treat as a large sample.



Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A company that sells mail order systems has been planning inventory and staffing based on an assumption that the mean of their weekly sales is 180. The weekly sales are normally distributed. The company selects 10 weeks at random from the past year and obtains the data shown below.
Weekly sales:
179 172 170 179 194 189 176 198 193 178
Set up the hypothesis test to test weather the sample mean is different from 185
At the 0.05 significance, what can you conclude about the company's assumption?
Treat as a large sample.
--------------------------------
x-bar = 182.8 ; s = 9.87477..
--------------------------------
Ho: u=185
Ha: u Is not 185
----------------------
Critical value for 2-tail test with alpha = 5%: z= +/- 1.96
------------------------
Test statistic: z(182.8) = (182.8-185)/[9.87477/sqrt(10)] = -0.704523
-----------------------------------------
Conclusion: Since test statistic is not in the reject interval, Fail
to reject Ho. The test gives statistical evidence the u=185
===========================================
Cheers,
Stan H.

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