Posted by **Henry** on Wednesday, April 4, 2012 at 4:45pm.

The manager at the YMCA believes members are staying longer at the Y due to what they feel was a very successful marketing plan. Studies show the previous mean time per visit was 36 minutes, with a standard deviation = 11 minutes. A simple random sample of n = 200 visits is selected, and the current sample mean = 36.8 minutes. Test the manager's claim at a =0.05 using the p-value.

- Statistics -
**MathGuru**, Wednesday, April 4, 2012 at 8:14pm
Try a one-sample z-test.

Hypotheses:

Ho: µ = 36 -->null hypothesis

Ha: µ > 36 -->alternate hypothesis

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

With your data:

z = (36.8 - 36)/(11/√200) = ?

Finish the calculation.

The p-value is the actual level of the test statistic. Find it using a z-table. Compare to 0.05 to determine whether or not to reject the null. If the null is rejected, you can conclude a difference.

I hope this will help get you started.

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