posted by Henry .
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.
Try a one-sample z-test.
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.