Question

. For a certain population of individuals the average cost of a trip to last year's Super Bowl was $875. The standard deviation was $50. This year 49 fans who scheduled the trip paid an average of $890 for the trip. Test the claim that the average cost of this year is greater than last year's cost. Should the null hypothesis be rejected at 0.01? Use the p-value method.

Answer 1

Here population standard deviation is known so we can use One sample z test:

Using Minitab:

1. Open the 1-Sample Z dialog box.
2. Statistics > 1-Sample Inference > Z
3. From the drop-down list, select summerize data.
4. 
   One-Sample Z for the Mean Summarized data 49 Sample size: Sample mean: 890 Known standard deviation: 50 vPerform hypothesis test Hypothesized mean: 875 Select Graphs. Cancel Help
   
5. In Known standard deviation, enter 50.
6. Select Perform hypothesis test.
7. In Hypothesized mean, enter 875.
8. Click OK.

Output:

One-Sample Z Test of μ-875 vs > 875 The assumed standard deviation50 N Mean SE Mean 99% Lower Bound 49 890.00 7.14 873.38 2.10 0.018

P-value 0.018 > 0.01 so we do not reject the null hypothesis.

Hope this will be helpful. Thanks and God Bless You :)