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Computing the P-Value of a Two-Tailed Test

Example: Use the data below to test Grant’s claim that the mean price of a three-year-old Corvette is different from $37,500 at the  level of significance using the P-value approach.

Enter the data from the above table into L1.  Because the sample size is less than 30, the data must be tested for normality and checked for outliers.

To set up the normal probability plot, press 2^nd [STAT PLOT].  Press ENTER to select Plot 1.  Highlight On and press ENTER.  Set Type to the normal probability plot which is the third selection in the second row.  Press ENTER.  Set Data List to L1 and Data Axis to X.  For Marks select the small square.

Press ZOOM and select 9:ZoomStat and ENTER.

This plot is fairly linear, indicating that the data generally follows a normal distribution. 

To set up the boxplot, press 2^nd [STAT PLOT].  Press ENTER to select Plot 1.  Highlight On and press ENTER.  Set Type to the boxplot with outliers which is the first selection in the second row.  Press ENTER.  Set XList to L1 and Freq to 1.  For Marks select the small square.

Press ZOOM and select 9:ZoomStat and ENTER.

There are no outliers indicated in the boxplot.  (Note:  Outliers would appear as *’s at the extreme left or right ends of the boxplot.)

This test is a two-tailed test for vs.  .  The population standard deviation, , is 4100.  The Z-Test is the appropriate test.  To run the test, press STAT, highlight TESTS and select 1:Z-Test.   Since you are using the actual data, which is stored in L1, for the analysis, select Data of Inpt and press ENTER.  For   enter 37500, the value for  in the null hypothesis.  For  enter 4100.  Enter L1 for List, and 1 for Freq.  On the next line, choose the appropriate alternative hypothesis and press ENTER.  For this example, it is , a two-tailed test.

Highlight Calculate and press ENTER.

Or, highlight Draw and press ENTER.

Notice the P-value is equal to .4805.  In this example, a is .01.  Since the P-value is greater than a, the correct conclusion is to Fail to Reject .  (Note:  the P-value calculated using the TI-83 slightly different from the P-value obtained using the Z-table.  That difference is simply due to rounding.)