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Inference for Regression Using Technology for AP Statistics (page 2)

By — McGraw-Hill Professional
Updated on Feb 4, 2011

All of the mechanics needed to do a t-test for the slope of a regression line are contained in this printout. You need only to quote the appropriate values in your write-up. Thus, for the problem given above, we see that t = 15.23 P-value = 0.000.

A confidence interval for the slope of a regression line follows the same pattern as all confidence intervals (estimate ± (critical value) × (standard error)): b ± t*sb , based on n – 2 degrees of freedom. A 99% confidence interval for the slope in this situation (df = 10 t* = 3.169 from Table B) is 0.9934 ± 3.169(0.06523) = (0.787, 1.200). (Note: The newest software for the TI-84 has a LinRegTInt built in. The TI-83/84 and earlier versions of the TI-84 do not have this program. The program requires that the data be available in lists and, unlike other confidence intervals in the STAT TESTS menu, there is no option to provide Stats rather than Data.)

To use the calculator to do the regression, enter the data in, say, L1 and L2. Then go to STAT TESTS LinRegTTest. Enter the data as requested (response variable in the Ylist:). Assuming the alternative is two sided (HA : β ≠ 0), choose β and ρ ≠ 0. Then Calculate. You will get the following two screens of data:

This contains all of the information in the computer printout except sb. It does give the number of degrees of freedom, which Mini- Tab does not, as well as greater accuracy. Note that the calculator lumps together the test for both the slope (β) and the correlation coefficient (ρ) because, as we noted earlier, the test statistics are the same for both.

If you have to do a confidence interval using the calculator and do not have a TI-84 with the LinRegTInt function, you first need to determine sb. Because you know that , it follows that , which agrees with the standard error of the slope ("St Dev" of "Number") given in the computer printout.

A 95% confidence interval for the slope of the regression line for predicting temperature from the number of chirps per minute is then given by 0.9934 ± 2.228(0.0652) = (0.848, 1.139). t* = 2.228 is based on C = 0.95 and df = 12 – 2 = 10. Using LinRegTInt, if you have it, results in the following (note that the "s" given in the printout is the standard error of the residuals, not the standard error of the slope):

Practice problems for these concepts can be found at:

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