Solutions
 r = 0.9817, height = 25.41 + 0.261(age)
 The line does appear to be a good model for the data.
 The residual pattern seems quite random. A line still appears to be a good model for the data.
 Height = 25.41 + 0.261(35) = 34.545 (Y1(35) = 34.54). You probably shouldn't be too confident in this prediction. 35 is well outside of the data on which the LSRL was constructed and, even though the line appears to be a good fit for the data, there is no reason to believe that a linear pattern is going to continue indefinitely. (If it did, a 25yearold would have a predicted height of 25.41 + 0.261(12 × 25) = 103.71", or 8.64 feet!)
 The slope of the regression line is 0.261. This means that, for an increase in age of 1 month, height is predicted to increase by 0.261 inches. You could also say, that, for an increase in age of 1 month, height will increase on average by 0.261 inches.

 Math = 185.77 + 0.6419(Verbal).
 b = 0.6419. For each additional point scored on the SAT Verbal test, the score on the SAT Math test is predicted to increase by 0.6419 points (or: will increase on average by 0.6419 points). (Very important on the AP exam: be very sure to say "is predicted" or "on average" if you'd like maximum credit for the problem!)
 The standard error of the slope is sb = 0.1420. This is an estimate of the variability of the standard deviation of the estimated slope for predicting SAT Verbal from SAT Math.
 The standard error of the residuals is s = 7.457. This value is a measure of variation in SAT Verbal for a fixed value of SAT Math.

 The hypotheses being tested are H_{o} : β = 0 (which is equivalent to H_{o} : r = 0) and H_{A} β ≠ 0, where β is the slope of the regression line for predicting SAT Verbal from SAT Math.
 The test statistic used in the analysis is = 4.52, df = 23 – 2 = 21.

 df = 23 – 2 = 21 t* = 2.080. The 95% confidence interval is: 0.6419 ± 2.080(0.1420) = (0.35, 0.94). We are 95% confident that, for each 1 point increase in SAT Verbal, the true increase in SAT Math is between 0.35 points and 0.94 points.
 The procedure used to generate the confidence interval would produce intervals that contain the true slope of the regression line, on average, 0.95 of the time.

 Let β = the true slope of the regression line for predicting SAT Math score from the percentage of graduating seniors taking the test.
 H_{0}: β = 0.
 H_{A}: β < 0.
 We use a linear regression t test with α = 0.01. The problem states that the conditions for doing inference for regression are met.
 We see from the printout that
 Because P < 0.01, we reject the null hypothesis. We have very strong evidence that there is a negative linear relationship between the proportion of students taking SAT math and the average score on the test.
based on 50 – 2 = 48 degrees of freedom. The Pvalue is 0.000. (Note: The Pvalue in the printout is for a twosided test. However, since the Pvalue for a onesided test would only be half as large, it is still 0.000.)
 Let β = the true slope of the regression line for predicting SAT Math score from the percentage of graduating seniors taking the test.

 , df = 20 – 2 = 18 Pvalue = 0.644.
 df = 18 t* = 2.878; 0.2647 ± 2.878(0.5687) = (–1.37, 1.90).
 No. The Pvalue is very large, giving no grounds to reject the null hypothesis that the slope of the regression line is 0. Furthermore, the correlation coefficient is only , which is very close to 0. Finally, the confidence interval constructed in Part (b) contains the value 0 as a likely value of the slope of the population regression line.
 The tratio would still be 0.47. The Pvalue, however, would be half of the 0.644, or 0.322 because the computer output assumes a twosided test. This is a lower Pvalue but is still much too large to infer any significant linear relationship between pulse rate and height.


 Let β = the true slope of the regression line for predicting the number of manatees killed by powerboats from the number of powerboat registrations.
 H_{0}: β = 0.
 H_{A}: β > 0.
 We use a ttest for the slope of the regression line. The problem tells us that the conditions necessary to do inference for regression are present.
 We will do this problem using the TI83/84 as well as Minitab.
 On the TI83/84, enter the number of powerboat registration in L1 and the number of manatees killed in L2. Then go to STAT TESTS LinRegTTest and set it up as shown below:
After "Calculate," we have the following two screens.
The Minitab output for this problem is:
 Because the Pvalue is very small, we reject the null. We have very strong evidence of a positive linear relationship between the number of powerboat registrations and the number of manatees killed by powerboats.
 Let β = the true slope of the regression line for predicting the number of manatees killed by powerboats from the number of powerboat registrations.
 Using the residuals generated when we did the linear regression above, we have:

 Using the TI83/84 results, df = 12 t*=1.782. We need to determine s_{b}. We have . The confidence interval is 0.125 ± 1.782(0.013) = (0.10, 0.15).
 Directly from the Minitab output: 0.125 ± 1.782(0.013) = (0.10, 0.15).
There appears to be no pattern in the residual plot that would cause us to doubt the appropriateness of the model. A line does seem to be a good model for the data.

(Assuming that you have put the age data in L1 and the height data in L2, remember that this can be done on the TI83/84 as follows: STAT CALC LinReg(a+bx) L1,L2,Y1.)
(After the regression equation was calculated on the TI83/84 and the LSRL stored in Y1, this was constructed in STAT PLOT by drawing a scatterplot with Xlist:L1 and Ylist:L2.)
(This scatterplot was constructed on the TI83/84 using STAT PLOT with Xlist:L1 and Ylist:RESID. Remember that the list of residuals for the most recent regression is saved in a list named RESID.)
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