Question

As part of a research program for a new cholesterol drug, a pharmaceutical company would like to investigate the relationship between the ages and LDL (low-density lipoprotein) cholesterol of men. The accompanying data set shows the ages and LDL cholesterol levels of seven randomly selected men. Construct a 95% prediction interval to estimate the LDL cholesterol level of a 28-year-old man. Click the icon to view the data table. Determine the upper and lower limits of the prediction interval UPLE LPL- (Round to one decimal place as needed.)

Answer 1

Obs X Y Y XY 1 23 143 529 20449 3289 2 38 177 1444 31329 6726 3 25 158 625 24964 3950 4 32 196 1024 38416 6272 5 43 158 1849 24964 6794 6 31 142 961 20164 4402 7 40 207 1600 42849 8280 Sum = 232 1181 8032 203135 39713 The sum of squares obtained from the table above are: 2 1 SSxx 3-(3x) 8032 – < (232) 342.8571 SSyy Šv! (EX) 203135 – (1181) 3883.4286 72 1 SSXY ΣΧΥ X 11 39713 - _ (232) (1181) 571.2857 The slope and y-intercept coefficients are computed using the following formulas SSXY SSXX 571.2857 342.8571 1.6663 Bo = Y - i * * 168.7143 - 1.6663 x 33.1429 113.49 Therefore, the regression equation is: Y = 113.49 + 1.6663X Now that we have the regression equation, we can compute the predicted value for X = 28, by simply plugging in the value of x = 28 in the regression equation found above Ý 113.49 +1.6663 x 28 160.145

Now, we are going to compute the 95% prediction interval for the predicted value Y = 160.145. The first step consists of finding the standard error of the estimate, for which we need to deal with the regression sum of squares and with the sum of squared errors. We know that the total sum of squares is: S STotal = SSyy = 3883.4286 Also, the regression sum of squares is computed as follows: SS Regression Å SSxr 1.6663 x 571.2857 951.9048 Since we know that S STotal = SS regression + SSError, then we can compute the sum of squared errors as follows: S SError SSTotal - SS Regression 3883.4286 - 951.9048 2931.5238 Hence, the means squared error is: SS Error MSE n - 2 2931.5238 7 - 2 586.3048 Now, taking square root, we get the standard error of the estimate: VMSE V586.3048 24.2137 Since we want to compute a 95% prediction in tervel for the predicted value Y = 160.145, the significance level used is a = 0.05. The critical t- value for df = n - 2 = 7 - 2 = 5 degrees of freedom, and a = 0.05 is to = 2.571. Now, with all this information, we are ready to compute the margin of error of the prediction interval: 1 E ta/2;n – 21 g? 1 + + (Xo-X SSxx n 1 2.571 x 586.3048 1+ IN + (28 - 33.14292 342.8571 66.1029 Hence, the 95% prediction in tervel for the predicted value Y = 160.145 is: PI (160.145 - E, 160.145 - E) (160.145 - 66.1029, 160.145 +66.1029) (94.0421, 226.2479)

UPL = 226.2

LPL = 94.0

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