Question

14.4.26-T A Question Help O 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 following data set shows the ages and LDL cholesterol levels of seven randomly selected men. Construct a 95% confidence interval to estimate the average LDL cholesterol level of a 33-year-old man. Click the icon to view the data table. Determine the upper and lower limits of the confidence interval. UCL = 1 LCL = 1 (Round to one decimal place as needed.) - X Data Table Age Cholesterol 23 144 38 26 177 155 32 199 43 160 32 142 40 210 Print Done

Answer 1

Obs. X Y Y? 1 23 144 529 20736 3312 2 38 177 1444 31329 6726 3 26 155 676 24025 4030 4 32 199 1024 39601 6368 5 43 160 1849 25600 6880 6 32 142 1024 20164 4544 7 40 210 1600 44100 8400 Sum = 234 1187 8146 205555 40260 The sum of squares obtained from the table above are: 2 SSxx 3*-:(**) 8146 – _ (234) 323.7143 2 1 72 SSyy Y? n 1 205555 – (1187) 4273.7143 11 SSXY Źxy-- (**) ( 40260 (234) (1187) 7 580.2857 The slope and y-intercept coefficients are computed using the following formulas: SSXY SSxx 580.2857 323.7143 1.7926 Bo = Y - XX 169.5714 - 1.7926 X 33.4286 = 109.6478 Therefore, the regression equation is: Y = 109.6478 + 1.7926X Now that we have the regression equation, we can compute the predicted value for X = 33, by simply plugging in the value of X = 33 in the regression equation found above: û = 109.6478 +1.7926 x 33 168.8032

Now, we are going to compute the 95% prediction interval for the predicted value Y = 168.8032. 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 SSTotal = SSyy = 4273.7143 Also, the regression sum of squares is computed as follows: SSRegression B, SS xy 1.7926 X 580.2857 1040.2121 Since we know that SSTotal = SSRegression + SSError, then we can compute the sum of squared errors as follows: SS Error = SSTotal - SS Regression 4273.7143 1040.2121 = 3233.5022 Hence, the means squared error is: SSError MSE n-2 3233.5022 7 - 2 = 646.7004 Now, taking square root we get the standard error of the estimate: o MSE 646.7004 25.4303 Since we compute a prediction in tervel for the predi value Y = 168.8032, 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 te = 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 22 (1+ (X0 - X2 SSxx n = 2.571 x 646.7004 1 + 1 7 + (33 33.4286)2 323.7143 69.8437 Hence, the 95% prediction in tervel for the predicted value Y = 168.8032 is: PI = (168.8032-E, 168.8032 - E) (168.8032-69.8437, 168.8032 +69.8437) (98.9595, 238.6469)

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