27) The total blood cholesterol concentrations, in mg/dL, are shown below for a randomsample of six people. In the table below, x is the age of the subject and y is the subject'stotal cholesterol concentration. Find the 95% prediction interval when x = 58 years.27)x (age)364958696371204256194257245y (mg/dL)209A) Regression line: y = 1.82x +122.5Standard error of the estimate: sest= 16.51184Prediction interval for y(58): 159

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Asked Dec 1, 2019
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27) The total blood cholesterol concentrations, in mg/dL, are shown below for a random
sample of six people. In the table below, x is the age of the subject and y is the subject's
total cholesterol concentration. Find the 95% prediction interval when x = 58 years.
27)
x (age)
36
49
58
69
63
71
204
256
194
257
245
y (mg/dL)
209
A) Regression line: y = 1.82x +122.5
Standard error of the estimate: s
est
= 16.51184
Prediction interval for y(58): 159 <y(58)< 259
B) Regression line: y = 1.720697x+ 122.506444
Standard error of the estimate: s
= 16.51184
est
Prediction interval for y(58): 159 <y(58) < 259
C) Regression line: y = 1.820697x + 122.506444
Standard error of the estimate: s
= 16.51184
est
Prediction interval for y(58): 179 <y(58) < 279
D) Regression line: y 1.72x + 122.5
Standard error of the estimate: st 16.51184
est
Prediction interval for y(58): 179 <y(58) < 279
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27) The total blood cholesterol concentrations, in mg/dL, are shown below for a random sample of six people. In the table below, x is the age of the subject and y is the subject's total cholesterol concentration. Find the 95% prediction interval when x = 58 years. 27) x (age) 36 49 58 69 63 71 204 256 194 257 245 y (mg/dL) 209 A) Regression line: y = 1.82x +122.5 Standard error of the estimate: s est = 16.51184 Prediction interval for y(58): 159 <y(58)< 259 B) Regression line: y = 1.720697x+ 122.506444 Standard error of the estimate: s = 16.51184 est Prediction interval for y(58): 159 <y(58) < 259 C) Regression line: y = 1.820697x + 122.506444 Standard error of the estimate: s = 16.51184 est Prediction interval for y(58): 179 <y(58) < 279 D) Regression line: y 1.72x + 122.5 Standard error of the estimate: st 16.51184 est Prediction interval for y(58): 179 <y(58) < 279

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Expert Answer

Step 1

27)

 

Excel Procedure for Regression:

 

  • Enter the data for x and y in Excel sheet.
  • Go to Data
  • Click on Data Analysis.
  • Select ‘Regression’ and click on ‘OK’
  • Select the column of y under ‘Input Y Range’.
  • Select the column of x under ‘Input X Range’.
  • Click on ‘OK’.

 

Excel Output:

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SUMMARY OUTPUT Regression Statistics Multiple R 0.853072998 R Square 0.72773354 Adjusted R Square 0.659666925 Standard Error 16.51183898 Observations 6 ANOVA Significance F df SS MS F Regression 2914.936694 2914.936694 10.69149009 0.030795419 Residual 4 1090.563306 272.6408264 Total 5 4005.5 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 122.5064443 31.41079703 213.6020915 32.81017409 3.7337944 0.020229953 1.820697498 0.556824937 3.269784411 0.030795419 0.274703628 3.366691368 X

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Step 2

From the output, the regression equation is Y=122.506444+1.820697x.

 

The standard error of the estimate is 16.51184 and it is calculated below:

 

From the Excel output, the residual mean sum of squares is 272.6408264.

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Step 3

The 95% prediction interval for x=58 is (179,279) and it is calculated below:

 

The pre...

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y 122.506444+1.820697x =122.506444 (1.820697 x 58) = 228.1069

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