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a) Construct a 95% confidence interval about the mean response at x = 46 using the data set Steam.

b) Calculate the R2 value and the coefficient of correlation for the data set Steam.

Steam Data Set

Month Temperature(x) Usage/1000 (y) Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec 21 24 32 43 50 59 70 74 62 50 41 30 185.79 214.47 288.03 424.84 454.58 539.03 621.55 675.06 562.03 452.93 369.95 273.98

math   Statistics-And-Probability   
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Answer #1

Calculations for Regression coefficients: bo and b x = temperature Y usage/1000 21 185.79 24 214.47 32 288.03 43 424.84 50 454.58 59 539.03 70 621.55 74 675.06 62 562.03 50 452.93 Sample Size: n:-length (x)-12 n-1 AIC Mean value of x: 46.333 n-1 Mean value of y: y-130_ Sample slope numberator:SSXY-Σ Sample slope denominator: SSX-Σ 4-2) Sample Slope: n-1 (z-z).(*-L)=30857.917 n-1 3410.667 に0 SSXY 9.047 SSXSSXY Sample Slope: 9.047 50 452.93 41 369.95 30 273.98 1 SSX Sample Y intercept: Predicted value of score on art expenditures Measurement of variation in y and prediçted y Total Sum of Squares(SST): y:-2.654+9.047-ar SST_ Σ (Y.-M.): 280620.865 Regression sum of squares(SSR) 20-u)=279156.9638 SSR n-1 Error sum of squares (SSE) SSE-: Σ (y,-%) - X -1434.6851 Total Sum of Squares(SST): SST-SSR + SSE-280591.649Calculating the coefficient of determination SSR =0.997 SST r0.9949 From r-0.9949 , that 99.49% of variation in usage/1000(yCritical value of the t distribution, with n -1 degrees of freedom :-lqt,dii-2.200985 n-1 For 95% confidence, the critical value from the distribution is 2.200985 S35.145 35.145 </H<57.521 n-1 Vn Thus, with 95% confidence, you conclude that the mean temperature of all the similar samples will be between 35.145 and 57.521.

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