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Using the Estimated Regression Equation for Estimation and Prediction - Please work our problem fully so that I can learn how to do the problem. This is my second time submitting. The first time the answers were wrong and they did not explain.

Given are five observations for two variables, x and y. y3 7 6 11 14 Round your answers to two decimal places. a. Using the following equation: Estimate the standard deviation of ý when x 4. b. Using the following expression: Develop a 95% confidence interval for the expected value of y when x-4. to c. Using the following equation: 1 nY Estimate the standard deviation of an individual value of y when x -4. d. Using the following expression a/2 pred Develop a 95% prediction interval for y when x-4. If your answer is negative, enter minus (-) sign. to

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(x-x) 4.0000 1.0000 0.0000 1.0000 4.0000 10.0000 (y-y) 27.0400 1.4400 4.8400 7.8400 33.6400 74.8000 (x-x)(y-y 10.4000 1.2000 0.0000 2.8000 11.6000 26.0000 S.No 1 2 1 2 4 14 Total 15 3.0000 Mean 8.2000 SSX SSY sample size: here 8.2000 3.0000 from above 10.0000 26.0000 slope- 2.6000 intercept- 0.4000 hence line equation: SSE-Syy-(Sxy)/Sxx- error variance σ2- 0.400 2.600 x 7.2000 s -SSE/(n-2) 2.400 std error σ 1.5492 4 predcited value at X=4; 10.800 std error confidence interval- s*/(1/n+(xo-XYSx.) 0.8485 for 95 % Cl value of t= margin of error E-t std error lower confidence bound-sample mean-margin of error- Upper confidence bound-sample mean+margin of error- 3.182 2.7000 8.10 13.50 std error prediction interval= s*/(1+1/n+(xgxY%) 1.7664 for 95 % CI value of t margin of error E-t std error lower prediction bound-sample mean-margin of error- Upper prediction bound-sample mean+margin of error- 3.182 5.6205 5.18 16.42

a)

from above:

std deviation of Yhat=0.85

b)95% confidence interval =8.10 ; 13.50

c) std deviation of individual value =1.77

95% prediction interval =5.17 ; 16.43 ( please try 5.18 ; 16.42 if this comes wrong)

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