1. Compute a 95% confidence interval for the mean response when x =10. A straigh
ID: 3180418 • Letter: 1
Question
1. Compute a 95% confidence interval for the mean response when x=10.
A straight line is fitted to some data using least squares. Summary statistics are below. n 10, T 5, y -12, ss. 141, SS 122, SS 155 The least squares intercept and slope are 7.65 and o.87, respectively, and the ANOVA table is below. Source DF SS MS Regression 1 105.56 105.56 Residual 8 49.44 6.18 Total 9 155 What is the estimated mean value for y when x-17? 13.7400 2 pt(s)] You are correct. Previous Tries Your receipt no. is 157-1432 2 If the fitted value for y is 18, what is the corresponding value for x? 11.8966 [2 pt(s You are correct. Previous Tries Your receipt no. is 157-3803 What proportion of the variation in y is explained by the regression in x? 0.6810 2 pt(s)] You are correct. Previous Tries Your receipt no. s 157-1958 What is the standard error of the slope estimate? 0.2094 2 p You are correct. Previous Tries Your receipt no. s 157-1836Explanation / Answer
all are correct and please find the required answer
confidence interval for an observed value of the response variable
corresponding to x=10
for x=10, y=7.65+0.85*10=16.15
confidence interval=y ± t(/2,error df)*se*sqrt(1/n+(x-x-)2/SSxx=16.15±t(0.05/2,8)*sqrt(MS(residual))*sqrt(1/10 + (10-5)*(10-5)/141)
=16.15±2.306*sqrt(6.18)*sqrt(1/10 + (10-5)*(10-5)/141)=16.15±3.0188=(13.1312,19.1688)
prediction interval=
=y ± t(/2,error df)*se*sqrt(1 + 1/n+(x-x-)2/SSxx
=16.15±t(0.05/2,8)*sqrt(MS(residual))*sqrt(1+1/10 + (10-5)*(10-5)/141)
=16.15±2.306*sqrt(6.18)*sqrt(1+1/10 + (10-5)*(10-5)/141)=16.15±6.4789=(9.6711,22.6304)
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