Construct a 90% prediction interval for a single value of y for the following da
ID: 3177369 • Letter: C
Question
Construct a 90% prediction interval for a single value of y for the following data; use x = 100. Construct a 90% prediction interval for a single value of y for the following data; use x = 130. Compare the results. Which prediction interval is greater? Why? 90% prediction interval for a single value of y with x = 100: 90% prediction interval for a single value of y with x = 130: The width of this confidence interval of y for x_0 = is wider that the confidence interval of y for x_0 = because x_0 = is nearer to the value of x than is x_0 .Explanation / Answer
Tabulate the calculations to determine the regression equation.
x x-xbar y y-ybar (x-xbar)(y-ybar) (x-xbar)^2 ycap (y-ycap)^2
141 59.6 25 -45.9 -2735.64 3552.16 17.590 54.9140
119 37.6 29 -41.9 -1575.44 1413.76 37.266 68.3334
103 21.6 44 -26.9 -581.04 466.56 51.577 57.4079
91 9.6 70 -0.9 -8.64 92.16 62.310 59.1423
63 -18.4 88 17.1 -314.64 338.56 87.353 0.4189
29 -52.4 112 41.1 -2153.64 2745.76 117.762 33.2053
24 -57.4 128 57.1 -3277.54 3294.76 122.234 33.2421
Using the above table, xbar=sigma x/n=570/7=81.4, ybar=496/7=70.9, sigma (x-xbar)(y-ybar)=-10647, sigma(x-xbar)^2=11904
Slope, beta1=sigma(x-xbar)(y-ybar)/sigma(x-xbar)^2=-10647/11904=-0.8944
Y intercept, beta0=ybar-beta1*xbar=70.9-(-0.8944)*81.4=143.7
Regression eqaution: yhat=143.7-0.8944x
Now, substitute x in the regression eqaution with given values of x and obtan ycap (see table). Subtract ycap from y and sqaure it to obtain SSE.
Therefore, standard error of residual, se=sqrt SSE/n-2=sqrt 306.66/7-2=sqrt 61.332=7.83
The 90% prediction interval is: ycapp+-talpha/2,df=n-2*se sqrt[1/n+(xp-xbar)^2/Sxx] [ycapp=143.7-0.8944*100=54.26, Sxx=sigma(x-xbar)^2]
=54.26+-2.015*7.83 sqrt[1/7+(100-81.4)^2/11904]
=47.7182, 60.8018
---
The 90% prediction interval is: ycapp+-talpha/2,df=n-2*se sqrt[1/n+(xp-xbar)^2/Sxx] [ycapp=143.7-0.8944*130=27.428, Sxx=sigma(x-xbar)^2]
=27.428+-2.015*7.83 sqrt[1/7+(130-81.4)^2/11904]
=18.2110, 36.6482