From q bank #8652: A variable y is regressed against a single variable x across 28 observations. The value of the slope is 1.89, and the constant is 1.1. The mean value of x is 1.50, and the mean value of y is 3.94. The standard deviation of the x variable is 0.96, and the standard deviation of the y variable is 2.85. The regression sum of squares is 79.07, and the total sum of squares is 195.24. For an x value of 2.0, what is the 95% confidence interval for the y value? A) 1.01 to 8.75. B) 1.21 to 8.65. C) 1.50 to 8.36. D) 0.41 to 9.35. I’ll beg assistance after people have had a chance!
i just laughed a little to myself at how i don’t remember quant at all, especially having not looked at it for months. my stab at it is D, but honestly, i will be reading the answer and learning when you post it up. note to self, review quant soon.
sqrt(79.07/28) = 1.68 t 0.05, 27 = 2.0518 y @ x=2 = 1.1 + 1.89 * 2 = 4.88 y conf interval = 4.88 +/- 2.0518 * 1.68 = 4.88 +/- 3.45 = [1.43, 8.33] So is it choice C? I am not sure if this is right! CP
I believe you have to calculate the standard error of the forecast, then the prediction interval should be y(predicted)+/- t(crit)*st. error of forecast. I get 4.88+/-4.44=.44 to 9.32 D? Note only A and D have a midpoint of 4.88…
From the 'bank (although it has since changed around the options) “The correct answer was D. First the standard error of the estimate must be calculated—it is equal to the square root of the mean squared error, which is equal to the residual sum of squares divided by the number of observations minus 2. The residual sum of squares is equal to the difference between the total sum of squares and the regression sum of squares, which is 195.24 − 79.07 = 116.17. The standard error of the estimate is equal to (116.17 / 26)^1/2 = 2.11. The standard deviation of the prediction is equal to the standard error of the estimate multiplied by {1 + 1/n + (X −X )^2/[(n − 1)s_x^2]}^1/2 = 2.11 × [1 + 1/28 + (2.0 − 1.5)^2/(27 × 0.962)]^1/2 = 2.17. The prediction value is 1.1 + (2.0 × 1.89) = 4.88. The t-value for 26 degrees of freedom is 2.056. The endpoints of the interval are 4.88 ± 2.056 × 2.17 = 0.41 and 9.35.” I got as far as 4.88 +/- t (26, 5%) * (standard error of forecast), but there was never a chance in hell of me remembering how to calculate that! On balance, I think I’ll leave remembering how to calculate the st error of the forecast to people who favour the beard, and hope they give me the number in the exam! I think I’ll have to try harder to remember standard error of estimate = (MSE)^.5 though.
SST = SSR + SSE so SSE = SST - SSR = 195.24 - 79.07 I did that originally – and took 27 degrees of freedom for the t-parameter – did not get any of the answers, then back tracked to the closest answer. and of course I forgot the (n-2) on the d of f. CP
seriously- i think maybe most of us could’ve gotten the t value there and then 4.88 part, but heck if i’m going to ever remember that massive formula for std error of the forecast (although i now am thumbing through my quant notecards and yes, i do have a notecard with this formula). dimes was right above in that you could’ve knocked 2 out right off the bat since 4.88 has to be the midpoint of whatever answer was right. So down to A or D, i’ll take that punt, leave it to the beards, and move on and keep studying FSA.