I included all of the relevant information. exVIGRX = b0 + b1(exS&P) + e n = 36 Coefficient - Coefficient Estimate - Standard Error b0 - 0.0023 - 0.0022 b1 - 1.1163 - 0.0624 RSS = 0.0228 SSE = 0.0024 Coldplay forecasts the excess return on the S&P 500 for June 2007 to be 5% and the 95% confidence interval for the predicted value of the excess return on the VIGRX for June 2007 to be 3.9% to 7.7%. The standard error of the forecast is closest to: A. 0.0080 B. 0.0083 C. 0.0111 Could you walk me through how you solve this problem? I want to see how you guys approach this and then I’ll give the answer and the explanation and, if I still need to, explain what I’m having an issue with. Thanks!
K I’ll give it a shot… so the forecast for S&P 500 is 5% and we have the regression equation. plugging in numbers, exVIGRX (june) = .0023 + 1.1163*.05 exVIGRX (june) = .058115 confidence interval using a t-test would be: exVIGRX (june) +/- critical t-value*standard error (SE) .058115 +/- 1.96 *SE we are given the interval… so .058115 + 1.96*SE = .077 SE = .0096 …and that isn’t close to any of your answers lol. fml. well i guestimated the critical t-value at 1.96 cause too lazy to look at the table, since its 36 obs it is probably ~2. if we use critical t-value of 2, SE = 0.0094. so still not right … dunno what else to do though? what is the answer
the above poster’s answer is correct. the original poster wrote down the answers wrong. should be answer (B) 0.0093 not 0.0083
bump… confused me My WAG (wild A$$ GUESS) - the assumption is the standard error of the forecast is the same as SEE = Standard error of the estimate. MSE = .0024/(36-2) SEE MSE^.5 (.0024/(36-2))^.5 =.008401 so B?
Ah yes, I’m sorry. The answer for B should have been 0.0093. Tiredofstudying, you were correct. Sorry for any confusion. I figured out the stupid thing I was doing. I think I was just tired last night when I was going over the problem. I’ll write out what the explanation here: This is a tricky question because you are given the confidence interval and its midpoint and asked to solve for the standard error of the forecast (sf). Remember to also convert the percentages to decimals. The critical two-tailed 5% t-value with 34 degrees of freedom is approximately 2.03. The midpoint, or predicted value is 0.0023+1.1163*0.05 = 0.058. Therefore 0.058 +/- (2.03)(sf) is equivalent to 0.039 to 0.077 and solving for sf yields, sf=0.0093.