Question

An economic theory is that the money flowing into and out of mutual funds​ (fund flows) is related to the performance of the stock market​ (market return). To the right is part of the regression​ analysis, where the response variable is Fund Flows​ ($ million) and the explanatory variable is Market Return​ (%). Complete a and b below. nbsp Dependent variable​ is: Fund Flows R squaredequals19.2​% s equals 11 comma 021 with 154 minus 2 equals 152 degrees of freedom Variable nbsp nbsp Coeff nbsp nbsp SE left parenthesis Coeff right parenthesis nbsp Intercept 9615.60 897.3 Market Return 1157.50 201.68 Mean market returnequals7.9​% ​a) Find the 99​% prediction interval for a month that reports a market return of 9​%. The 99​% prediction interval is ​($ nothing ​million, ​$ nothing ​million). ​(Round to the nearest whole number as​ needed.) ​b) Do you think predictions made by this regression will be very​ accurate? A. ​Yes, because the intercept is not significantly different than zero. B. ​Yes, because the slope standard error is small and the value of R squared is large. C. ​No, because the intercept is not zero. D. ​No, because the slope standard error is large and the value of R squared is small.

Answer 1

a) 99% prediction interval

y^= 9615 + 1157.50 *x

= 9615 + 1157.50 *9

= 20032.5

df =n-2 = 152

t = =T.INV.2T(0.01,152) = 2.6086

standard error of prediction interval = Se* sqrt(1 + 1/n + (Xbar - Xi)^2 /SSxx)

= 11021 * sqrt(1 + 1/154 + (9 - 7.9)^2/ 2986.182455)

=11021 * 1.003443

= 11058.945303

note SSxx = Se^2/Sb^2 = (11021/201.68)^2 = 2986.182455

hence

( 20032.5 - 2.6086*11058.945303 ,20032.5 + 2.6086*11058.945303)

= (     -8815.8647 , 48880.8647 )

b)

D. ​No, because the slope standard error is large and the value of R squared is small.