Ch 13
Ch 13.
(37).

It's so close to 1 that it really doesn't need testing, but just for drill:

H0: rho <= 0
Ha: rho >0

Critical value for t (0.05, 1 tail, n-2=25-2=23 degrees of freedom) = 1.714

Test statistic: t = r/sqrt((1-r^2)/(n-2)) t = 0.94/sqrt((1-.094^2)/(n-2)) t = 0.94/sqrt(0.1164/23) t = 13.21

Since the test statistic exceeds the critical value, we can reject the null hypothesis in favor of the alternative and presume that there is a positive correlation between passengers and luggage (duh!).

(40.)
1.The breakfast revenue increases as the number of rooms occupied increases. This is seen from the positive slope (1.478) of the regression line in the scatter graph.
2. R = 0.4230. R is positive but sufficiently far away from 1. This implies the correlation between breakfast revenue and number of rooms occupied is weak.
3.
t = R * sqrt(n - 2) / sqrt(1 - R^2) | | t = 0.4230 * sqrt(25 - 2) / sqrt(1 - 0.4230^2) = 2.2388
Right-tailed p- value for t = 2.2388 and 23 degrees of freedom is 0.0176

4.Since 0.0176 < 0.10, it appears that the correlation between revenue and number of rooms occupied is significant.

5.R^2 = 0.178. This implies that about 17.8% of the variation in revenue is accounted for by the number of rooms occupied.
CH. 14
(17).
a. What are the estimated sales for the Bryne store, which has four competitors, a regional population of 0.4 (400,000), and advertising expense of 30 ($30,000)?
Use the regression:
14 - 1*4 + 30*0.4 + 0.2*30
= 28
So the sales would be $28000, since the regression gives the value in thousands

b. Compute the value of r^2. ssr/sst = 3050/5250
= 0.58095

c. Compute the multiple standard error of estimate. sqrt(mse) = sqrt(84.62)
=