Question

23. Given that SSR = 432.189 and SSE = 113.456 for a multiple regression model, compute R²: *

(A) 0.2079

(B) 0.2625

(C) 0.6274

(D) 0.7921

24. The salary of junior executives in a large retailing firm is normally distributed with population standard deviation $1,500. If a random sample of 25 junior executives yields an average salary of $16,400, find a 95% confidence interval for the population mean salary: *

(A) ($12,900 , $28,048)

(B) ($12,078 , $19,500)

(C) ($10,800 , $18,988)

(D) ($15,812 , $16,988)

25. The lifetime of a particular brand of light bulb is normally distributed with unknown mean and known standard deviation 75 hours. What is the probability that in a random sample of 49 bulbs, the sample average lifetime is within 21 hours of the population average lifetime? (This means anything from 21 hours below to 21 hours above the population mean.): *

(A) 0.90

(B) 0.95

(C) 0.96

(D) 0.99

Answer 1

23)

since R² =SSR/(SSR+SSE )=432.189/(432.189+113.456)=0.7921

24)

sample mean 'x̄=	16400.000
sample size    n=	25.00
std deviation σ=	1500.000
std error ='σx=σ/√n=	300.0000

for 95 % CI value of z=		1.960
margin of error E=z*std error =		587.99
lower bound=sample mean-E=		15812.0108
Upper bound=sample mean+E=		16987.9892
from above 95% confidence interval for population mean =(15812 , 16988)

25)

standard deviation σ=	75
sample size       =n=	49
std error=σx̅=σ/√n=	10.7143

probability =P((-21)/10.714)<Z<(21)/10.714)=P(-1.96<Z<1.96)=0.975-0.025=0.95