Question

Bayus (1991) studied the mean numbers of auto dealers visited by early and late replacement buyers. Letting μ be the mean number of dealers visited by all late replacement buyers, set up the null and alternative hypotheses needed if we wish to attempt to provide evidence that μ differs from 4 dealers. A random sample of 100 late replacement buyers yields a mean and a standard deviation of the number of dealers visited of x⎯⎯x¯ = 4.42 and s = .65. Using a critical value and assuming approximate normality to test the hypotheses you set up by setting α equal to .10, .05, .01, and .001. Do we estimate that μ is less than 4 or greater than 4? (Round your answers to 3 decimal places.)

t=_____

tα/2 = 0.05

tα/2 = 0.025

tα/2 = 0.005

tα/2 = 0.0005

Answer 1

One-Sample T

Descriptive Statistics

N	Mean	StDev	SE Mean	90% CI for μ
100	4.4200	0.6500	0.0650	(4.3121, 4.5279)

μ: mean of Sample

Test

Null hypothesis	H₀: μ = 4
Alternative hypothesis	H₁: μ ≠ 4

= 6.46

T-Value	P-Value
6.46	0.000

critical values are

tα/2 = 0.05 = 1.96

tα/2 = 0.025 = 1.984

tα/2 = 0.005 = 2.626

tα/2 = 0.0005 = 3.390

for all the cases we reject null hypothesis and conclude that the mean is significantly differ by 4.