Question

# Bayus (1991) studied the mean numbers of auto dealers visited by early and late replacement buyers....

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

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.

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