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¯ = 4.38 and s = .58. 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.)

H0 : μ (Click to select)≠= 4 versus Ha : μ (Click to select)=≠ 4.

t

tα/2 = 0.05
tα/2 =0.025
tα/2 =0.005
tα/2 =0.0005

There is (Click to select)weakvery strongnoextremely strongstrong evidence.

μ is (Click to select)less thangreater than 4.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 4
Alternative Hypothesis, Ha: μ ≠ 4

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (4.38 - 4)/(0.58/sqrt(100))
t = 6.55

for 0.05
Rejection Region
This is two tailed test, for α = 0.1 and df = 99
Critical value of t are -1.66 and 1.66.
Hence reject H0 if t < -1.66 or t > 1.66

reject the null hypothesis.

for 0.025

Rejection Region
This is two tailed test, for α = 0.05 and df = 99
Critical value of t are -1.984 and 1.984.
Hence reject H0 if t < -1.984 or t > 1.984

reject the null hypothesis.

for 0.005

Rejection Region
This is two tailed test, for α = 0.01 and df = 99
Critical value of t are -2.626 and 2.626.
Hence reject H0 if t < -2.626 or t > 2.626

reject the null hypothesis.

for 0.0005

Rejection Region
This is two tailed test, for α = 0.001 and df = 99
Critical value of t are -3.392 and 3.392.
Hence reject H0 if t < -3.392 or t > 3.392

reject the null hypothesis.

There is extremely strong evidence. μ is less than or greater than 4.