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¯ = 4.26 and s = .52. 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 H_a : 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)noextremely strongvery strongstrongweak evidence.

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

Answer 1

To Test :-

H₀ :-  

H₁ :-  

For

Test Statistic :-

t = 5

Test Criteria :-

Reject null hypothesis if

Result :- Reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to support the claim that that μ differs from 4 dealers.

For

Test Statistic :-

t = 5

Test Criteria :-

Reject null hypothesis if

Result :- Reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to support the claim that that μ differs from 4 dealers.

For

Test Statistic :-

t = 5

Test Criteria :-

Reject null hypothesis if

Result :- Reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to support the claim that that μ differs from 4 dealers.

For

Test Statistic :-

t = 5

Test Criteria :-

Reject null hypothesis if

Result :- Reject null hypothesis

Conclusion :- Accept Alternative Hypothesis

There is sufficient evidence to support the claim that that μ differs from 4 dealers.