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

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Bayus (1991) studied the mean numbers of auto dealers visited byearly and late replacement buyers. Letting Î¼ be the meannumber of dealers visited by all late replacement buyers, set upthe null and alternative hypotheses needed if we wish to attempt toprovide evidence that Î¼ differs from 4 dealers. A randomsample of 100 late replacement buyers yields a mean and a standarddeviation of the number of dealers visited of xâŽ¯âŽ¯xÂ¯ = 4.26 ands = .52. Using a critical value and assuming approximatenormality to test the hypotheses you set up by setting Î± equal to.10, .05, .01, and .001. Do we estimate that Î¼ is lessthan 4 or greater than 4? (Round your answers to 3 decimalplaces.)
H₀ : Î¼ (Click to select)=â‰ 4 versusH_a : Î¼ (Click to select)â‰ =4.
t Â Â Â Â Â Â Â Â
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t_{Î±/2 = 0.05}
t_{Î±/2 =0.025}
t_{Î±/2 =0.005}
t_{Î±/2 =0.0005}

There is (Click to select)noextremely strongvery strongstrongweakevidence.
Î¼ is (Click to select)less thangreater than4.

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To Test H0 H1 For Test Statistic t 5Test Criteria Reject null hypothesis if Result Reject null hypothesisConclusion Accept Alternative HypothesisThere is sufficient evidence See Answer

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t_{Î±/2 = 0.05}
t_{Î±/2 =0.025}
t_{Î±/2 =0.005}
t_{Î±/2 =0.0005}