Bayus (1991) studied the mean numbers of auto dealers visited byearly and late replacement buyers. Letting μ 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 μ differs from 4 dealers. A randomsample of 100 late replacement buyers yields a mean and a standarddeviation of the number of dealers visited of x⎯⎯x¯ = 4.26 ands = .52. Using a critical value and assuming approximatenormality to test the hypotheses you set up by setting α equal to.10, .05, .01, and .001. Do we estimate that μ is lessthan 4 or greater than 4? (Round your answers to 3 decimalplaces.)
H0 : μ (Click to select)=≠4 versusHa : μ (Click to select)≠=4.
t        Â
| Â Â | |
| tα/2 = 0.05 | |
| tα/2 =0.025 | |
| tα/2 =0.005 | |
| tα/2 =0.0005 | |
|
There is (Click to select)noextremely strongvery strongstrongweakevidence.
μ is (Click to select)less thangreater than4.
