In: Statistics and Probability
1.Consider the e-billing case. The mean and the standard deviation of the sample of n = 65 payment times are x¯x¯ = 18.4730 and s = 3.9045. Test H0: μ = 18.6 versus Ha: μ < 18.6 by setting α equal to .01 and using a critical value rule and assume normality of the population. (Round your "t" and "t0.01" answers to 3 decimal places and p-value answer to 4 decimal places. Negative value should be indicated by a minus sign. Use a statistical software package - e.g., Minitab, MegaStat, etc., to derive the p-value.)
t | |
t0.01 | |
p-value | |
2. 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.)
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)extremely strongnovery strongstrongweak
evidence.
μ is (Click to select)greater thanless than 4.