(from Table B).(When read directly from Table B, t = –1.48 with df = 10 is between tail probabilities of
0.05 and 0.10. However, those are one-sided values and must be doubled for the two-sided
alternative since we are interested in the probability of getting 1.48 standard deviations
away from the mean in any direction. Using the TI-83/84, P -value =
2tcdf(-100,-1.48,10)= 0.170. Using the 2SampTTest from the STAT TESTS menu, P -
value = 0.156 with df = 18.99.)
IV . The P -value is too large to be strong evidence against the null hypothesis that there is no
difference between the machines. We do not have strong evidence that the types of machines
actually differ in the number of defects produced.
Using a two-sample t -test, Steps I and II would not change. Step III would change to
based on df = min{30 – 1,30 – 1} = 29. Step IV would probably arrive at a different conclusion—
reject the null because the P -value is small. Large sample sizes make it easier to detect statistically
significant differences.11.
This P -value is quite low and provides evidence against the null and in favor of the alternative that
security procedures actually detect less than the claimed percentage of banned objects.I . Let μ (^) d = the mean of the differences between the scores of husbands and wives.
H 0 : μ (^) d = 0.
H (^) A : μ (^) d < 0.
II . We are told to assume that σ = 1.77 (Note : This is a reasonable assumption, given the large
sample size). This is a matched-pairs situation and we will use a one-sample z -test for a
population mean. We assume that this is a random sample of married couples.
III.
(If you are bothered by using z rather than t —after all, we really don’t know σ (^) D —note that