I . Let p = the true proportion of Wednesday night television viewers who are watching “I
Want to Marry a Statistician.”H 0 : p = 0.55.H (^) A : p < 0.55.
II . We want to use a one-proportion z-test at α = 0.01. 500(0.55) = 275 > 5 and 500(1 – 0.55)
= 225 > 5. Thus, the conditions needed for this test have been met.
III .
(On the TI-83/84, normalcdf(-100,-1.80)=0.0359 , or you can use STAT TESTS 1-
PropZTest. )
IV . Because P > 0.01, we do not have sufficient evidence to reject the null hypothesis. The
evidence is insufficient to conclude at the 0.01 level that the proportion of viewers has
dropped since the program was moved to a different night.
Let’s suppose that the Stats class constructed a 95% confidence interval for the true proportion of
students who plan to vote for Harvey (we are assuming that this a random sample from the population
of interest, and we note that both n and n (1 – ) are greater than 10). (as Harvey
figured). Then a 95% confidence interval for the true proportion of votes Harvey can expect to get is
. That is, we are 95% confident that between
35.5% and 71.2% of students plan to vote for Harvey. He may have a majority, but there is a lot of
room between 35.5% and 50% for Harvey not to get the majority he thinks he has. (The argument is
similar with a 90% CI: (0.384, 0.683); or with a 99% CI: (0.299, 0.768).)I . Let μ 1 = the true mean number of defects produced by machine A.
Let μ 2 = the true mean number of defects produced by machine B.
H 0 : μ 1 – μ 2 = 0.H (^) A : μ 1 – μ 2 ≠ 0.
II . We use a two-sample t -test for the difference between means. The conditions that need to
be present for this procedure are given in the problem: both samples are simple random
samples from independent, approximately normal populations.
III .