Solutions to H7

1) (19.38) The population of interest is heterosexuals from high-risk cities who have multiple sex partners. Let p be the true proportion from this population who never use condoms. We have  a sample of size n=803, with 304 never using condoms, so the sample proportion is phat= 304/803 = .379. Then a 95% confidence interval for p is
.379 ± 1.96 root( .379(1-.379)/803) = .379 ± .0335 = (.345,.413). We claim p is in this interval, the method used to make the claim gives correct claims in 95% of samples. Now as far as the question goes, we can see that 1/3 is not in the interval, so we reject this as a reasonable value of p, and conclude p is greater than 1/3.

2) (20.22) Let p be the true proportion of papers that have statistical assistance. We have  a sample of size n=704, with 514 using statisticians, so the sample proportion is phat= 514/704 = .730. Then a 95% confidence interval for p is
.730 ± 1.96 root( .73(1-.73)/704) = .730 ± .032 = (.698,.762). We claim p is in this interval, the method used to make the claim gives correct claims in 95% of samples.

3) (20.23) Let p1=true proportion of papers rejected when a statistician is not used. Let p2= true proportion of papers rejected when a statistician is used. Now we have n1=190 with p1hat=135/190=0.711 and n2= 514 with p2hat=293/514=0.570.
Now p1hat-p2hat = .711-.570 =.141. And the standard error of this is
root( .711(1-.711)/190 + .57(1-.57)/514) = .0395 So a 95% confidence interval for p1-p2 is
.141 ± 1.96(.0395) = (.064, .218).
We claim that p1-p2 is in this interval. The method used gives correct claims in 95% of samples. Note the interval does not contain 0, and hence we can conclude the true proportions are really different from one another, and that papers that do not use a statistician are more likely to be rejected.

4) (20.32)

(a) Observational study. Nothing is being manipulated here.

(b) Let p1=true proportion of female Hispanic drivers using seat belts in NY . Let p2= true proportion of female Hispanic drivers using seat belts in Boston. Now we have n1=220 with p1hat=183/220=0.832 and n2= 117 with p2hat=68/117=0.581.
Now p1hat-p2hat = .832-.581 =.251. And the standard error of this is
root( .832(1-.832)/220 + .581(1-.581)/117) = .0521 So a 95% confidence interval for p1-p2 is
.251 ± 1.96(.0521) = (.149, .353).
We claim that p1-p2 is in this interval. The method used gives correct claims in 95% of samples. Note the interval does not contain 0, and hence we can conclude the true proportions are really different from one another, and that female Hispanic drivers are more likely to be belted in NY than Boston.