Answers - Millersville University

(a) Do a higher proportion of the subjects in the experimental group experience dry
mouth than the subjects in the control group at the α = 0.05 level of significance?
We are being asked to perform a hypothesis test regarding two population proportions.
The samples are independent. There are some conditions we must
check before we can conduct the hypothesis test using the z-distribution. Note that
n1 = 553 and x1 = 77. Thus the sample proportion for Group 1 is
ˆp1 = 77
553 = 0.1392 and n1ˆp1(1 − ˆp1) = 553(0.1392)(1 − 0.1392) = 66.3 ≥ 10.
Similarly we note that n2 = 373 and x2 = 34. Thus the sample proportion for
Group 2 is
ˆp2 = 34
373 = 0.0912 and n2ˆp2(1 − ˆp2) = 373(0.0912)(1 − 0.0912) = 30.9 ≥ 10.
Assuming that each sample is less than 5% of the population, we can conduct the
hypothesis test.
The hypotheses are
The pooled estimate is
H0 : p1 = p2
H1 : p1 > p2 (right-tailed test).
ˆp = x1 + x2
n1 + n2
= 77 + 34
111
= = 0.1199.
553 + 373 926
Since α = 0.05 the critical value of z is zα = z0.05 = 1.645. The test statistic is
ˆp1 − ˆp2
z0 =
ˆp(1 − ˆp) 0.1392 − 0.0912
=
1 1 + 0.1199(1 − 0.1199) n1 n2
= 2.21.
1 1 + 553 373
Since the test statistics lies to the right of the critical value we reject the null
hypothesis.
Conclusion: the sample data support the claim that a higher proportion of the
experimental group experience dry mouth than in the control group at the α = 0.05
level of significance.
(b) Construct a 90% confidence interval for the difference between the two population
proportions, p1 − p2.
When constructing a 90% confidence interval α = 0.10 which implies α/2 = 0.05
and zα/2 = 1.645. The margin of error is
E = zα/2
= 1.645
ˆp1(1 − ˆp1)
n1
+ ˆp2(1 − ˆp2)
n2
0.1392(1 − 0.1392)
553
+ 0.0912(1 − 0.0912)
373
= 0.0345.