Question

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained physical activity. What sample size should be obtained if she wishes the estimate to be within four percentage points with 95% confidence, assuming that (a) she uses the estimates of 22.6% male and 18.9% female from a previous year? (b) she does not use any prior estimates?

c.detirmine the test statistic

Answer 1

1)

a)

p1          =	0.226
q1=1-p1=	0.774
p2          =	0.189
q2=1-p2=	0.811
here margin of error E =	0.04
for95% CI crtiical Z          =	1.960	from excel:normsinv((1+0.95)/2)
sample size n (p1q1+p2q2)*(Z/E)2=	788

b)

p1          =	0.5
q1=1-p1=	0.5
p2          =	0.5
q2=1-p2=	0.5
here margin of error E =	0.04
for95% CI crtiical Z          =	1.960	from excel:normsinv((1+0.95)/2)
sample size n (p1q1+p2q2)*(Z/E)2=	1201

2)

a)

	Pop 1	Pop 2
sample mean x =	10.500	11.700
std deviation s=	2.800	2.700
sample size n=	19	19
Point estimate =x1-x2=		-1.200
std error =√(S21/n1+S22/n2)=		0.8924
test stat t =(x1-x2-Δo)/Se =		-1.34
p value : =	0.1954		from excel: tdist(1.345,18,2)

b)

Point estimate of differnce =x1-x2     =		-1.200
for 95 % CI & 18 df value of t=		2.101	from excel: t.inv(0.975,18)
margin of error E=t*std error                   =		1.875
lower bound=mean difference-E     =		-3.0749
Upper bound=mean differnce +E      =		0.6749
from above 95% confidence interval for population mean =(-3.075,0.675)

(please try -3.07 , 0.67 if required to 2 decimal places)