help 3 A sample survey interviews an SRS of 267 college wome

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3- A sample survey interviews an SRS of 267 college women. Suppose that 70% of all college women have been on diet within the past 12 months.

a) What are the mean and standard deviation of the sampling distribution for samples of 267 women?

b) If we selected 267 college women and found 200 of them are on diet, construct a 95% confidence interval for the population proportion.

c) Is the true population proportion (0.7) included in the confidence interval you constructed in b or not?

d) If you plan to do another study about the proportion of college women on diet such that the margin equals to 0.06 with 99% confidence level, what is the sample size you need?

Solution

n=267

p = 0.7

mean = m = n*p = 267 * 0.7 = 186.9

standard deviation = s = sqrt(n*p*q) = 7.48

2.

out of 267, 200 found out on diet

so = p = 200/267 = 0.749

q = 0.251

n = 267

m = n*p = 200

s = sqrt(nqp) = 7.08

P(-x1<x<x1) = 0.95

P(-z1<z<z1) = 0.95

2*P(0<z<z1) = 0.95

P(z<z1) - P(z<0) = 0.475

P(z<z1) = 0.975

z = 1.96

so,

range = (m-1.96s , m+1.96s)

range=(200-1.96*7.08, 200+1.96*7.08)

range = (186.11,213.89)

3.

no, this is not true population confidence interval. this interval is made with p=0.749.

4.

margn = 0.06

for 99%,

z value = 2.57

z*s/n = 0.06

2.57*sqrt(n*p*q)/n = 0.06

0.7*0.3/n=(0.06/2.57)^2

n = 385