Posted by **Tracy** on Friday, June 28, 2013 at 12:20am.

In a survey of families in which both parents work, one of the questions asked was, "Have you refused a job, promotion, or transfer because it would mean less time with your family?" A total of 200 men and 200 women were asked this question. "Yes" was the response given by 34% of the men and 26% of the women. Based on this survey, can we conclude that there is a difference in the proportion of men and women responding "yes" at the 0.05 level of significance? (Use Men - Women.)

(a) Find z. (Give your answer correct to two decimal places.)

(ii) Find the p-value. (Give your answer correct to four decimal places.)

- Math -
**Kuai**, Friday, June 28, 2013 at 1:22am
z = 1.75

p-value =.0809

If p-value is greater than alpha

Fail to reject Ho

If p-value is less than alpha

Reject Ho

- Math -
**Kuai**, Friday, June 28, 2013 at 2:31am
1. Men 2. Women

n1 = 200. n2 = 200

x1 = .34(200) =64. x2 = .26(200) = 52

phat1 = .34. phat2 = .26

pbar = (x1 + x2)/(n1+n2)

pbar = (68+52)/(200+200)

pbar = .30

qbar = 1-pbar = .70

z = (phat1-phat2)/sqrt(phar*qbar/n1 + pbar*qbar/n2)

z = (.34-.26)/sqrt(.3*.7/200 + .3*.7/200)

z = -+ 1.7457

ti-83

2nd vars

2-normalcdf(-10, -1. 7457) = = .04043

two tail so you multiply 2

2( .04043) = 0.08086

Four decimal places = .0809

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