TAKE IT HOME When we think about the differences in sample proportions, we often
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TAKE IT HOME When we think about the differences in sample proportions, we often assume that there is no difference in population proportions. we can use what we know about the normality of the distribution of PI- Pr,to calculate the probability of observing particular values of the sample proportions. In this way we can draw conclusions about the reasonableness of the assumption that the true proportions are equal, or that In a recent Gallup poll, researchers asked husbands and wives who pays the bills in their household. Let's assume that 50% of husbands pay the bills in their homes (ph 0.50) and 50% of wives pay the bills in their homes (pa so). i The article on Gallup's website reported that 35% of husbands say they pay the bills in their home and 45%of wives say they pay the bills in their home. Suppose this poll was based on sample sizes of n,. 15 for the husbands and n, z 25 for the wives. Would a normal distribution be an appropriate model for the differences in sample proportions? Why or why not? appropriate 2 Suppose this poll was based on sample sizes of n, 120 for the husbands and na 1000 for the wives. Would a normal distribution be an appropriate model for the differences in sample proportions? Why or why not? 3 suppose 35% of the 120 husbands sampled pay the bills in their home, and 45% of the 100 wives sampled say they pay the bills in their home. What is the value of p 1 p 2? Frank Newport, wives Still Do Laundry, Men Do Yard work Gallup.com, April 4, 2008, accessed November 30, 2014 http//www.gallup.com/pol/106249/wives still Iau work aspx Statway" Version 3.0, o 2016 The Carnegie Foundation for the Advancement of Teaching 125Explanation / Answer
4. Mean of sampling distribution of the difference is sample proportions, p1hat-p2hat=0.35-0.45=-0.10.
5. Since, the results come from same underlying population, therefore, pool them to get a better estimate.
Standard error, SEpooled(p1hat-p2hat)=sqrt[phatpooled(1-phatpooled)/n1+phatpooled(1-phatpooled)/n2], where, phat is sample proportion, and n is sample size. phatpooled is as follows:
phatpooled=(Success1+success2)/(n1+n2)
=(42+45)/(120+100) [120*0.35=42, and 100*0.45=45]
=0.3955
SEpooled(p1hat-p2hat)=sqrt[0.3955(1-0.3955)/120+0.3955(1-0.3955)/100]
=0.0662 (ans)
6. Z=(p1hat-p2hat)/SEpooled
=(0.35-0.45)/0.0662
=-1.51
7. The Z score results into p value: 0.1310, which is not less than 0.05. Therefore, conclude that there is insufficient sample evidence to conclude that there exists significant difference in proportion of husbands and wives who paid their bills at home. This is not unusual.
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