2 Let\'s suppose that there another poll to measure the proportions of husbands

Question

2 Let's suppose that there another poll to measure the proportions of husbands and wives who pay the random sample of wives, we observe a sample proportion of 0.45. In an independent random sample of 773 husbands, a sample proportion of 0.35 is observed. A What is the estimated standard error for the differences of proportions, P, pa, of wives and husbands who pay the bills at home? B What is the margin of error corresponding to a 90% confidence interval? C What is the 90% confidence interval? D Interpret this confidence interval in the context of the problem of who pays bills at home. Explain your interpretation using information about proportions. E Think about your confidence interval. Does it allow for proportions wives and husbands who the of pay the bills at home to be equal? Why or why not? If not, which proportion is likely to be greater? Statway Version 3.0, o 2016 The Carnegie Foundation for the Advancement of Teathinr

Explanation / Answer

a) pw =0.45 ; n1=580 ; ph=0.35 ; n2=773

std error =(pw(1-pw)/n1+ph(1-ph)/n2)1/2 =0.02685

b) for 90% CI, z=1.64485

hence margin of error =z*std error =0.0442

c)confidence interval =pw-pm +/-margin of error =0.0558 ; 0.1442

d)for above confidence interval gives 90% probabilty to contain population proprtion difference of people who pay bill at home b/w wifes and husbands

e)No as confidence interval values are above 0, they should not be equal, from above wifes' proportion is higher

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