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I need PART A, B AND C PLEASE. I\'m literally begging A study is conducted to de

ID: 3154690 • Letter: I

Question

I need PART A, B AND C PLEASE. I'm literally begging

A study is conducted to determine whether there is a significant difference in union membership based on sex of the person. Random samples of 5000 factory-employed men were polled, and of this group, 785 were members of a union. A random of 3000 factory-employed women were also polled, and of this group, 327 were members of a union. Use the p-value at a level of significance of 0.05 to test the hypothesis that the there is a significant difference in union membership based on sex of the person. Construct the confidence interval that is relevant to this problem. Use the same level of significance, and draw conclusion (explain) about the hypothesis using the C.I. If p1 = .15 and p2 = 0.12, the analyst wishes to have a probability of Type II error of 0.10, and same type I error as in pans (a) and (b). How many union members of each sex should be surveyed?

Explanation / Answer

a) H0: Pu1-Pu2=0 (There is no difference in union membership of men and women)

H1:Pu1-Pu2 not equal to 0. (There is difference in union membership of men and women)

Where, Pu1 and PU2 denote proportion of union membership of men and women respectively.

Distribution: Z(Large sample)

Alpha=0.05.

Test statistic:

Compute pooled proportion.

Pu=(N1Ps1+N2Ps2)/(N1+N2)

=(5000*785/5000+3000*327/3000)/(5000+3000)

=0.139

Compute SE(Ps1-Ps2)=sqrt [Pu(1-Pu)]sqrt [(N1+N2)/N1*N2]

=sqrt [0.139(1-0.139)sqrt [(5000+3000)/5000*3000]

=0.007989

Z(Obtained)=(Ps1-Ps2)/SE(Ps1-Ps2)=(785/5000-327/3000)/0.007989

=6.00

p value: 0.0000

The p value is less tha alpha=0.05. Therefore, reject null hypothesis to conclude that there i ssignificant differnce in union membership proportion of men and women.

b) 95% c.i=(Ps1-Ps2)+-Z SE(Ps1-Ps2), where Z correspond sto z score at alpha=0.05

=(785/5000-327/30000+-1.96*0.007989

=0.03234 to 0.06366

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