In a certain election, 57% of the women voters voted for Candidate A, while only

Question

In a certain election, 57% of the women voters voted for Candidate A, while only 52% of men voters voted for Candidate A. An exit poll of 163 women and 212 men was conducted. Let [^(p)]1 and [^(p)]2 denote sample proportions of women and men voters respectively who voted for Candidate A.
Answer all the questions below (where appropriate) as a fraction not as a percentage.

What is the expected value of [^(p)]1? [Answer to two decimal places.] 0.57

What is the standard deviation of [^(p)]1? [Answer to four decimal places.] 0.0388

What is the expected value of [^(p)]2? [Answer to two decimal places.] 0.52

What is the standard deviation of [^(p)]2? [Answer to four decimal places.] 0.0343

What is the expected value of [^(p)]1 [^(p)]2? [Answer to two decimal places.] 0.05

What is the standard deviation of [^(p)]1 [^(p)]2? [Answer to four decimal places.] 0.0518

What is the probability that the difference [^(p)]1 [^(p)]2 will be larger than 0.0900000000000001? [Answer to four decimal places.]

Explanation / Answer

We are given the P1= sample proportions of women voters respectively who voted for Candidate A.=0.57 & P2= sample proportions of men voters respectively who voted for Candidate A.=0.52

We know that , p1~N[ P1, (P1)(1-P1)/n1 ] & p2~N[ P2, (P2)(1-P2)/n2 ]

Also p1-p2 ~ N[ P1-P2 , (P1)(1-P1)/n1+ (P2)(1-P2)/n2 ]

1)

We know E[p1]=P1= 0.57= 57/100

2)

Also Var[p1]= (P1)(1-P1)/n1= (0.57)(0.43) / 163 = 0.001504

So Standard deviation[p1]= (0.001504)^0.5 = 0.0388

3)

We know E[p2]=P2= 0.52= 52/100

4)

Also Var[p2]= (P2)(1-P2)/n2= (0.52)(0.48) / 212 = 0.001177

So Standard deviation[p2]= (0.001177)^0.5 = 0.0343

5)

Now E[p1-p2] =E[p1] - E[p2] = P1 - P2 = 0.57 - 0.52 = 0.05 =5/100

6)

We know that Variance[p1-p2] = (P1)(1-P1)/n1+ (P2)(1-P2)/n2 = (0.57*0.43)/163 + (0.52*0.48)/212 = 0.002681

So Standard deviation of [p1-p2] = (0.002681)^0.5 = 0.0518

7)

Now P[(p1 - p2) > 0.0900000000000001] = P[ Z >(0.0900000000000001 - 0.05) / 0.0518 ] = P[ Z>0.7722] = 0.2207

(Using meean and standard deviation from the 5th and 6th part )

(Using Normal probability tables)

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