Question

Two random samples are taken, with each group asked if they support a particular candidate. A summary of the sample sizes and proportions of each group answering yes'' are given below: Pop. 1 and pop. 2 :

?1=92,?2=96,

?̂ 1=0.788?̂ 2=0.632

Suppose that the data yields (0.0069, 0.3051) for a confidence interval for the difference ?1−?2 of the population proportions. What is the confidence level? (Give your answer as a percentage, without the % symbol, rounded to one decimal place.) Hint: You know the margin of error and the SE. Solve for z-star from the formula for margin of error. Now find the value of alpha knowing that z-star leaves an area of alpha/2 to its right. The confidence is 1 minus alpha.

Confidence Level = _____ %

Answer 1

Solution:

We have

Confidence interval for difference between two population proportions:

Confidence interval = (P1_hat – P2_hat) ± Z*sqrt[(P1_hat*(1 – P1_hat)/N1) + (P2_hat*(1 – P2_hat)/N2)]

Confidence interval = Difference ± Margin of error

WE are given

N1 = 92

N2 = 96

P1_hat = 0.788

P2_hat = 0.632

Lower limit = 0.0069

Upper limit = 0.3051

Difference = 0.788 – 0.632

Difference = 0.156

Lower limit = Difference - Margin of error = 0.0069

Upper limit = Difference + Margin of error = 0.3051

Margin of error = E

0.156 - E = 0.0069

E = 0.156 - 0.0069

E = 0.1491

Margin of error = Z*Standard error

Standard error = sqrt[(P1_hat*(1 – P1_hat)/N1) + (P2_hat*(1 – P2_hat)/N2)]

Standard error = sqrt((0.788*(1 - 0.788)/92) + (0.632*(1 - 0.632)/96))

Standard error =0.065104

Margin of error = Z*Standard error

0.1491 = Z*0.065104

Z = 0.1491/0.065104

Z = 2.290182

P(Z<2.290182) = 0.988995

(by using z-table)

(1 – C)/2 = 1 - 0.988995 = 0.011005

1 – C = 2*0.011005

C = 1 - 2*0.011005 = 0.97799

C = 97.8%

Answer: 97.8