Question

A telephone service representative believes that the proportion of customers completely satisfied with their local telephone service is different between the Midwest and the Northeast. The representative's belief is based on the results of a survey. The survey included a random sample of 1400 midwestern residents and 1320 northeastern residents. 50% of the midwestern residents and 38% of the northeastern residents reported that they were completely satisfied with their local telephone service. Find the 98% confidence interval for the difference in two proportions. Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval.

Answer 1

(0.1361 ; 0.2239)

Explanation:

n1 = 1400 ; p1 = 0.50 ; n2 = 1320 ; p2 = 0.38

Confidence interval = mean ± margin of error

Mean = p1 - p2

Point estimate :

p1 - p2 = 0.50 - 0.38 = 0.18

Margin of Error (MOE) = Zcritical * S. E

Standard Error(S. E) :

Sqrt[(p1(1-p1)/n1) + (p2(1-p2)/n2)]

(p1(1-p1)/n1) = 0.50(0.50) / 1400 = 0.0001785

(p2(1-p2)/n2) = 0.38(0.62) / 1320 = 0.0001784

S. E = sqrt(0.0001785 + 0.0001784)

S. E = sqrt(0.0003569)

S. E = 0.0188917

Zcritical at 98% = 2.326

MOE = Zcritical * S. E

MOE = 2.326 * 0.0188917

MOE = 0.0439420942

MOE = 0.0439

Confidence interval = mean ± margin of error

Confidence interval = 0.18 ± 0.0439

Lower boundary = (0.18 - 0.0439) = 0.1361

Upper boundary = (0.18 + 0.0439) = 0.2239

(0.1361 ; 0.2239)