Question

A telephone service representative believes that the proportion of customers completely satisfied with their local telephone service is different between the Northeast and the Midwest. The representative's belief is based on the results of a survey. The survey included a random sample of 1200 northeastern residents and 1280 midwestern residents. 44% of the northeastern residents and 36% of the midwestern residents reported that they were completely satisfied with their local telephone service. Find the 80% 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

a) The 80 % of confidence interval for the difference in two proportions.

(0.055 ,0.105)

b)The critical value that should be used in constructing intervals.

z₀.₂ = 1.28 ( from z-table) at 80 % of level of significance

Explanation:

Step :1

Given data the survey included a random sample of 1200 northeastern residents and 1280 midwestern residents.

n₁ = 1200

n₂ = 1280

Given 44% of the northeastern residents reported that they were completely satisfied with their local telephone service.

The first proportion 'p₁' = 0.44

q₁ = 1- p₁ = 1- 0.44 = 0.56

Given 36% of the midwestern residents reported that they were completely satisfied with their local telephone service.

The second proportion 'p₂' = 0.36

q₂ = 1- p₂ = 1- 0.36 = 0.64

Step 2:-

The 80% confidence interval of p₁- p₂ are

(p₁- p₂ ± zα S.E (p₁- p₂)

where standard error of p₁- p₂

`S.E (p_(1)-p_(2) ) = \sqrt{(p_(1)q_(1) )/(n_(1) )+(p_(2)q_(2) )/(n_(2) ) }`

substitute all values ,

`S.E (p_(1)-p_(2) ) = \sqrt{(0.44X0.56 )/(1200 )+(0.36X0.64 )/(1280 ) }`

on calculation , we get

standard error of (p₁- p₂ ) = 0.0196

Step 3:-

a) The critical value that should be used in constructing intervals.

z₀.₂ = 1.28 ( from z-table) at 80 % of level of significance

Now the 80 % of confidence interval for the difference in two proportions.

(p₁- p₂ ± zα S.E (p₁- p₂)

((0.44-0.36) - 1.28 (0.0196) , 0.44-0.36 + 1.28 (0.0196))

(0.08 - 0.0250 ,0.08 + 0.0250)

(0.055 ,0.105)

Conclusion:-

the 80 % of confidence interval for the difference in two proportions.

(0.055 ,0.105)

Answer 2

80% confidence interval for the difference in two proportions is [0.055 , 0.105].

Explanation:

We are given that a telephone service representative believes that the proportion of customers completely satisfied with their local telephone service is different between the Northeast and the Midwest.

The survey included a random sample of 1200 northeastern residents and 1280 mid western residents. 44% of the northeastern residents and 36% of the mid western residents reported that they were completely satisfied with their local telephone service.

Firstly, the pivotal quantity for 80% confidence interval for the difference in two proportions is given by;

P.Q. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of northeastern residents who were completely satisfied with their local telephone service = 0.44

`\hat p_2` = sample proportion of mid western residents who were completely satisfied with their local telephone service = 0.36

`n_1` = sample of northeastern residents = 1200

`n_2` = sample of mid western residents = 1280

`p_1` = true proportion of northeastern residents who were completely satisfied with their local telephone service

`p_2` = true proportion of mid western residents who were completely satisfied with their local telephone service

Here for constructing 80% confidence interval we have used Two-sample z proportion statistics.

So, 80% confidence interval for the difference in two proportions,
`(p_1-p_2)` is ;

P(-1.2816 < N(0,1) < 1.2816) = 0.80 {As the critical value of z at 10% level of

significance are -1.2816 & 1.2816}

P(-1.2816 <
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` < 1.2816) = 0.80

P(
`-1.2816 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` <
`{(\hat p_1-\hat p_2)-(p_1-p_2)}` <
`1.2816 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` ) = 0.80

P(
`(\hat p_1-\hat p_2)-1.2816 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` <
`(p_1-p_2)}` <
`(\hat p_1-\hat p_2)+1.2816 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` ) = 0.80

80% confidence interval for
`(p_1-p_2)}` =

[
`(\hat p_1-\hat p_2)-1.2816 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }` ,
`(\hat p_1-\hat p_2)+1.2816 * {\sqrt{(\hat p_1(1-\hat p_1))/(n_1) +(\hat p_2(1-\hat p_2))/(n_2) } }`]

= [
`(0.44-0.36)-1.2816 * {\sqrt{(0.44(1-0.44))/(1200) +(0.36(1-0.36))/(1280) } }` ,
`(0.44-0.36)+1.2816 * {\sqrt{(0.44(1-0.44))/(1200) +(0.36(1-0.36))/(1280) } }` ]

= [0.055 , 0.105]

Therefore, 80% confidence interval for the difference in two proportions is [0.055 , 0.105].