Answer :
Answer:
80% confidence interval for the difference in two proportions is [0.055 , 0.105].
Step-by-step 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. = [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ~ N(0,1)
where, [tex]\hat p_1[/tex] = sample proportion of northeastern residents who were completely satisfied with their local telephone service = 0.44
[tex]\hat p_2[/tex] = sample proportion of mid western residents who were completely satisfied with their local telephone service = 0.36
[tex]n_1[/tex] = sample of northeastern residents = 1200
[tex]n_2[/tex] = sample of mid western residents = 1280
[tex]p_1[/tex] = true proportion of northeastern residents who were completely satisfied with their local telephone service
[tex]p_2[/tex] = 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, [tex](p_1-p_2)[/tex] 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 < [tex]\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < 1.2816) = 0.80
P( [tex]-1.2816 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < [tex]{(\hat p_1-\hat p_2)-(p_1-p_2)}[/tex] < [tex]1.2816 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ) = 0.80
P( [tex](\hat p_1-\hat p_2)-1.2816 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] < [tex](p_1-p_2)}[/tex] < [tex](\hat p_1-\hat p_2)+1.2816 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ) = 0.80
80% confidence interval for [tex](p_1-p_2)}[/tex] =
[[tex](\hat p_1-\hat p_2)-1.2816 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex] ,[tex](\hat p_1-\hat p_2)+1.2816 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2(1-\hat p_2)}{n_2} } }[/tex]]
= [ [tex](0.44-0.36)-1.2816 \times {\sqrt{\frac{0.44(1-0.44)}{1200} +\frac{0.36(1-0.36)}{1280} } }[/tex] , [tex](0.44-0.36)+1.2816 \times {\sqrt{\frac{0.44(1-0.44)}{1200} +\frac{0.36(1-0.36)}{1280} } }[/tex] ]
= [0.055 , 0.105]
Therefore, 80% confidence interval for the difference in two proportions is [0.055 , 0.105].
Answer:
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
Step-by-step 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
[tex]S.E (p_{1}-p_{2} ) = \sqrt{\frac{p_{1}q_{1} }{n_{1} }+\frac{p_{2}q_{2} }{n_{2} } }[/tex]
substitute all values ,
[tex]S.E (p_{1}-p_{2} ) = \sqrt{\frac{0.44X0.56 }{1200 }+\frac{0.36X0.64 }{1280 } }[/tex]
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)