Answer: [tex](20.0\%,24.0\%)[/tex]
Step-by-step explanation:
Given, A 2015 Gallup poll of 1,627 adults found that only 22% felt fully engaged with their mortgage provider.
Here , 1,627 adults are determining the sample.
Thus, sample size : n = 1627
Also, the sample proportion of adults felt fully engaged with their mortgage provider are [tex]\hat{p}=22\%=0.22[/tex]
for 95% confidence level , critical z-value =1.96
Then , the 95% confidence interval would be :-
[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
Substituting values , we get
[tex]0.22\pm 1.96(\sqrt{\dfrac{0.22(1-0.22)}{1627}})\\\\=0.22\pm 1.96\sqrt{\dfrac{0.22\times0.78}{1627}}\\\\\approx0.22\pm0.0201\\\\=(0.22-0.0201,\ 0.22+0.0201)\\\\=(0.1999,\ 0.2401)\approx(20.0\%,24.0\%)[/tex]
Hence, the required 95% confidence interval using only 1 decimal : [tex](20.0\%,24.0\%)[/tex] .
