Answer :
Answer:
95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].
Step-by-step explanation:
We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.
Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of residents 65-years or older who had obtained a flu shot = [tex]\frac{28}{111}[/tex] = 0.25
n = sample of residents 65-years or older = 111
p = population proportion of residents who were getting the flu shot
Here for constructing 95% confidence interval we have used One-sample z test for proportions.
So, 95% confidence interval for the population proportion, p is ;
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
95% confidence interval for p = [ [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex],[tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ]
= [ [tex]0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }[/tex] , [tex]0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }[/tex] ]
= [0.169 , 0.331]
Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].