Question

g 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. The researcher wants to calculate a 95% confidence interval for the percent of the 65-plus population that were getting the flu shot. Historically in this county, 30% of residents typically obtain a flu shot each year.

Answer 1

95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

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. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of residents 65-years or older who had obtained a flu shot =
`(28)/(111)` = 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 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

P(
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

95% confidence interval for p = [
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }`,
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ]

= [
`0.25-1.96 * {\sqrt{(0.25(1-0.25))/(111) } }` ,
`0.25+1.96 * {\sqrt{(0.25(1-0.25))/(111) } }` ]

= [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].