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

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

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

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$,$\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

   = [ $0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ , $0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ ]

   = [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].