Question

Teen obesity:

The 2013 National Youth Risk Behavior Survey (YRBS) reported that 13.7% of U.S. students in grades 9 through 12 who attend public and private schools were obese. [Source: Kann, L., Kinchen, S., Shanklin, S.L., Flint, K.H., Hawkins, J., Harris, W.A., et. al.(2013) YRBS 2013]
Suppose that 15% of a random sample of 300 U.S. public high school students were obese. Using the estimate from the 2013 YRBS, we calculate a standard error of 0.020. Since the data allows the use of the normal model, we can determine an approximate 95% confidence interval for the percentage of all U.S. public high school students who are obese.
Which interval is the approximate 95% confidence interval?

A) 0.097 to o.177
B) 0.117 to 0.157
C) 0.110 to 0.190
D) 0.013 to o.170

Answer 1

95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

Explanation:

We are given that 15% of a random sample of 300 U.S. public high school students were obese.

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 % of U.S. public high school students who were obese = 15%

n = sample of U.S. public high school students = 300

p = population percentage of all U.S. public high school students

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

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) } }`]

= [
`0.15-1.96 * {\sqrt{(0.15(1-0.15))/(300) } }` ,
`0.15+1.96 * {\sqrt{(0.15(1-0.15))/(300) } }` ]

= [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].