Question

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 0.177b. 0.117 to 0.157c. 0.110 to 0.190d. 0.013 to 0.170

Answer 1

Answer:

The right answer is

c) 0.110 to 0.190

Step-by-step explanation:

The question is incomplete:

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

Kann, L., Kinchen, S., Shanklin, S.L., Flint, K.H., Hawkins, J., Harris, W.A., et. al.(2013) YRBS 2013 Report. http://www.cdc.gov/healthyyouth/yrbs/index.htm

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 0.177

b) 0.117 to 0.157

c) 0.110 to 0.190

d) 0.013 to 0.170

We have to calculate a 95% confidence interval for the proportion.

The sample size is n=300 and the sample proportion is p=0.15.

 

The standard error of the proportion is calculated from the 2013 YRBS and has a value of σp=0.020.

The critical z-value for a 95% confidence interval is z=1.96.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.96 \cdot 0.020=0.040$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.150-0.040=0.110\\\\UL=p+z \cdot \sigma_p = 0.150+0.040=0.190$

The 95% confidence interval for the population proportion is (0.110, 0.190).