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29.01.2020 •
Mathematics
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Ответ:
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:
Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the population proportion is (0.110, 0.190).