Answer:
The right answer is
c) 0.110 to 0.190
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Step-by-step explanation:
<em>The question is incomplete:</em>
<em>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.
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<em>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
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<em>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?
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<em>a) 0.097 to 0.177
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<em>b) 0.117 to 0.157
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<em>c) 0.110 to 0.190
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<em>d) 0.013 to 0.170</em>
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<em>
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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](https://tex.z-dn.net/?f=MOE%3Dz%5Ccdot%20%5Csigma_p%3D1.96%20%5Ccdot%200.020%3D0.040)
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](https://tex.z-dn.net/?f=LL%3Dp-z%20%5Ccdot%20%5Csigma_p%20%3D%200.150-0.040%3D0.110%5C%5C%5C%5CUL%3Dp%2Bz%20%5Ccdot%20%5Csigma_p%20%3D%200.150%2B0.040%3D0.190)
The 95% confidence interval for the population proportion is (0.110, 0.190).