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pogonyaev
2 years ago
10

Teen obesity: The 2013 National Youth Risk Behavior Survey (YRBS) reported that 13.7% of U.S. students in grades 9 through 12 wh

o attend public and private school were obese. After seeing extensive efforts to educate parents and teens on healthy lifestyles, we want to know if the proportion has decreased this year. We select a random sample of 200 U.S. students and find that 11% are obese.
After performing the hypothesis test for p = 0.137 versus p < 0.137, we obtain a P‑value of 0.133. Which of the following interpretations of the P‑value is correct?

a. There is a 13.3% chance that 11% of U.S. students in grades 9 through 12 who attend public and private school were obese this year.
b. There is a 13.3% chance that 13.7% of U.S. students in grades 9 through 12 who attend public and private school were obese this year.
c. There is a 13.3% chance that 11% of U.S. students in grades 9 through 12 who attend public and private school were obese this year if 13.7% were obese in 2013.
d. There is a 13.3% chance that a sample of 200 U.S. students in grades 9 through 12 who attend public and private school will have 11% or fewer obese students if 13.7% of the population is obese this year.
Mathematics
1 answer:
djyliett [7]2 years ago
8 0

There is a 13.3% chance that a sample of 200 U.S. students in grades 9 through 12 who attend public and private school will have 11% or fewer obese students if 13.7% of the population is obese this year. (Option D)

<u>Explanation:</u>

From the given information it is observed that the 2013 national Youth Risk Behavioural Survey reported that 13.7 percent of US students in grades 9 through the grade 12 who attended the public school were obese.

Consider the random sample of 200 US students and find that 11 percent are obesed. The Null and alternative hypothesis are

Null hypothesis, = 0.137, Alternative hypothesis, = p less than 0.137

Thus, the value of p is 0.133

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The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034

The estimated standard error is:

\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316

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

Then, the lower and upper bounds are:

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<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

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This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

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