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Answer to Question #350957 in Statistics and Probability for SimplyJhoyz

Question #350957

A nutritionist wants to know the populatiin proportion of grade 11 students who eat vegetables. Pegged at a confidence of 95% survey among 1200 respondents was conducted and 200 said that they eat vegetables.

a.What is the margin of error?

b.Construct the confidence interval

c.Find the length of the confidence interval

1
2022-06-16T10:01:13-0400

a.

The critical value for "\\alpha = 0.05" is "z_c = z_{1-\\alpha\/2} = 1.96."

"ME=z_c\\times\\sqrt{\\dfrac{\\hat{p}(1-\\hat{p}}{n}}"

"\\hat{p}=\\dfrac{x}{n}=\\dfrac{200}{1200}=\\dfrac{1}{6}"

"ME=1.96\\times\\sqrt{\\dfrac{\\dfrac{1}{6}(1-\\dfrac{1}{6})}{1200}}\\approx0.021"

b. The corresponding confidence interval is computed as shown below:

"CI(Proportion)=(\\hat{p}-ME, \\hat{p}+ME)"

"=(\\dfrac{1}{6}-0.021, \\dfrac{1}{6}+0.021)"

"=(0.146, 0.188)"

Therefore, based on the data provided, the 95% confidence interval for the population proportion is "0.146 < p < 0.188," which indicates that we are 95% confident that the true population proportion "p" is contained by the interval "(0.146, 0.188)."

c. The length of the confidence interval is "2ME=0.042."

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