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Suppose you want to estimate the true proportion of U.S. women with a gluten sensitivity. In...

Suppose you want to estimate the true proportion of U.S. women with a gluten sensitivity. In a large study using a random sample of 2500 U.S. women, 163 women, 163 women had a gluten sensitivity.

1. Calculate a 98% confidence interval for the true proportion of women with a gluten-sensitivity. As in the last assignment, show all the steps.

2. Interpret the confidence LEVEL.

3. How many women need to be examined in order to estimate the true proportion to within 0.5% (0.005) with the same level of confidence. Show the calculations.

math   Statistics-And-Probability   
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Answer #1

1. z value for 98% CI is 2.326 as P(-2.326<z<2.326)=0.98

Now 163 p = 2509

So Margin of the Error is E = 2*1pq = 2.326*10.935*0.065. = 0.011 2500

Hence CI is CI = p E = 0.065 +0.011 = (0.054,0.076)

Population proportion will lie in the range of 0.054 to 0.076 with 98% CI

Now for E=0.005

So 2.326, 2)2*0.065*0.935 = 13152.38 = 13153 0.005

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