Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the...

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 20 who smoke. Step 2 of 2 : Suppose a sample of 966 Americans over 20 is drawn. Of these people, 783 don't smoke. Using the data, construct the 98% confidence interval for the population proportion of Americans over 20 who smoke. Round your answers to three decimal places.

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Answer #1

Solution :

=> given that a sample of 966 Americans over 20 is drawn. Of these people, 783 don't smoke

=> x = 966 - 783 = 183 will smoke

=> p = P(smoke) = x/n = 183/966 = 0.189

=> q = 1 - p = 0.811

=> For 98% confidence interval , Z = 2.326

=> The 98% confidence interval of the population proportion of Americans over 20 who smoke is

=> p +/- Z*sqrt(p*q/n)

=> 0.189 +/- 2.326*sqrt(0.189*0.811/966)

=> (0.1597 , 0.2183)

=> (0.160 , 0.218) (rounded)

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