The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 47 who smoke. <

Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 47 who smoke.

Step 1 of 2: Suppose a sample of 861 Americans over 47 is drawn. Of these people, 577 don’t smoke. Using the data, estimate the proportion of Americans over 47 who smoke. Enter your answer as a fraction or a decimal number rounded to three decimal places.

Step 2 of 2: Suppose a sample of 861 Americans over 47 is drawn. Of these people, 577 don’t smoke. Using the data, construct the 80% confidence interval for the population proportion of Americans over 47 who smoke. Round your answers to three decimal places.

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Sophia 2 weeks 2021-09-13T13:43:14+00:00 1 Answer 0

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    2021-09-13T13:44:32+00:00

    Answer:

    80% confidence interval for the population proportion of Americans over 47 who smoke is [0.308 , 0.349].

    Step-by-step explanation:

    We are given that a sample of 861 Americans over 47 is drawn. Of these people, 577 don’t smoke.

    Let \hat p = sample proportion of Americans who smoke =  1 - \frac{577}{861} = 0.329

    Firstly, the pivotal quantity for 80% confidence interval for the population proportion is given by;

                                  P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

    where, \hat p = sample proportion of Americans who smoke = 0.329

               n = sample of Americans = 861

               p = population proportion of Americans over 47 who smoke

    Here for constructing 80% confidence interval we have used One-sample z proportion statistics.

    So, 80% confidence interval for the population proportion, p is ;

    P(-1.282 < N(0,1) < 1.282) = 0.80  {As the critical value of z at 10% level

                                                         of significance are -1.282 & 1.282}  

    P(-1.282 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.282) = 0.80

    P( -1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.80

    P( \hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.80

    80% confidence interval for p = [\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } },\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

    = [ 0.329-1.282 \times {\sqrt{\frac{0.329(1-0.329)}{861} } } , 0.329+1.282 \times {\sqrt{\frac{0.329(1-0.329)}{861} } } ]

     = [0.308 , 0.349]

    Therefore, 80% confidence interval for the population proportion of Americans over 47 who smoke is [0.308 , 0.349].

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