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.

Answer 1

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

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 - (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{(\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{(\hat p(1-\hat p))/(n) } }` < 1.282) = 0.80

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

P(
`\hat p-1.282 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.282 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.80

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

= [
`0.329-1.282 * {\sqrt{(0.329(1-0.329))/(861) } }` ,
`0.329+1.282 * {\sqrt{(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].