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

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.

## Answers ( )

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 = sample proportion of Americans who smoke = = 0.329

Firstly, the pivotal quantity for 80% confidence interval for the population proportion is given by;P.Q. = ~ N(0,1)

where, = 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 < < 1.282) = 0.80

P( < < ) = 0.80

P( < p < ) = 0.80

80% confidence interval for p= [,]= [ , ]

= [0.308 , 0.349]

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