Respuesta :
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 [tex]\hat p[/tex] = sample proportion of Americans who smoke = [tex]1 - \frac{577}{861}[/tex] = 0.329
Firstly, the pivotal quantity for 80% confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = 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 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.282) = 0.80
P( [tex]-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.80
P( [tex]\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.80
80% confidence interval for p = [[tex]\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex],[tex]\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]
= [ [tex]0.329-1.282 \times {\sqrt{\frac{0.329(1-0.329)}{861} } }[/tex] , [tex]0.329+1.282 \times {\sqrt{\frac{0.329(1-0.329)}{861} } }[/tex] ]
= [0.308 , 0.349]
Therefore, 80% confidence interval for the population proportion of Americans over 47 who smoke is [0.308 , 0.349].
