Answer:
99% confidence interval for the proportion of Americans that were not confident about their food is [0.06 , 0.10].
Step-by-step explanation:
We are given that in a survey of 1,040 American adults, 83 said they were not confident that the food they eat in the United States is safe.
Firstly, the pivotal quantity for 99% confidence interval for the proportion of Americans that were not confident about their food is given by;
P.Q. =
~ N(0,1)
where,
= proportion of adults who said that they were not confident that the food they eat in the United States is safe in a sample of 1,040 adults = ![\frac{83}{1040}](https://tex.z-dn.net/?f=%5Cfrac%7B83%7D%7B1040%7D)
n = sample of American adults = 1,040
p = population proportion of Americans
<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>
So, 99% confidence interval for the population proportion, p is ;
P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%
significance level are -2.5758 & 2.5758}
P(-2.5758 <
< 2.5758) = 0.99
P(
<
<
) = 0.99
P(
< p <
) = 0.99
<u>99% confidence interval for p</u>=[
,
]
= [
,
]
= [0.06 , 0.10]
Therefore, 99% confidence interval for the proportion of Americans that were not confident about their food is [0.06 , 0.10].