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. Construct a 99% confidence interval for the proportion of Americans that were not confident about their food. Round to the nearest tenth of a percent. *

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 =

           n = sample of American adults  = 1,040

           p = population proportion of Americans

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

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

99% confidence interval for p=[ , ]

                      = [ , ]

                      = [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].

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