Question

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. *

Answer 1

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. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = 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 = $\frac{83}{1040}$

           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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.99

99% confidence interval for p=[ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

                      = [ $\frac{83}{1040}-2.5758 \times {\sqrt{\frac{\frac{83}{1040}(1-\frac{83}{1040})}{1040} }$ , $\frac{83}{1040}+2.5758 \times {\sqrt{\frac{\frac{83}{1040}(1-\frac{83}{1040})}{1040} }$ ]

                      = [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].