Question

15 con in

In a survery of 101 households, a Food Marketing Institute found that 41 households spend more than $125 a week on groceries. Please find the 99.9% confidence interval for the true proportion of the households that spend more than $125 a week on groceries.

1. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

   Confidence interval =

2. 
   
3. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

   < p <

4. 
   
5. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

   p =  ±±

Answer 1

a)

Sample proportion $\hat{p}$ = 41 / 101 = 0.406

99.9% confidence interval for p is

$\hat{p}$ - Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$) / n ] < p < $\hat{p}$ + Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$) / n ]

0.406 - 3.2905* sqrt [ 0.406 * ( 1 - 0.406) / 101] < p < 0.406 - 3.2905* sqrt [ 0.406 * ( 1 - 0.406) / 101]

0.245 < p < 0.567

99.9% C I is (0.245 , 0.567 )

b)

Sample proportion $\hat{p}$ = 41 / 101 = 0.406

99.9% confidence interval for p is

$\hat{p}$ - Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$) / n ] < p < $\hat{p}$ + Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$) / n ]

0.406 - 3.2905* sqrt [ 0.406 * ( 1 - 0.406) / 101] < p < 0.406 - 3.2905* sqrt [ 0.406 * ( 1 - 0.406) / 101]

0.245 < p < 0.567

c)

$\hat{p}$$\pm$ E = $\hat{p}$$\pm$ Z$\alpha$/2 * sqrt [ $\hat{p}$ ( 1 - $\hat{p}$) / n ]

= 0.406 $\pm$ 3.2905* sqrt [ 0.406 * ( 1 - 0.406) / 101]

= 0.406 $\pm$ 0.161