1 answer

15 con in In a survery of 101 households, a Food Marketing Institute found that 41...

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 =

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

    < p <

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

    p =  ±±

Math Statistics And Probability
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Answers

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

.

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