Question

A) Express the confidence interval 66.4%<p<81.2%66.4%<p<81.2% in the form of ˆp±Ep^±E.

_____% ± ± %_____

B) Express the confidence interval 0.69±0.0510.69±0.051 in open interval form (i.e., (0.155,0.855)).

C) In a survery of 181 households, a Food Marketing Institute found that 141 households spend more than $125 a week on groceries. Please find the 98% 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 =  ±±

D) In our Condor Cafe, the Oxnard College Culinary Program offers the most delicious and affortable breakfast and lunch choices among the 152 Community Colleges in California. This week, a random sample of 333 orders were recorded, and 134 orders were for the $5-meals. Please find the 80% confidence interval for the true proportion of the orders for the $5-meals.

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