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In: Statistics and Probability

In a survery of 352 households, a Food Marketing Institute found that 148 households spend more...

In a survery of 352 households, a Food Marketing Institute found that 148 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. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval = Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. < p < Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. p = ±

A toy store would like to open a new store in the city of Thousand Oaks, and they would like to know what proportion of households have kids. Out of 600 households sampled, 372 had kids. Based on this, construct a 95% confidence interval for the true population proportion of Thousand Oaks households with kids.

  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 =  ±

We wish to estimate what percent of adult residents in a certain county are parents. Out of 400 adult residents sampled, 252 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county.

Express your answer in tri-inequality form. Give your answers as decimals, to three places.

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

p =  ±

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orchestra answered 1 year ago
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