In a survery of 259 households, a Food Marketing Institute found that 174 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 = ±±

Q: If n=100 and X = 80, construct a 95% confidence interval. Enter your answer as an open-interval…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 300…

A: sample size (n) =300favorable outcomes (x) = 24095% ci p.

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A: The given sample size is 370, sample mean is 15.9 and sample standard deviation is 9.7.

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A: Critical value t* = 2.583 , x¯ = 33.7 , s = 13.3 , n = 737

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 200…

A: Givensample size(n)=200x=16confidence level=90%

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 300…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 300…

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Q: If n=200 and X = 160, construct a 99% confidence interval. Enter your answer as an open-interval…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

A: The question is about confidence interval Given : Randomly selected no. of adults ( n ) = 400 No. of…

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 500…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 100…

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Q: In a survery of 313 households, a Food Marketing Institute found that 131 households spend more than…

A: Given information Total Household(n) = 313 No of Households spend more than $125(x) = 131

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 500…

A:

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 500…

A:

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…

A: Givenx=360n=600p^=xn=360600=0.60confidence interval = 99%

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 200…

A:

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Q: Assume that a sample is used to estimate a population proportion p. Find the 99.5% confidence…

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A: Here we need to find a 90% confidence interval for the population proportion.

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

A: Given,n=400x=116

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 400…

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Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…

A: Given that, x=120,n=600 The sample proportion is, Critical values: By using the standard normal…

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…

A: Point estimate for population proportion (p) is the sample proportion. Z-critical value. Margin of…

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.