- It was observed that 4 out of 10 randomly selected students went to the cinema. Find the 95% confidence interval estimate for the proportion of moviegoers in the population, \pi. a) P( 0.30 < π < 0.50 ) = 0.95 B) P( 0.60 < π < 0.75 ) = 0.95 NS) P( 0.20 < π < 0.30) = 0.95 D) P( 0.10 < π < 0.25 ) = 0.95 TO) P( 0.30 < π < 0.50 ) = 0.05

- It was observed that 4 out of 10 randomly selected students went to the cinema. Find the 95% confidence interval estimate for the proportion of moviegoers in the population, \pi. a) P( 0.30 < π < 0.50 ) = 0.95 B) P( 0.60 < π < 0.75 ) = 0.95 NS) P( 0.20 < π < 0.30) = 0.95 D) P( 0.10 < π < 0.25 ) = 0.95 TO) P( 0.30 < π < 0.50 ) = 0.05

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter11: Data Analysis And Probability

Section: Chapter Questions

Problem 8CR

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7 -

It was observed that 4 out of 10 randomly selected students went to the cinema. Find the 95% confidence interval estimate for the proportion of moviegoers in the population, \pi.

a)	P( 0.30 < π < 0.50 ) = 0.95

B)	P( 0.60 < π < 0.75 ) = 0.95

NS)	P( 0.20 < π < 0.30) = 0.95

D)	P( 0.10 < π < 0.25 ) = 0.95

TO)	P( 0.30 < π < 0.50 ) = 0.05

Definition Definition Method in statistics by which an observation’s uncertainty can be quantified. The main use of interval estimating is for describing a range that is made by transforming a point estimate by determining the range of values, or interval within which the population parameter is likely to fall. This range helps in measuring its precision.

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