- 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

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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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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$G i v e n d a t a, n = 10 x = 4 p = \frac{x}{n} = \frac{4}{10} = 0.4 F i n d t h e 95 % c o n f i d e n c e i n t e r v a l$

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