Question

In: Statistics and Probability

A survey of 1100U.S. adults found that 31%of people said that they would get no work...

A survey of 1100U.S. adults found that 31%of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the

90%confidence interval of the true proportion. Round intermediate answers to at least five decimal places. Round your final answers to at least three decimal places.

Expert Solution

Solution =

Given,

n = 1100 ....... Sample size

Let denotes the sample proportion.

= 31% = 0.31

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

/2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645 ......(use z table)

Now , the margin of error is given by

E = Z/2 * [ *(1 - ) / n]

= 1.645 * [ 0.31 *(1 - 0.31 ) / 1100 ]

= 0.02294

Now the confidence interval is given by

( - E) ( + E)

( 0.31 - 0.02294 ) ( 0.31 + 0.02294 )

0.287 0.333

Required 90% Confidence Interval is ( 0.287 , 0.333 )

orchestra answered 2 years ago

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