A study is conducted to determine how long a person will wait on hold on a telephone call before hanging up. A sample of 1000 people found that 569 hung up the phone in the first minute of waiting. It is of interest to estimate a range of plausible values for the true proportion of people that will not wait more than one minute on hold. Round all answers to three decimal places.(a) Construct a two-sided 95% confidence interval for p, the true proportion of people that will not wait on hold more than one minute. This confidence interval is

CI = ( 0,539 ; 0,599)

Step-by-step explanation:

The proportion p₀ of person that hung up the phone in the first minute of waiting is:

p₀ = 569/1000         p₀ = 0,569       p₀  =  56,9 %

Then q₀  = 1 - 0,569                          q₀  = 0,431       q₀ = 43,1 %

Sample size is  n = 1000

A 95 % CI  is:     α  = 5 %    α = 0,05      a/2 = 0,025

In z-table we look  z ( score ) for  0,025      z  = ± 1,96

Now CI for proportion is:

p₀ - z (α/2) * √ (p₀*q₀ )/n   < p < p₀ + z (α/2) * √ (p₀*q₀ )/n

we need to find out

z(α/2) * √ (p₀*q₀ )/n   =   1,96*√ ( 0,569*0,431)/1000 = 1,96 *0,01566 =0,03

Then

CI ( 95%) =  0,569 ± 0,03  =  ( 0,569 - 0,03 ;  0,569 + 0,03)

CI = ( 0,539 ; 0,599)

answer: i think it's a

step-by-step explanation: