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)

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.