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 ______

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jtellezd jtellezd · Answer 1 · 2021-03-28T16:05:05+02:00

Answer:

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)