An Internet service provider sampled 540 customers and found that 75 of them experienced an interruption in high-speed service during the previous month. (a) Find a point estimate for the population proportion of all customers who experienced an interruption. DO NOT ROUND the answer. (b) Construct a 90% confidence interval for the proportion of all customers who experienced an interruption. Round your answer to 3 decimal places. (c) Estimate the sample size needed so that a 95% confidence interval will have a margin of error of 0.03.

a) point estimate for the population proportion of all customers who experienced an interruption

E = 0.029

b)

90% confidence interval for the proportion of all customers who experienced an interruption

(0.11441 , 0.16319)

C)

Size of the sample 'n' = 509

Step-by-step explanation:

Step(i):-

a)

An Internet service provider sampled 540 customers and found that 75 of them experienced an interruption in high-speed service during the previous month.

Sample proportion

The point estimate of all customers who experienced an interruption

  Level of significance

                          Z₀.₀₅ = 1.96

                         E  =  0.029

b)

90% confidence interval for the proportion of all customers who experienced an interruption

( 0.1388 - 0.02439 , 0.1388 + 0.02439)

(0.11441 , 0.16319)

c)

Margin of error

                   √n = 22.58

Squaring on both sides, we get

                 n = 509

yep

step-by-step explanation:

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