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 580 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.

The estimate of a range of plausible values for the true proportion of people that will not wait more than one minute on hold at a 95% confidence level is

Confidence interval = (549.404, 610.596)

Step-by-step explanation:

We will be finding the confidence interval at 95% confidence level.

The proportion of people that hang up the phone in the first minute of waiting

= P = (580/1000) = 0.58

We can then calculate the standard deviation of the distribution of sample means = σₓ = √[np(1-p)]

where n = sample size = 1000

σₓ = √[np(1-p)] = √[1000×0.58×0.42] = 15.61

Confidence interval = (Sample mean) ± (Margin of error)

Sample mean = 580

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Critical value = 1.960

Even though we do not have information on the population mean and standard deviation, we can use the z-distribution's z-score for 95% confidence interval instead of the t-distribution's t-score since the sample size is 1000.

Margin of error = 1.960 × 15.61 = 30.596

Confidence interval = (Sample mean) ± (Margin of error)

Confidence interval = 580 ± 30.596

Confidence interval = (549.404, 610.596)

Hope this Helps!!!

a: 1/6, j: 1/4

step-by-step explanation: you are welcome