PROBLEM 1. Suppose an internet service provider sampled 2050 customers and found

Question

PROBLEM 1. Suppose an internet service provider sampled 2050 customers and found that 305 of them experienced an interruption in high-speed service during the a) Report a point estimate for the population proportion of all customers who b) Compute a 95% confidence interval for the proportion of all customers who c) Compute a 99% confidence interval for the proportion of all customers who previous month. experienced an interruption. Round your answer to 4 decimal places. experienced an interruption. Report the margin of error of your point estimate associated with a 95% confidence level. experienced an interruption. Report the margin of error of your poirnt estimate associated with a 99% confidence level. confidence interval? confidence. What sample size do you need to use in order to meet this d) How does the increase in the confidence level affect the width of the e) Now you want the margin of error not to exceed 0.015 with a 99% specification regardless of a p value?

Explanation / Answer

Solutiona:

point estimate =sample proportion=p^=x/n=305/2050=0.1488

Solutionb:

Z crit=1.96

margin of error=zcrit*sqrt(p^(1-p^)/n)

=1.96*sqrt(0.1488*1-0.1488)/2050

MOE =0.0154

lower limit=sample prop-MOE

=0.1488-0.0154

=0.1334

upper limit=sample prop+MOE

=0.1488+0.0154

=0.1642

Solutionc:

Z crit for 99=2.576

margin of error=zcrit*sqrt(p^(1-p^)/n)

=2.576*sqrt(0.1488*1-0.1488)/2050

MOE =0.0202

lower limit=0.1488-0.0202

=0.1286

upper limit=0.1488+0.0202

=0.169

Solutiond:

Increasing confidence interval increases z crit there by increasing the range.Confidence intervla becomes wider from 95 to 99%

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