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Chapter 8: Problem 80

Determine the most conservative sample size for the estimation of the population proportion for the following. a. \(E=.025\), confidence level \(=95 \%\) b. \(E=.05, \quad\) confidence level \(=90 \%\) c. \(E=.015\), confidence level \(=99 \%\)

Short Answer

Expert verified
The most conservative sample sizes for the given conditions are: \n a. 1537 \n b. 271 \n c. 11189

Step by step solution

01

Look up z-scores for given confidence levels

The z-scores corresponding to the given confidence levels are as follows: \n For 95% confidence, it's 1.96. \n For 90% confidence, it's 1.645. \n For 99% confidence, it's 2.576.
02

Substitute values into formula for part (a)

Substitute \( P=Q=0.5 \), \( Z=1.96 \), and \( E=0.025 \) into the formula. The resulting calculation is \( n = \frac {(1.96^2)(0.5)(0.5)}{(0.025^2)} \). Solving this, we get an approximate sample size of 1537.
03

Substitute values into formula for part (b)

Substitute \( P=Q=0.5 \), \( Z=1.645 \), and \( E=0.05 \) into the formula. The resulting calculation is \( n = \frac {(1.645^2)(0.5)(0.5)}{(0.05^2)} \). Solving this, we get an approximate sample size of 271.
04

Substitute values into formula for part (c)

Substitute \( P=Q=0.5 \), \( Z=2.576 \), and \( E=0.015 \) into the formula. The resulting calculation is \( n = \frac {(2.576^2)(0.5)(0.5)}{(0.015^2)} \). Solving this, we get an approximate sample size of 11189.

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Most popular questions from this chapter

What is the point estimator of the population proportion, \(p\) ?

A computer company that recently developed a new software product wanted to estimate the mean time taken to learn how to use this software by people who are somewhat familiar with computers. A random sample of 12 such persons was selected. The following data give the times taken (in hours) by these persons to learn how to use this software. $$ \begin{array}{llllll} 1.75 & 2.25 & 2.40 & 1.90 & 1.50 & 2.75 \\ 2.15 & 2.25 & 1.80 & 2.20 & 3.25 & 2.60 \end{array} $$ Construct a \(95 \%\) confidence interval for the population mean. Assume that the times taken by all persons who are somewhat familiar with computers to learn how to use this software are approximately normally distributed.

Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32\). Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.

a. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\) b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\)

Calculating a confidence interval for the proportion requires a minimum sample size. Calculate a confidence interval, using any confidence level of \(90 \%\) or higher, for the population proportion for each of the following. a. \(n=200 \quad\) and \(\quad \hat{p}=.01\) b. \(n=160 \quad\) and \(\quad \hat{p}=.9875\) Explain why these confidence intervals reveal a problem when the conditions for using the normal approximation do not hold.
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