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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

A sample of 11 observations taken from a normally distributed population produced the following data. \(\begin{array}{lllllllllll}-7.1 & 10.3 & 8.7 & -3.6 & -6.0 & -7.5 & 5.2 & 3.7 & 9.8 & -4.4 & 6.4\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(95 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for \(\mu\) in part b?

The principal of a large high school is concerned about the amount of time that his students spend on jobs to pay for their cars, to buy clothes, and so on. He would like to estimate the mean number of hours worked per week by these students. He knows that the standard deviation of the times spent per week on such jobs by all students is \(2.5\) hours. What sample size should he choose so that the estimate is within \(.75\) hour of the population mean? The principal wants to use a \(98 \%\) confidence level.

A hospital administration wants to estimate the mean time spent by patients waiting for treatment at the emergency room. The waiting times (in minutes) recorded for a random sample of 35 such patients are given below. \(\begin{array}{lrrrrrr}30 & 7 & 68 & 76 & 47 & 60 & 51 \\ 64 & 25 & 35 & 29 & 30 & 35 & 62 \\ 96 & 104 & 58 & 32 & 32 & 102 & 27 \\ 45 & 11 & 64 & 62 & 72 & 39 & 92 \\ 84 & 47 & 12 & 33 & 55 & 84 & 36\end{array}\) Construct a \(99 \%\) confidence interval for the corresponding population mean. Use the \(t\) distribution.

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For a data set obtained from a sample, \(n=20\) and \(\bar{x}=24.5 .\) It is known that \(\sigma=3.1\). The population is normally distributed. a. What is the point estimate of \(\mu ?\) b. Make a \(99 \%\) confidence interval for \(\mu\). c. What is the margin of error of estimate for part b?
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