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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 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.

You are interested in estimating the mean commuting time from home to school for all commuter students at your school. Briefly explain the procedure you will follow to conduct this study. Collect the required data from a sample of 30 or more such students and then estimate the population mean at a \(99 \%\) confidence level. Assume that the population standard deviation for such times is \(5.5\) minutes.

Briefly explain the meaning of the degrees of freedom for a \(t\) distribution. Give one example.

\(8.104\) A random sample of 25 life insurance policyholders showed that the average premium they pay on their life insurance policies is \(\$ 685\) per year with a standard deviation of \(\$ 74\). Assuming that the life insurance policy premiums for all life insurance policyholders have a normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).

What is the point estimator of the population mean, \(\mu ?\) How would you calculate the margin of error for an estimate of \(\mu\) ?
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