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Chapter 8: Problem 60

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 size for the estimation of population proportion for a margin of error .025 and a 95% confidence level is approximately 1537. For a margin of error .05 and a 90% confidence level, it is around 272. And for a margin of error .015 with a 99% confidence level, it is about 7411.

Step by step solution

01

Identify Given Variables and Find Z-Score

From the problem, we have the margin of error (E) and the confidence level. Use a standard normal distribution table or a Z-score calculator to find the Z-score corresponding to the given confidence level. For a 95% confidence level, the Z-score is 1.96. For a 90% confidence level, the Z-score is 1.645. For a 99% confidence level, the Z-score is 2.576.
02

Substituting the values in the Formula for Part a

For the first part, we have E = .025 and the Z-score is 1.96 for a 95% confidence level. Assuming \(P = 0.5\) (most conservative estimate), substitute these values in the population proportion formula \(n = \frac{{Z^2 * P * (1-P)}}{{E^2}}\). This simplifies to \(n = \frac{{1.96^2 * 0.5 * (1-0.5)}}{(.025)^2}\), which calculates approximately to 1536.64.
03

Rounding up for Part a

Since sample size can't be a fraction, round up the calculated n to the next whole number. So the smallest conservative sample size required for part a is 1537.
04

Substituting the values in the Formula for Part b

In the second part, E = .05 and the Z-score is 1.645 for a 90% confidence level. Substitute these values in the formula, then the equation becomes \(n = \frac{{1.645^2 * 0.5 * (1-0.5)}}{(.05)^2}\), which calculates approximately to 271.24.
05

Rounding up for Part b

Rounding up the number, you find that the smallest conservative sample size required for part b is 272.
06

Substituting the values in the Formula for Part c

In the last part, E = .015 and the Z-score is 2.576 for a 99% confidence level. When you put these values in the formula, the equation becomes \(n = \frac{{2.576^2 * 0.5 * (1-0.5)}}{(.015)^2}\), which calculates approximately to 7410.31.
07

Rounding up for Part c

After rounding up, the smallest conservative sample size required for part c is 7411.

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Most popular questions from this chapter

A large city with chronic economic problems is considering legalizing casino gambling. The city council wants to estimate the proportion of all adults in the city who favor legalized casino gambling. What is the most conservative estimate of the minimum sample size that would limit the margin of error to be within \(.05\) of the population proportion for a \(95 \%\) confidence interval?

A company that produces detergents wants to estimate the mean amount of detergent in 64 -ounce jugs at a \(99 \%\) confidence level. The company knows that the standard deviation of the amounts of detergent in all such jugs is \(.20\) ounce. How large a sample should the company select so that the estimate is within \(.04\) ounce of the population mean?

a. Find the value of \(t\) for the \(t\) distribution with a sample size of 21 and area in the left tail equal to \(.10 .\) b. Find the value of \(t\) for the \(t\) distribution with a sample size of 14 and area in the right tail equal to \(.025 .\) c. Find the value of \(t\) for the \(t\) distribution with 45 degrees of freedom and \(.001\) area in the right tail. d. Find the value of \(t\) for the \(t\) distribution with 37 degrees of freedom and \(.005\) area in the left tail.

A random sample of 36 mid-sized cars tested for fuel consumption gave a mean of \(26.4\) miles per gallon with a standard deviation of \(2.3\) miles per gallon. a. Find a \(99 \%\) confidence interval for the population mean, \(\mu\). b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

a. How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(p\) is \(.035\) when the value of the sample proportion obtained from a preliminary sample is \(.29 ?\) b. Find the most conservative sample size that will produce the margin of error for a \(99 \%\) confidence interval for \(p\) equal to \(.035\).
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