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Chapter 8: Problem 17

Determine the sample size for the estimate of \(\mu\) for the following. a. \(E=2.3, \quad \sigma=15.40\), confidence level \(=99 \%\) b. \(E=4.1, \quad \sigma=23.45\), confidence level \(=95 \%\) c. \(E=25.9, \quad \sigma=122.25, \quad\) confidence level \(=90 \%\)

Short Answer

Expert verified
The required sample sizes for part a, b and c are obtained after evaluating the calculations in step 1, 2 and 3. The calculated sample sizes need to be rounded up as sample sizes are always positive integers.

Step by step solution

01

Calculate Sample Size for Part A

Given \(E=2.3\), \(\sigma=15.40\), and a confidence level of 99%. The Z score is approximately 2.33 for a 99% confidence level. So we use the formula \(n = \left(\frac{Z*\sigma}{E}\right)^2\) to calculate the required sample size \(n = \left(\frac{2.33*15.40}{2.3}\right)^2\). The result is then rounded up to the nearest whole number as sample sizes are always positive integers.
02

Calculate Sample Size for Part B

Given \(E=4.1\), \(\sigma=23.45\), and a confidence level of 95%. The Z score is approximately 1.96 for a 95% confidence level. So we use the formula \(n = \left(\frac{Z*\sigma}{E}\right)^2\) to calculate the required sample size \(n = \left(\frac{1.96*23.45}{4.1}\right)^2\). The rounded result will be the sample size.
03

Calculate Sample Size for Part C

Given \(E=25.9\), \(\sigma=122.25\), and a confidence level of 90%. The Z score is approximately 1.645 for a 90% confidence level. So we use the formula \(n = \left(\frac{Z*\sigma}{E}\right)^2\) to calculate the required sample size \(n = \left(\frac{1.645*122.25}{25.9}\right)^2\). Again, the result should be rounded up to the nearest whole number to obtain the required sample size.

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Most popular questions from this chapter

A survey of 500 randomly selected adult men showed that the mean time they spend per week watching sports on television is \(9.75\) hours with a standard deviation of \(2.2\) hours. Construct a \(90 \%\) confidence interval for the population mean, \(\mu\).

a. A random sample of 400 observations taken from a population produced a sample mean equal to \(92.45\) and a standard deviation equal to \(12.20\). Make a \(98 \%\) confidence interval for \(\mu\). b. Another sample of 400 observations taken from the same population produced a sample mean equal to \(91.75\) and a standard deviation equal to \(14.50 .\) Make a \(98 \%\) confidence interval for \(\mu\). c. A third sample of 400 observations taken from the same population produced a sample mean equal to \(89.63\) and a standard deviation equal to \(13.40 .\) Make a \(98 \%\) confidence interval for \(\mu\). d. The true population mean for this population is \(90.65\). Which of the confidence intervals constructed in parts a through \(\mathrm{c}\) cover this population mean and which do not?

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 an approximate normal distribution, make a \(99 \%\) confidence interval for the population mean, \(\mu\).

A random sample of 300 female members of health clubs in Los Angeles showed that they spend, on average, \(4.5\) hours per week doing physical exercise with a standard deviation of . 75 hour. Find a \(98 \%\) confidence interval for the population mean.

A random sample of 200 observations selected from a population produced a sample proportion equal to \(.91\). a. Make a \(90 \%\) confidence interval for \(p\). b. Construct a \(95 \%\) confidence interval for \(p\). c. Make a \(99 \%\) confidence interval for \(p\). d. Does the width of the confidence intervals constructed in parts a through \(\mathrm{c}\) increase as the confidence level increases? If yes, explain why.
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