91Ó°ÊÓ

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

What assumptions must hold true to use the \(t\) distribution to make a confidence interval for \(\mu ?\)

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?

A random sample of 20 acres gave a mean yield of wheat equal to \(41.2\) bushels per acre with a standard deviation of 3 bushels. Assuming that the yield of wheat per acre is normally distributed, construct a \(90 \%\) confidence interval for the population mean \(\mu .\)

Check if the sample size is large enough to use the normal distribution to make a confidence interval for \(p\) for each of the following cases. a. \(n=80\) and \(\hat{p}=.85\) b. \(n=110\) and \(\hat{p}=.98\) c. \(n=35\) and \(\hat{p}=.40\) d. \(n=200\) and \(\hat{p}=.08\)

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 \(.035\).

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 \(=99 \%\)

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.

\(8.104\) 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, \(\mu\).

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.