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Chapter 8: Problem 24

Find a \((1-\alpha) 100 \%\) confidence interval for a population mean \(\mu\) for these values: a. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\) b. \(\alpha=.10, n=65, \bar{x}=1049, s^{2}=51\) c. \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

Short Answer

Expert verified
a. The 99% confidence interval for the population mean, 渭, in case a is (32.61, 35.39). b. The 90% confidence interval for the population mean, 渭, in case b is (1046.50, 1051.50). c. The 95% confidence interval for the population mean, 渭, in case c is (65.97, 66.63).

Step by step solution

01

a. Case \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\)

First, we need to calculate the standard error by dividing the sample standard deviation, s, by the square root of the sample size, n: $$SE = \frac{s}{\sqrt{n}} =\frac{\sqrt{12}}{\sqrt{38}} \approx 0.564$$ Then, we find the critical value from the t-distribution table (or using statistical software): $$t_{\frac{\alpha}{2}, n-1} = t_{0.005, 37} \approx 2.840$$ Now, we can plug the values into the confidence interval formula: $$(34 - 2.840 \cdot 0.564, 34 + 2.840 \cdot 0.564) \approx (32.61, 35.39)$$ The \(99\%\) confidence interval for the population mean, \(\mu\), in case a is \((32.61, 35.39)\).
02

b. Case \(\alpha=.10, n=65, \bar{x}=1049, s^{2}=51\)

First, calculate the standard error: $$SE = \frac{s}{\sqrt{n}} = \frac{\sqrt{51}}{\sqrt{65}} \approx 0.890$$ Then, find the critical value: $$t_{\frac{\alpha}{2}, n-1} = t_{0.05, 64} \approx 1.669$$ Plug the values into the confidence interval formula: $$(1049 - 1.669 \cdot 0.890, 1049 + 1.669 \cdot 0.890) \approx (1046.50, 1051.50)$$ The \(90\%\) confidence interval for the population mean, \(\mu\), in case b is \((1046.50, 1051.50)\).
03

c. Case \(\alpha=.05, n=89, \bar{x}=66.3, s^{2}=2.48\)

First, calculate the standard error: $$SE = \frac{s}{\sqrt{n}} = \frac{\sqrt{2.48}}{\sqrt{89}} \approx 0.166$$ Then, find the critical value: $$t_{\frac{\alpha}{2}, n-1} = t_{0.025, 88} \approx 1.989$$ Plug the values into the confidence interval formula: $$(66.3 - 1.989 \cdot 0.166, 66.3 + 1.989 \cdot 0.166) \approx (65.97, 66.63)$$ The \(95\%\) confidence interval for the population mean, \(\mu\), in case c is \((65.97, 66.63)\).

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Most popular questions from this chapter

Refer to the Interpreting Confidence Intervals applet. a. Suppose that you have a random sample of size \(n=50\) from a population with unknown mean \(\mu\) and known standard deviation \(\sigma=35 .\) Calculate the half width of a \(95 \%\) confidence interval for \(\mu\). What would the width of this interval be? b. Use the button to create a single confidence interval for \(\mu\). What is the width of this interval? Compare your results to the calculation you did in part a.

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To compare the effect of stress in the form of noise on the ability to perform a simple task, 70 subjects were divided into two groups. The first group of 30 subjects acted as a control, while the second group of 40 were the experimental group. Although each subject performed the task in the same control room, each of the experimental group subjects had to perform the task while loud rock music was played. The time to finish the task was recorded for each subject and the following summary was obtained: $$\begin{array}{lll} & \text { Control } & \text { Experimental } \\\\\hline n & 30 & 40 \\\\\bar{x} & 15 \text { minutes } & 23 \text { minutes } \\ s & 4 \text { minutes } & 10 \text { minutes }\end{array}$$ a. Find a \(99 \%\) confidence interval for the difference in mean completion times for these two groups. b. Based on the confidence interval in part a, is there sufficient evidence to indicate a difference in the average time to completion for the two groups? Explain.
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