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Chapter 10: Problem 2

A random sample of \(n=15\) observations was selected from a normal population. The sample mean and variance were \(\bar{x}=3.91\) and \(s^{2}=.3214 .\) Find a \(90 \%\) confidence interval for the population variance \(\sigma^{2}\).

Short Answer

Expert verified
Answer: The 90% confidence interval for the population variance is (0.2004, 0.7698).

Step by step solution

01

Identify the given values

In this problem, we are given: - Sample size (\(n\)): 15 - Sample mean (\(\bar{x}\)): 3.91 - Sample variance (\(s^2\)): 0.3214
02

Find the degrees of freedom

The degrees of freedom for a chi-square distribution is equal to the sample size minus 1: $$df = n - 1$$ Substituting the given values: $$df = 15 - 1 = 14$$
03

Find the critical chi-square values

The confidence level is given as 90%. Therefore, we need to find the critical chi-square values at the lower and upper limits of the 90% confidence interval. From the chi-square table, we can find these values for 14 degrees of freedom: - \(蠂_{1-0.45}^2\): 22.362 - \(蠂_{0.95}^2\): 6.571
04

Calculate the confidence interval for the variance

The confidence interval for the population variance can be calculated using the following formula: $$\left(\frac{(n-1)s^2}{蠂_{1-伪/2}^2}, \frac{(n-1)s^2}{蠂_{伪/2}^2}\right)$$ Where: - \(n\) is the sample size, - \(s^2\) is the sample variance, - \(伪\) is the significance level, and - \(蠂_{1-伪/2}^2\) and \(蠂_{伪/2}^2\) are the critical chi-square values. We can plug in the given values and the critical chi-square values from Step 3: $$\text{Confidence interval for } 蟽^2 = \left(\frac{(15 - 1)(0.3214)}{22.362}, \frac{(15 - 1)(0.3214)}{6.571}\right)$$
05

Calculate the confidence interval

Perform the calculations in the confidence interval formula: $$\text{Confidence interval for } 蟽^2 = \left(\frac{14(0.3214)}{22.362}, \frac{14(0.3214)}{6.571}\right) = (0.2004, 0.7698)$$ So, the 90% confidence interval for the population variance is (0.2004, 0.7698).

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Most popular questions from this chapter

A paired-difference experiment consists of \(n=18\) pairs, \(\bar{d}=5.7,\) and \(s_{d}^{2}=256 .\) Suppose you wish to detect \(\mu_{d}>0\). a. Give the null and alternative hypotheses for the test. b. Conduct the test and state your conclusions.

Car manufacturers try to design the bumpers of their automobiles to prevent costly damage in parking-lot type accidents. To compare repair costs of front versus back bumpers for several brands of cars, the cars were subject to a front and rear impacts at \(5 \mathrm{mph},\) and the repair costs recorded. \({ }^{15}\) \begin{tabular}{lrr} Vehicle & Front & Rear \\ \hline WW Jetta & \(\$ 396\) & \(\$ 602\) \\ Daewoo Nubira & 451 & 404 \\ Acura 3.4 RL & 1123 & 968 \\\ Dodge Neon & 687 & 748 \\ Nissan Sentra & 583 & 571 \end{tabular} Do the data provide sufficient evidence to indicate that there is a significant difference in average repair costs for front versus rear bumper repairs costs? Test using \(\alpha=.05 .\)

Exercise 10.24 describes a dental experiment conducted to investigate the effectiveness of an oral rinse used to inhibit the growth of plaque on teeth. Subjects were divided into two groups: One group used a rinse with an anti plaque ingredient, and the control group used a rinse containing inactive ingredients. Suppose that the plaque growth on each person's teeth was measured after using the rinse after 4 hours and then again after 8 hours. If you wish to estimate the difference in plaque growth from 4 to 8 hours, should you use a confidence interval based on a paired or an unpaired analysis? Explain.

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Refer to Exercise 10.91 . Suppose that the word-association experiment is conducted using eight people as blocks and making a comparison of reaction times within each person; that is, each person is subjected to both stimuli in a random order. The reaction times (in seconds) for the experiment are as follows: \begin{tabular}{ccc} Person & Stimulus 1 & Stimulus 2 \\ \hline 1 & 3 & 4 \\ 2 & 1 & 2 \\ 3 & 1 & 3 \\ 4 & 2 & 1 \\ 5 & 1 & 2 \\ 6 & 2 & 3 \\ 7 & 3 & 3 \\ 8 & 2 & 3 \end{tabular} Do the data present sufficient evidence to indicate a difference in mean reaction times for the two stimuli? Test using \(\alpha=.05 .\)
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