/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 The following problem is based o... [FREE SOLUTION] | 91Ó°ÊÓ

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The following problem is based on information taken from Academe, Bulletin of the American Association of University Professors. Let \(x\) represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of \(x\) is approximately \(\sigma^{2}=47.1 .\) However, a random sample of 15 colleges and universities in Kansas showed that \(x\) has a sample variance \(s^{2}=83.2 .\) Use a \(5 \%\) level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a \(95 \%\) confidence interval for the population variance.

Short Answer

Expert verified
The variance for colleges in Kansas is greater than 47.1, with a confidence interval of (44.61, 207.15).

Step by step solution

01

State the Hypotheses

To test the claim, we set up our null and alternative hypotheses. The null hypothesis is that the variance in Kansas is equal to the population variance, \(H_0: \sigma^2 = 47.1\). The alternative hypothesis is that the variance in Kansas is greater, \(H_1: \sigma^2 > 47.1\).
02

Determine the Test Statistic

For testing the population variance, we use the chi-square test statistic: \( \chi^2 = \frac{(n-1) s^2}{\sigma^2} \). Plugging in the values, \( n = 15 \), \( s^2 = 83.2 \), and \( \sigma^2 = 47.1 \), we calculate \( \chi^2 = \frac{(15-1) \times 83.2}{47.1} \approx 24.706 \).
03

Determine the Critical Value

We need the critical chi-square value for \( n - 1 = 14 \) degrees of freedom at a \(5\%\) level of significance for a one-tailed test. Using chi-square tables or software, we find \( \chi^2_{0.05, 14} \approx 23.685 \).
04

Decision Rule

If the calculated chi-square \( 24.706 \) is greater than the critical value \( 23.685 \), we reject the null hypothesis. Otherwise, we fail to reject the null.
05

Make a Decision

Since \( 24.706 > 23.685 \), we reject the null hypothesis. This suggests that the variance for colleges and universities in Kansas is greater than 47.1.
06

Find the 95% Confidence Interval

The 95% confidence interval for the population variance is given by \( \left( \frac{(n-1) s^2}{\chi^2_{0.025, 14}}, \frac{(n-1) s^2}{\chi^2_{0.975, 14}} \right) \). Using \( \chi^2_{0.025, 14} \approx 26.119 \) and \( \chi^2_{0.975, 14} \approx 5.629 \), the confidence interval is \( \left( \frac{14 \times 83.2}{26.119}, \frac{14 \times 83.2}{5.629} \right) \approx (44.61, 207.15) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-Square Test
The Chi-Square Test is a statistical method used to compare observed data with expected data to validate hypotheses about population variance. It's a powerful tool when the underlying population distribution is normal. In this test, we utilize the chi-square statistic formula:
  • The formula is: \( \chi^2 = \frac{(n-1) s^2}{\sigma^2} \)
  • \( n \) represents the sample size.
  • \( s^2 \) is the sample variance.
  • \( \sigma^2 \) symbolizes the known population variance.
With our given data, we calculated this statistic to determine if there's a significant discrepancy between sample and population variance. If the chi-square statistic exceeds the critical value, we reject the null hypothesis, suggesting a departure from the expected variance.
Population Variance
Population variance, represented as \( \sigma^2 \), measures how much individual data points in a population deviate from the population mean. It is a crucial concept in statistics, providing insight into data variability and distribution.A high population variance indicates that data points are spread out over a wider range of values, whereas a low variance signifies that data points are clumped closely around the mean. Understanding population variance helps in assessing risk, variability, and distribution normality across various data sets.In our exercise, we compared the sample variance from a Kansas study to the known population variance of professors' salaries across the U.S. to investigate if there is significant deviation.
Confidence Interval
A confidence interval gives a range of values within which we expect the true population parameter to lie. In hypothesis testing, it's a critical measure that provides insight into the reliability of the sample estimates.For a 95% confidence interval of a population variance, we use the following bounds:
  • Lower bound: \( \frac{(n-1) s^2}{\chi^2_{0.025, \text{df}}} \)
  • Upper bound: \( \frac{(n-1) s^2}{\chi^2_{0.975, \text{df}}} \)
Given the calculations, our confidence interval indicates where 95% of sample variances will likely fall, ensuring precision in determining if our sample reflects the population variance.
Level of Significance
The level of significance, denoted by \( \alpha \), represents the probability of rejecting the null hypothesis when it's true. Commonly set at 0.05, it allows researchers to define what constitutes a statistically significant result.In our scenario, a 5% level of significance implies a 95% confidence level, where we accept a 5% risk of incorrectly rejecting a true null hypothesis. This threshold helps balance the risk of Type I errors (false positives) while providing enough leeway to uncover true patterns or deviations in the data.By setting this level, we structured our hypothesis test around a meaningful benchmark, determining if observed variance in the Kansas sample indeed surpasses the national parameters.

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Most popular questions from this chapter

Jim Mead is a veterinarian who visits a Vermont farm to examine prize bulls. In order to examine a bull, Jim first gives the animal a tranquilizer shot. The effect of the shot is supposed to last an average of 65 minutes, and it usually does. However, Jim sometimes gets chased out of the pasture by a bull that recovers too soon, and other times he becomes worried about prize bulls that take too long to recover. By reading journals, Jim has found that the tranquilizer should have a mean duration time of 65 minutes, with a standard deviation of 15 minutes. A random sample of 10 of Jim's bulls had a mean tranquilized duration time of close to 65 minutes but a standard deviation of 24 minutes. At the \(1 \%\) level of significance, is Jim justified in the claim that the variance is larger than that stated in his journal? Find a \(95 \%\) confidence interval for the population standard deviation.

Wild irises are beautiful flowers found throughout the United States, Canada, and northern Europe. This problem concerns the length of the sepal (leaf-like part covering the flower) of different species of wild iris. Data are based on information taken from an article by \(\mathrm{R} .\) A. Fisher in Annals of Eugenics (Vol. 7, part 2, pp. 179-188). Measurements of sepal length in centimeters from random samples of Iris setosa (I), Iris versicolor (II), and Iris virginica (III) are as follows: \(\begin{array}{ccc}\text { I } & \text { II } & \text { III } \\ 5.4 & 5.5 & 6.3 \\ 4.9 & 6.5 & 5.8 \\ 5.0 & 6.3 & 4.9 \\ 5.4 & 4.9 & 7.2 \\ 4.4 & 5.2 & 5.7 \\ 5.8 & 6.7 & 6.4\end{array}\) \(5.5\) Shall we reject or not reject the claim that there are no differences among the population means of sepal length for the different species of iris? Use a \(5 \%\) level of significance.

A new fuel injection system has been engineered for pickup trucks. The new system and the old system both produce about the same average miles per gallon. However, engineers question which system (old or new) will give better consistency in fuel consumption (miles per gallon) under a variety of driving conditions. A random sample of 31 trucks were fitted with the new fuel injection system and driven under different conditions. For these trucks, the sample variance of gasoline consumption was 58.4. Another random sample of 25 trucks were fitted with the old fuel injection system and driven under a variety of different conditions. For these trucks, the sample variance of gasoline consumption was \(31.6 .\) Test the claim that there is a difference in population variance of gasoline consumption for the two injection systems. Use a \(5 \%\) level of significance. How could your test conclusion relate to the question regarding the consistency of fuel consumption for the two fuel injection systems?

How productive are U.S. workers? One way to answer this question is to study annual profits per employee. A random sample of companies in computers (I), aerospace (II), heavy equipment (III), and broadcasting (IV) gave the following data regarding annual profits per employee (units in thousands of dollars). (Source: Forbes Top Companies, edited by J. T. Davis, John Wiley and Sons.) \(\begin{array}{rrrr}\text { I } & \text { II } & \text { III } & \text { IV } \\\ 27.8 & 13.3 & 22.3 & 17.1 \\ 23.8 & 9.9 & 20.9 & 16.9 \\ 14.1 & 11.7 & 7.2 & 14.3 \\ 8.8 & 8.6 & 12.8 & 15.2 \\ 11.9 & 6.6 & 7.0 & 10.1 \\ & 19.3 & & 9.0\end{array}\) Shall we reject or not reject the claim that there is no difference in population mean annual profits per employee in each of the four types of companies? Use a \(5 \%\) level of significance.

A random sample of leading companies in South Korea gave the following percentage yields based on assets (see reference in Problem 7): \(\begin{array}{lllllll}2.5 & 2.0 & 4.5 & 1.8 & 0.5 & 3.6 & 2.4\end{array}\) \(\begin{array}{llllllll}0.2 & 1.7 & 1.8 & 1.4 & 5.4 & 1.1\end{array}\) Use a calculator to verify that \(s^{2}=2.247\) for these South Korean companies. Another random sample of leading companies in Sweden gave the following percentage yields based on assets: \(\begin{array}{lllllllll}2.3 & 3.2 & 3.6 & 1.2 & 3.6 & 2.8 & 2.3 & 3.5 & 2.8\end{array}\) Use a calculator to verify that \(s^{2}=0.624\) for these Swedish companies. Test the claim that the population variance of percentage yields on assets for South Korean companies is higher than that for companies in Sweden. Use a \(5 \%\) level of significance. How could your test conclusion relate to an economist's question regarding volatility of corporate productivity of large companies in South Korea compared with those in Sweden?

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