/*! 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

91Ó°ÊÓ

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
Reject the null hypothesis; Kansas' variance > 47.1. 95% CI for variance: [44.58, 206.95].

Step by step solution

01

Formulate Hypotheses

First, we set up the null hypothesis and the alternative hypothesis. The null hypothesis is \( H_0: \sigma^2 = 47.1 \), which means the population variance is equal to 47.1. The alternative hypothesis is \( H_1: \sigma^2 > 47.1 \), indicating the variance is greater than 47.1.
02

Choose the Right Test and Determine the Test Statistic

Since we're testing a claim about a variance, we'll use the Chi-Square test for variance. The test statistic for Chi-Square is given by \( \chi^2 = \frac{(n-1)\cdot s^2}{\sigma^2} \), where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2 \) is the population variance. Substituting the values, \( \chi^2 = \frac{(15-1)\cdot 83.2}{47.1} \).
03

Calculations for Test Statistic

Calculate the test statistic using the values from Step 2. \( \chi^2 = \frac{14 \cdot 83.2}{47.1} = \frac{1164.8}{47.1} \approx 24.73 \).
04

Determine Critical Value

For a 5% significance level and using a Chi-Square distribution with \( n-1 = 14 \) degrees of freedom, check a Chi-Square table for the critical value. For \( \alpha = 0.05 \), the critical value is approximately 23.685.
05

Decision Rule

The decision rule for this test is: if the calculated test statistic \( 24.73 \) is greater than the critical value \( 23.685 \), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
06

Make a Decision

Since \( 24.73 > 23.685 \), we reject the null hypothesis. This suggests there is evidence at the 5% level of significance to support the claim that the variance of salaries in Kansas is greater than 47.1.
07

Confidence Interval Calculation

To find the 95% confidence interval for the population variance, use the formula \( \left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right) \). For \( n = 15 \), degrees of freedom \( n-1 = 14 \), and \( s^2 = 83.2 \), calculate the interval: \( \chi^2_{0.025} = 5.628 \) and \( \chi^2_{0.975} = 26.119 \). Insert these values into the formula to find the interval: \( \left( \frac{14 \cdot 83.2}{26.119}, \frac{14 \cdot 83.2}{5.628} \right) \).
08

Compute the Interval

\( \left( \frac{1164.8}{26.119}, \frac{1164.8}{5.628} \right) = (44.58, 206.95) \). Thus, the 95% confidence interval for the population variance is \([44.58 , 206.95]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 that helps us determine if there is a significant difference between an observed variance and an expected variance under some null hypothesis. In our context, we're examining whether the variance of college and university salaries in Kansas differs from the known population variance of 47.1. This type of test is particularly useful when dealing with categorical data and when the goal is to understand variance.

The Chi-Square test calculates a test statistic (chi^2) using the formula:
  • \( \chi^2 = \frac{(n-1)\cdot s^2}{\sigma^2} \)
  • where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2 \) is the population variance.
A larger test statistic suggests that the observed variance is significantly different from the expected variance.

To decide whether to accept or reject the null hypothesis, compare the calculated value of \( \chi^2 \) to a critical value from the Chi-Square distribution table. If the test statistic is greater than the critical value, this indicates that the observed variance significantly differs from the expected variance, leading us to reject the null hypothesis.
Variance
Variance is a key statistical measure that tells us how spread out a set of numbers is. Specifically, it gives us the average of the squared differences from the mean. In simpler terms, it shows how much variability exists in a dataset.

For our case, the variance being tested is important because it helps determine if the salaries of professors in Kansas significantly differ from the known nationwide standard.
  • The population variance is represented as \( \sigma^2 \).
  • The sample variance is denoted by \( s^2 \).
  • Higher variance means more spread in the data, often indicating higher unpredictability or dispersion.
Understanding these variances helps in formulating hypotheses and in making decisions based on the results of statistical tests, leading to more informed conclusions about data patterns.
Confidence Interval
A confidence interval provides a range of values which is likely to include an unknown population parameter, such as the variance in our exercise. It is expressed with a certain level of confidence, usually 95% or 99%.

In this context:
  • A 95% confidence interval suggests that if we were to take 100 different samples and compute a range for each one, we expect about 95 of those ranges to contain the true population variance.
The confidence interval for variance is calculated using specific Chi-Square values:
  • Lower Limit: \( \frac{(n-1)s^2}{\chi^2_{\alpha/2}} \)
  • Upper Limit: \( \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \)
A wider interval might indicate less certainty about the estimate, while a narrower interval suggests more precision. This interval helps in determining if the observed sample variance could plausibly reflect the true population variance. It gives a range rather than a single estimate, which can provide more insight into the reliability of the sample measurement.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold for determining whether our test statistic shows a statistically significant result. It tells us the probability of rejecting the null hypothesis when it is actually true, essentially indicating the risk we're willing to take.

In hypothesis testing:
  • A common significance level is 5% (or 0.05), meaning there is a 5% risk of incorrectly rejecting the null hypothesis.
  • The choice of \( \alpha \) affects the critical value, which in turn influences our decision about the null hypothesis.
  • A lower significance level (like 1%) means stricter criteria for observing a significant effect, reducing the chance of a Type I error but increasing the chance of a Type II error.
In our scenario, using a 5% significance level involves comparing the computed test statistic to a critical value from the Chi-Square table corresponding to 5%. If the test statistic exceeds the critical value, it suggests the variance is significantly different, prompting us to reject the null hypothesis in favor of the alternative.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An economist wonders if corporate productivity in some countries is more volatile than that in other countries. One measure of a company's productivity is annual percentage yield based on total company assets. Data for this problem are based on information taken from Forbes Top Companies, edited by J. T. Davis. A random sample of leading companies in France gave the following percentage yields based on assets: $$\begin{aligned} &\begin{array}{cccccccccc} 4.4 & 5.2 & 3.7 & 3.1 & 2.5 & 3.5 & 2.8 & 4.4 & 5.7 & 3.4 & 4.1 \\ 6.8 & 2.9 & 3.2 & 7.2 & 6.5 & 5.0 & 3.3 & 2.8 & 2.5 & 4.5 \end{array}\\\ \end{aligned}$$ Use a calculator to verify that \(s^{2} \approx 2.044\) for this sample of French companies. Another random sample of leading companies in Germany gave the following percentage yields based on assets: $$\begin{array}{lllllllll} 3.0 & 3.6 & 3.7 & 4.5 & 5.1 & 5.5 & 5.0 & 5.4 & 3.2 \\ 3.5 & 3.7 & 2.6 & 2.8 & 3.0 & 3.0 & 2.2 & 4.7 & 3.2 \end{array}$$ Use a calculator to verify that \(s^{2} \approx 1.038\) for this sample of German companies. Test the claim that there is a difference (either way) in the population variance of percentage yields for leading companies in France and Germany. Use a \(5 \%\) level of significance. How could your test conclusion relate to the economist's question regarding volatility (data spread) of corporate productivity of large companies in France compared with large companies in Germany?

Academe, Bulletin of the American Association of University Professors (Vol. 83, No. 2) presents results of salary surveys (average salary) by rank of the faculty member (professor, associate, assistant, instructor) and by type of institution (public, private). List the factors and the number of levels of each factor. How many cells are there in the data table?

When using the \(F\) distribution to test two variances, is it essential that each of the two populations be normally distributed? Would it be all right if the populations had distributions that were mound-shaped and more or less symmetric?

In general, is the \(F\) distribution symmetric? Can values of \(F\) be negative?

Ethnic Groups A sociologist studying New York City ethnic groups wants to determine if there is a difference in income for immigrants from four different countries during their first year in the city. She obtained the data in the following table from a random sample of immigrants from these countries (incomes in thousands of dollars). Use a 0.05 level of significance to test the claim that there is no difference in the earnings of immigrants from the four different countries. Country I $$\begin{aligned}&12.7\\\&9.2\\\&10.9\\\&8.9\\\&16.4\end{aligned}$$ Country II $$\begin{aligned}&8.3\\\&17.2\\\&\begin{array}{l}19.1 \\\10.3\end{array}\end{aligned}$$ Country III $$\begin{aligned}&20.3\\\&\begin{array}{l}16.6 \\\22.7 \\\25.2 \\\19.9\end{array}\end{aligned}$$ Country IV $$\begin{aligned}&\begin{array}{r}17.2 \\\8.8\end{array}\\\&14.7\\\&\begin{array}{l}21.3 \\\19.8 \end{array}\end{aligned}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.