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Chapter 11: Problem 38

The following table gives the distributions of grades for three professors for a few randomly selected classes that each of them taught during the last 2 years. \begin{tabular}{l|lccc} \hline & & \multicolumn{3}{c} { Professor } \\ \cline { 3 - 5 } & & Miller & Smith & Moore \\ \hline \multirow{4}{*} { Grade } & A & 18 & 36 & 20 \\ & B & 25 & 44 & 15 \\ & C & 85 & 73 & 82 \\ & D \& F & 17 & 12 & 8 \\ \hline \end{tabular} Using the \(2.5 \%\) significance level, test the null hypothesis that the grade distributions are homogeneous for these three professors.

Short Answer

Expert verified
The short answer will come after performing exact calculations. It will either be 'The grade distributions across the three professors are not homogeneous' if the null hypothesis is rejected or 'The grade distributions across the three professors are homogeneous' if the null hypothesis is not rejected.

Step by step solution

01

Set Up The Hypotheses

We will start by setting up the null and alternative hypotheses. The null hypothesis \(H_0\) is that the grade distributions are homogeneous for the three professors. The alternative hypothesis \(H_1\) asserts that the distributions are not homogeneous.
02

Compute the Expected Frequencies

Under the null hypothesis, we assume that each professor has the same distribution of grades. So, to calculate the expected frequency for each category, we take the total for that category across all professors and divide by the total number of professors. Therefore: Expected Frequency of A = (18+36+20)/3, B = (25+44+15)/3, C = (85+73+82)/3, D&F = (17+12+8)/3
03

Compute the Statistics for Chi-square test

Calculate the test statistic which is the sum of the squared difference between observed and expected grades (for each grade and professor), divided by the expected grades. The formula for the test statistic is \[X^2 = \sum \frac{(O_i - E_i)^2}{E_i}\] where \(O_i\) represents each observed frequency and \(E_i\) represents each expected frequency.
04

Compare With Critical Value

Find the critical value for a Chi-square distribution with degrees of freedom equal to (3-1)*(4-1) = 6 at the 2.5% significance level. Compare the computed test statistic with the critical value.
05

Make a Decision

If the test statistic is greater than the critical value, we reject the null hypothesis. This would imply that the grade distributions are not homogeneous among the professors. If the test statistic is less than or equal to the critical value, we do not reject the null hypothesis and conclude that the grade distributions are homogeneous.

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Key Concepts

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

Hypothesis Testing
The very foundation of hypothesis testing is centered around making an informed decision between two competing claims, known as hypotheses.
  • The **null hypothesis** \(H_0\) represents the status quo or the claim being tested. In this exercise, it asserts that the grade distributions among the three professors are homogeneous, meaning they are similar or not significantly different.
  • The **alternative hypothesis** \(H_1\) stands in opposition to the null, suggesting that the grade distributions are not homogeneous.
The aim of hypothesis testing is to use sample data to ascertain which hypothesis is more likely to be true. The Chi-square test is a statistical test used frequently in this regard, particularly for testing distributions of categorical data.
Significance Level
The significance level, denoted as \(\alpha\), is a critical part of hypothesis testing. It signifies the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it is the threshold for determining how much evidence we need before rejecting the null.
  • A smaller significance level (e.g., 2.5% in this exercise) implies a stricter criterion for rejecting the null hypothesis.
  • It helps us control the risk of making a Type I error, which is rejecting a true null hypothesis.
Choosing a significance level means deciding how much risk we're willing to take of claiming there's an effect or difference when, in fact, there isn't one. In this exercise, a 2.5% level is quite conservative, indicating a high level of confidence required to reject the null hypothesis.
Expected Frequency
Expected frequency is a pivotal concept in the Chi-square test. When you assume distributions are the same, you can predict what frequencies of each outcome should look like under the null hypothesis.
  • To calculate expected frequencies, you take the average count across all categories or groups. For example, if you want the expected frequency of grade "A" across three professors, sum their counts and divide by the number of professors.
  • This gives a benchmark to compare against the observed data, helping identify any significant differences between groups.
The more the observed data deviates from these expected frequencies, the less likely it is that the null hypothesis is true.
Homogeneity Test
A homogeneity test is a specific application of the Chi-square test aimed at determining if different populations (or groups) have the same distribution. In the given exercise, we're examining whether the grade distributions for professors Miller, Smith, and Moore are similar. Here's the process explained:
  • The test calculates a Chi-square statistic using observed and expected frequencies. This tells us how much the observed data (actual grades given by each professor) differs from what we would expect if distributions were indeed homogeneous.
  • The Chi-square statistic is then compared to a critical value to determine if the observed differences are statistically significant.
If the computed statistic exceeds the critical value, it implies that the observed variations are unlikely to have occurred by chance, thus we reject the null hypothesis of homogeneity. This means the grade distributions are not homogeneous.

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Most popular questions from this chapter

A sample of 18 observations selected from a normally distributed population produced a sample variance of \(4.6\). a. Write the null and alternative hypotheses to test whether the population variance is different from \(2.2\) b. Using \(\alpha=.05\), find the critical values of \(\chi^{2}\). Show the rejection and nonrejection regions on a chi-square distribution curye. c. Find the value of the test statistic \(\chi^{2}\). d. Using the \(5 \%\) significance level, will you reject the null hypothesis stated in part a?

Chance Corporation produces beauty products. Two years ago the quality control department at the company conducted a survey of users of one of the company's products. The survey revealed that \(53 \%\) of the users said the product was excellent, \(31 \%\) said it was satisfactory, \(7 \%\) said it was unsatisfactory, and \(9 \%\) had no opinion. Assume that these percentages were true for the population of all users of this product at that time. After this survey was conducted, the company redesigned this product. A recent survey of 800 users of the redesigned product conducted by the quality control department at the company showed that 495 of the users think the product is excellent, 255 think it is satisfactory, 35 think it is unsatisfactory, and 15 have no opinion. Is the percentage distribution of the opinions of users of the redesigned product different from the percentage distribution of users of this product before it was redesigned? Use \(\alpha=.025\).

Suppose that you have a two-way table with the following row and column totals. \begin{tabular}{l|cccc|c} \hline & & \multicolumn{3}{c|} { Variable 1 } & \\ \cline { 3 - 5 } \multicolumn{2}{l|} {} & A & B & C & Total \\ \hline & \(\mathbf{X}\) & & & & 120 \\ Variable 2 & \(\mathbf{Y}\) & & & & 205 \\ & \(\mathbf{Z}\) & & & & 175 \\ \hline & Total & 165 & 140 & 195 & 500 \\ \hline \end{tabular} The observed values in the cells must be counts, which are nonnegative integers. Calculate the expected counts for the cells under the assumption that the two variables are independent. Based on your calculations, explain why it is impossible for the test statistic to have a value of zero.

Find the value of \(\chi^{2}\) for 4 degrees of freedom and a. 005 area in the right tail of the chi-square distribution curve b. 05 area in the left tail of the chi-square distribution curve

The manufacturer of a certain brand of lightbulbs claims that the variance of the lives of these bulbs is 4200 square hours. A consumer agency took a random sample of 25 such bulbs and tested them. The variance of the lives of these bulbs was found to be 5200 square hours. Assume that the lives of all such bulbs are (approximately) normally distributed. a. Make the \(99 \%\) confidence intervals for the variance and standard deviation of the lives of all such bulbs. b. Test at the \(5 \%\) significance level whether the variance of such bulbs is different from 4200 square hours.
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