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91Ó°ÊÓ

The grades in a statistics course for a particular semester were as follows: \begin{tabular}{l|lllll} Grade & \(\mathrm{A}\) & \(\mathrm{B}\) & \(\mathrm{C}\) & \(\mathrm{D}\) & \(\mathrm{F}\) \\ \hline\(/\) & 14 & 18 & 32 & 20 & 16 \end{tabular} Test the hypothesis, at the 0.05 level of significance, that the distribution of grades is uniform.

Short Answer

Expert verified
Based on the Chi-Square Goodness-of-Fit Test, the computed test statistic (2.6) is less than the critical value (9.49) at the .05 significance level with 4 degrees of freedom. Therefore, there is not enough evidence to reject the Null Hypothesis. Hence, it can be concluded that the distribution of grades in the statistics course follows a uniform distribution.

Step by step solution

01

Identify Given Data

It's mentioned in the problem that the grades given were as follows: A = 14, B = 18, C = 32, D = 20, F = 16. The total number of students, \(N\), is the sum, 100.
02

Defining Hypotheses

The null hypothesis (\(H_0\)) assumes that the distribution of grades is uniform. The alternative hypothesis (\(H_a\)) assumes that the distribution is not uniform.
03

Compute Expected Counts

In a uniform distribution, all classes are equally likely. Thus, the expected count for each grade is \( \frac{N}{5} = \frac{100}{5} = 20 \), where 5 is the number of different grades.
04

Compute Test Statistic

The test statistic for a test of uniform goodness-of-fit is calculated by the chi-square formula \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \(O_i\) are observed counts and \(E_i\) are expected counts. Computing the test statistic with the provided data, we get \( \chi^2 = \frac{(14-20)^2}{20} + \frac{(18-20)^2}{20} + \frac{(32-20)^2}{20} + \frac{(20-20)^2}{20} + \frac{(16-20)^2}{20} = 2.6. \)
05

Compute Degrees of Freedom and Compare to Critical Value

Degrees of Freedom (DF) for this test is calculated as \(k - 1 = 5 - 1 = 4\), where \(k\) is the number of categories or classes. Looking into the Chi-Square table, at the .05 significance level (alpha) with 4 degrees of freedom, the critical value is approximately 9.49. Since the computed test statistic (2.6) is less than the critical value (9.49), there is not enough evidence to reject the Null Hypothesis.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population parameter based on sample data. The process involves setting up two opposing hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis represents a default position that there is no effect or no difference, while the alternative hypothesis represents what we are trying to find evidence for. In our exercise, the null hypothesis is that the distribution of grades is uniform – implying that each grade is equally likely to be awarded. The alternative hypothesis is that the grades are not uniformly distributed. To determine which hypothesis the data supports, we calculate a test statistic that measures how well the observed data fits the expected distribution under the null hypothesis. If this test statistic is extreme enough (compared to a threshold known as a critical value), we may reject the null hypothesis in favor of the alternative.
Uniform Distribution
A uniform distribution is a type of probability distribution where all outcomes are equally likely. In a discrete uniform distribution, which pertains to our example with grades, each category or class (A, B, C, D, or F) has the same probability of occurring because there are a finite number of outcomes. This means if the grades are indeed uniformly distributed, we would expect to see the same number of A's, B's, C's, D's, and F's. Since the total number of students is 100, we would expect, under the null hypothesis of uniformity, 20 students to receive each grade. Understanding uniform distribution helps us set our expected counts, which are critical for calculating the chi-square statistic.
Expected Counts
In the context of the chi-square goodness-of-fit test, expected counts refer to the number of observations we would expect in each category if the null hypothesis were true. They serve as a benchmark to compare against actual observed counts. We calculate expected counts based on the assumptions of our null hypothesis. In our example, since the null hypothesis is that grades are awarded uniformly, the expected count for each grade is the total number of students divided by the number of grade categories, which is 20. It is important to ensure that expected counts are sufficiently large, typically 5 or more, for the chi-square test to be valid. Small expected counts can lead to inaccurate chi-square values and can invalidate the results of the test.
Degrees of Freedom
Degrees of freedom (DF) in statistics indicate the number of values that are free to vary when calculating a statistic. In a chi-square goodness-of-fit test, the degrees of freedom are determined by the number of categories minus one (\( k - 1 \)). This subtraction accounts for the constraint placed on the categories by the fixed total number of observations. In this particular case, with five grade categories (A, B, C, D, and F), there are four degrees of freedom (\(5 - 1 = 4\)). The degrees of freedom are used to determine the critical value from the chi-square distribution table. This critical value is then compared with the test statistic to decide whether to reject the null hypothesis. The number of degrees of freedom reflects the number of independent ways the data can vary while still conforming to a given distribution, which in this scenario is the hypothesized uniform distribution of grades.

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