Question

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salaries of secondary school teachers. A random sample of the salaries of 30 elementary school teachers has a variance of 8324 and a random sample of the salaries of 30 secondary school teachers has a variance of \(2862 .\) At \(\alpha=0.05\) can the researcher conclude that the variation in the elementary school teachers' salaries is greater than the variation in the secondary school teachers' salaries? Use the \(P\) -value method.

Short answer

The variation in elementary school teachers' salaries is significantly greater than that of secondary school teachers.

Step-by-step solution

State the Hypotheses

In hypothesis testing, we first state the null and alternative hypotheses. - Null hypothesis (\( H_0 \)) : The variance of the elementary school teachers’ salaries is equal to that of the secondary school teachers’. Mathematically, \( H_0 : \sigma_1^2 = \sigma_2^2 \).- Alternative hypothesis (\( H_a \)) : The variance of the elementary school teachers’ salaries is greater than that of the secondary school teachers’. Mathematically, \( H_a : \sigma_1^2 > \sigma_2^2 \). The claim is in the alternative hypothesis, meaning the researcher claims that the variance of elementary school teachers' salaries is greater.

Determine the Critical Value

This problem involves comparing variances, so we use the F-distribution. The degrees of freedom for the numerator and denominator are calculated by \( df_1 = n_1 - 1 \) and \( df_2 = n_2 - 1 \), respectively. For both samples, \( df_1 = df_2 = 30 - 1 = 29 \). We are conducting a one-tailed test (right-tail) at \( \alpha = 0.05 \).Using an F-distribution table or calculator, the critical value for \( F \) with \( df_1 = 29 \) and \( df_2 = 29 \) at \( \alpha = 0.05 \) is approximately 1.855.

Compute the Test Value

We calculate the test statistic using the formula for the F-ratio: \[ F = \frac{s_1^2}{s_2^2} \]where \( s_1^2 \) is the variance of the elementary school teachers’ salaries and \( s_2^2 \) is the variance of the secondary school teachers’ salaries. Substituting the given values:\[ F = \frac{8324}{2862} \approx 2.908 \]

Make the Decision

Now, compare the calculated test value \( F = 2.908 \) with the critical value \( 1.855 \). Since \( 2.908 > 1.855 \), we reject the null hypothesis \( H_0 \) in favor of the alternative hypothesis \( H_a \). Thus, there is sufficient evidence to support the researcher's claim.

Summarize the Results

Based on the statistical test, we have sufficient evidence at \( \alpha = 0.05 \) to conclude that the variance in the salaries of elementary school teachers is greater than the variance in the salaries of secondary school teachers, supporting the researcher's claim.

Key concepts

F-distribution

The F-distribution, a fundamental concept in hypothesis testing, plays a crucial role when comparing two variances, especially in an analysis like this one. It is a ratio of two independent chi-squared distributions divided by their degrees of freedom. This unique distribution is often used to test the equality of variances from two different populations.

The shape of the F-distribution depends greatly on the degrees of freedom for both the numerator and denominator. In this exercise, each sample has 29 degrees of freedom, making the F-distribution the right tool for analysis. This method allows us to determine if the observed variances differ significantly.

Variance Comparison

Variance comparison is essential when evaluating variability between two datasets. In this context, variance reflects how much salaries deviate from the average. We have the variances of elementary school teachers (\(8324\)) and secondary school teachers (\(2862\)). The larger the variance, the greater the dispersion of data points.

The researcher's hypothesis that elementary school teachers have more salary variability poses a scientific question that variance comparison helps resolve. By calculating the F-ratio, we can objectively assess whether the difference in these variances is statistically significant or could be due to random sampling.

Critical Value

The critical value is a key element in hypothesis testing, acting as a threshold against which the calculated test statistic is compared. To find this value, one must consider the level of significance (\(\alpha = 0.05\)) and the degrees of freedom, in this case, 29 for both samples.

The F-distribution table provides this critical value, approximately \(1.855\). If our computed F-statistic exceeds this critical value, the variance difference is deemed significant enough to reject the null hypothesis. In this exercise, the test statistic \(2.908\) surpasses the critical value, indicating that the variability in salaries is indeed greater for elementary school teachers.

P-value Method

The P-value method in hypothesis testing offers an alternative to critical values. It provides the probability of observing test results as extreme as those given the null hypothesis is true. When the P-value is less than the significance level (\(\alpha = 0.05\)), we reject the null hypothesis.

In practice, a lower P-value suggests stronger evidence against the null hypothesis. By calculating the F-statistic and corresponding P-value, we concretely determine the outcome of the hypothesis test. For this exercise, although the exact P-value isn't mentioned, the conclusion remains the same: evidence supports the claim that elementary teachers' salary variability exceeds that of secondary teachers.