Question

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers in a large school district. The mean of the salaries of a random sample of 26 elementary school teachers is \(\$ 48,256,\) and the sample standard deviation is \(\$ 3,912.40 .\) The mean of the salaries of a random sample of 24 secondary school teachers is \(\$ 45,633\). The sample standard deviation is \(\$ 5533 .\) At \(\alpha=0.05,\) can it be concluded that the mean of the salaries of the elementary school teachers is greater than the mean of the salaries of the secondary school teachers? Use the \(P\) -value method.

Short answer

The p-value is compared against a alpha of 0.05; if less, reject H_0 and conclude elementary teacher salaries are higher.

Step-by-step solution

Define Hypotheses

In this step, define the null and alternative hypotheses for the study. The null hypothesis (H_0) states that there is no difference in the means of the salaries of elementary and secondary school teachers. The alternative hypothesis (H_1) states that the mean salary of elementary school teachers is greater than that of secondary school teachers.\[H_0: \mu_1 \leq \mu_2\]\[H_1: \mu_1 > \mu_2\]

Identify the Test Statistic

In order to compare the means, use a two-sample t-test for independent samples. The test statistic for this scenario when variances are not assumed equal is calculated using the formula for the two-sample t-test:\[t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]Where:\(\bar{x}_1 = 48,256\), \(s_1 = 3,912.40\), \(n_1 = 26\), \(\bar{x}_2 = 45,633\), \(s_2 = 5,533\), \(n_2 = 24\).

Calculate the Test Statistic Value

Plug the given values into the test statistic formula and calculate:\[t = \frac{48,256 - 45,633}{\sqrt{\frac{3,912.40^2}{26} + \frac{5,533^2}{24}}}\]Calculate the value inside the square root first, followed by solving for the t-value.

Find the Degrees of Freedom

Use the formula for degrees of freedom in a two-sample t-test with unequal variances:\[df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\]Calculate the effective degrees of freedom for the t-distribution.

Determine the P-value

With the calculated t-value and degrees of freedom, use a t-distribution table or software to find the p-value. This p-value indicates the probability of observing such a result if the null hypothesis is true.

Compare P-value with Significance Level

The significance level is \(\alpha = 0.05\). Compare the p-value found in Step 5 with \(\alpha = 0.05\). If the p-value is less than \(\alpha\), reject the null hypothesis.

Key concepts

Two-sample t-test

When comparing two groups to see if there is a significant difference between their means, the two-sample t-test is commonly used. This method helps us determine if the average values of two different groups are statistically different from one another. In the example given, we compare salary data from elementary and secondary school teachers.

• The two-sample t-test is appropriate because it compares the means of two independent samples.
• It does not assume that the groups have the same variance, which suits cases like ours, where the salary standard deviations differ.
• We use the sample means, standard deviations, and sizes to calculate the test statistic, which tells us how far apart these means are in standardized units.

Understanding and applying the two-sample t-test allows us to draw conclusions about populations based on sample data.

P-value method

The p-value method in hypothesis testing helps us decide whether to reject the null hypothesis. A p-value is the probability of observing test results at least as extreme as the observed results, assuming the null hypothesis is true.

• In our scenario, a low p-value (usually less than 0.05) indicates that the null hypothesis is likely false, leading us to support the alternative hypothesis.
• We calculate this by finding the probability that the calculated t-value deviates from zero by more than it actually does, based on the effective degrees of freedom.

If the computed p-value is lower than the predetermined significance level (alpha), we conclude that there is a significant difference between the two group means.

Independent samples

Samples are considered independent when the data from one sample does not influence or is not influenced by the data from the other. This is a critical assumption for the validity of a two-sample t-test.

• In the exercise, elementary and secondary teachers are examples of independent samples because each teacher’s salary in one group does not affect a teacher’s salary in the other group.
• This independence ensures that the statistical tests remain valid without the biases that can occur if samples were related.

Understanding the independence of samples ensures accurate analysis and reliable results when performing hypothesis tests.

Degrees of freedom

In hypothesis testing, degrees of freedom are values in the final calculation of a statistic that are free to vary. This concept is crucial in estimating the underlying variability of the data.

• For the two-sample t-test, degrees of freedom can be calculated using a special formula when variances are not assumed equal, also known as Welch's t-test.
• This involves calculating how variability within both independent samples contributes to the overall variability of the comparison.
• Accurate degrees of freedom are vital as they influence the t-distribution used for finding p-values.

Essentially, degrees of freedom give a measure of how many independent ways a dynamic system can move, and their correct computation is fundamental for statistical inference.