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Teacher Salaries 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 dollar 48,256, and the sample standard deviation is dollar 3,912.40 . The mean of the salaries of a random sample of 24 secondary school teachers is dollar 45,633 . The sample standard deviation is dollar 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

Expert verified
Yes, there is enough evidence to conclude that elementary school teachers earn more.

Step by step solution

01

State the Hypotheses

We begin by defining the null and alternative hypotheses for a one-tailed test. The null hypothesis \(H_0\) is that the mean salary of elementary school teachers is less than or equal to the mean salary of secondary school teachers. The alternative hypothesis \(H_a\) is that the mean salary of elementary school teachers is greater than the mean salary of secondary school teachers. Mathematically, this is represented as: \(H_0: \mu_1 \leq \mu_2\) and \(H_a: \mu_1 > \mu_2\), where \(\mu_1\) is the mean salary of elementary school teachers, and \(\mu_2\) is the mean salary of secondary school teachers.
02

Gather Sample Data and Calculate the Test Statistic

We have the sample means, sample standard deviations, and sample sizes: \(\bar{x}_1 = 48256\), \(s_1 = 3912.40\), \(n_1 = 26\) for elementary teachers, and \(\bar{x}_2 = 45633\), \(s_2 = 5533\), \(n_2 = 24\) for secondary teachers. We will use a two-sample t-test for the means. The test statistic \(t\) is calculated using the formula:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute the values:\[ t = \frac{48256 - 45633}{\sqrt{\frac{3912.40^2}{26} + \frac{5533^2}{24}}} \]
03

Simplify the Test Statistic Equation

First, calculate the individual variances: \(s_1^2\) and \(s_2^2\), then plug those into the denominator:\[ \text{Variance for elementary: } \frac{3912.40^2}{26} \approx 588152.3 \ \text{Variance for secondary: } \frac{5533^2}{24} \approx 1278739.34 \]Therefore, the denominator evaluates to:\[ \sqrt{588152.3 + 1278739.34} \]Next, compute the difference in means, \(2583\), and complete the calculation of \(t\).
04

Calculate the P-Value

Using the computed test statistic, we compare it with the t-distribution for the given degrees of freedom. Estimate the degrees of freedom \(df\) using the formula for unequal variances (Welch's t-test):\[ df \approx \left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2 \, / \, \left(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}\right) \]This simplifies to approximately 44.81, rounding down gives 44. Use a t-distribution table or calculator to find the P-value for this t-score with 44 degrees of freedom.
05

Compare the P-Value with Significance Level

Compare the calculated P-value to the significance level of \(\alpha = 0.05\). If the P-value is less than 0.05, we reject the null hypothesis \(H_0\) and conclude that there is significant evidence that the mean salary of elementary school teachers is greater than that of secondary school teachers.

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Key Concepts

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

Two-Sample T-Test
The two-sample t-test is a powerful statistical tool that allows us to compare the means of two independent groups to determine if there is a statistically significant difference between them. In this context, we are using it to compare the average salaries of elementary and secondary school teachers. By conducting this test, we aim to determine if elementary school teachers, on average, earn more than their secondary school counterparts.

To proceed, we first define our hypotheses. The null hypothesis ( H_0 ) assumes that the means are equal, or in our case, that the mean salary of elementary school teachers is less than or equal to that of secondary school teachers. The alternative hypothesis ( H_a ) posits that the mean salary of elementary school teachers is greater. This setup indicates a one-tailed test because we are looking for a difference in one specific direction.
  • Calculate the test statistic using sample means, standard deviations, and sample sizes from both teacher groups.
  • Use the formula for the t-statistic to determine if the observed difference in means is statistically significant.
When applying this method, it's crucial to ensure that the underlying assumptions of the t-test are met. These include the assumptions of normality and that the samples have similar variances, though the latter can be adjusted for through methods like Welch's t-test if the variances differ considerably.
P-Value Method
The P-value method is a key component of hypothesis testing, allowing researchers to make informed decisions based on statistical evidence. In simple terms, the P-value quantifies the strength of evidence against the null hypothesis. It tells us how likely we are to observe the data we saw if the null hypothesis were true.

Here's how it works with our educational salaries scenario. After computing the test statistic from the two-sample t-test, we determine the P-value by considering the t-distribution. Ideally, we compare this P-value to a significance level, denoted by \( \alpha \), set at 0.05 for many studies.
  • If the P-value is less than \( \alpha \), we reject the null hypothesis.
  • This suggests that there is strong evidence to support the alternative hypothesis, which claims higher average salaries for elementary teachers.
The beauty of the P-value method lies in its simplicity and directness, offering a clear threshold to guide decision-making in hypothesis testing.
Mean Comparison
Understanding mean comparison is central in our analysis of teacher salaries. By examining the average salaries of two groups (elementary and secondary teachers), we can gain insights into salary structure differences within educational institutions.

To perform a mean comparison, we first gather the sample means and standard deviations for each group. The sample mean is simply the average salary within each sample group, providing an estimate of the central tendency for that dataset.

In our case study:
  • For elementary teachers, the sample mean salary is \( \\(48,256 \).
  • For secondary teachers, it's \( \\)45,633 \).
We then assess whether the difference of \( \$2,583 \) between these means is statistically significant through the t-test and P-value, providing a robust conclusion beyond mere observational comparison.
Educational Salaries
Analyzing educational salaries is a significant part of understanding broader economic and social dynamics within educational systems. By focusing on the salaries of teachers at different educational levels, we can glean important insights into career incentives, funding allocations, and policy effectiveness.

In this analysis, we explored whether elementary school teachers earn, on average, more than secondary school teachers. Such findings have practical implications for stakeholders, from policymakers to educational institutions aiming to retain talent.
  • Higher salaries could indicate more value placed on early education professionals.
  • Salary differences might reflect supply and demand dynamics within localized educational job markets.
Understanding these salary patterns can drive improvements in workforce management and resource distribution across schools, ultimately impacting educational outcomes.

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Most popular questions from this chapter

For Exercises 9 through \(24,\) 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. Museum Attendance A metropolitan children's museum open year-round wants to see if the variance in daily attendance differs between the summer and winter months. Random samples of 30 days each were selected and showed that in the winter months, the sample mean daily attendance was 300 with a standard deviation of \(52,\) and the sample mean daily attendance for the summer months was 280 with a standard deviation of \(65 .\) At \(\alpha=0.05,\) can we conclude a difference in variances?

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Leisure Time In a sample of \(150 \mathrm{men}, 132\) said that they had less leisure time today than they had 10 years ago. In a random sample of 250 women, 240 women said that they had less leisure time than they had 10 years ago. At \(\alpha=0.10,\) is there a difference in the proportions? Find the \(90 \%\) confidence interval for the difference of the two proportions. Does the confidence interval contain \(0 ?\) Give a reason why this information would be of interest to a researcher.

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Obstacle Course Times An obstacle course was set up on a campus, and 8 randomly selected volunteers were given a chance to complete it while they were being timed. They then sampled a new energy drink and were given the opportunity to run the course again. The "before" and "after" times in seconds are shown. Is there sufficient evidence at \(\alpha=0.05\) to conclude that the students did better the second time? Discuss possible reasons for your results. $$ \begin{array}{l|ccccccc}{\text { Student }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} \\ \hline \text { Before } & {67} & {72} & {80} & {70} & {78} & {82} & {69} & {75} \\ \hline \text { After } & {68} & {70} & {76} & {65} & {75} & {78} & {65} & {68}\end{array} $$

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Overweight Dogs A veterinary nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Six overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? $$ \begin{array}{l|cccccc}{\text { Before }} & {42} & {53} & {48} & {65} & {40} & {52} \\ \hline \text { After } & {39} & {45} & {40} & {58} & {42} & {47}\end{array} $$

Commuting Times for College Students The mean travel time to work for Americans is 25.3 minutes. An employment agency wanted to test the mean commuting times for college graduates and those with only some college. Thirty-five college graduates spent a mean time of 40.5 minutes commuting to work with a population variance of 67.24 . Thirty workers who had completed some college had a mean commuting time of 34.8 minutes with a population variance of \(39.69 .\) At the 0.05 level of significance, can a difference in means be concluded?

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