91Ó°ÊÓ

How do the average weekly incomes of electricians and carpenters compare? A random sample of 17 regions in the United States gave the following information about average weekly income (in dollars) (Reference: U.S. Department of $$\begin{array}{|l|ccccccccc} \hline \text { Region } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text { Electricians } & 461 & 713 & 593 & 468 & 730 & 690 & 740 & 572 & 805 \\ \hline \text { Carpenters } & 540 & 812 & 512 & 473 & 686 & 507 & 785 & 657 & 475 \\ \hline \\ \hline \text { Region } & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ \hline \text { Electricians } & 593 & 593 & 700 & 572 & 863 & 599 & 596 & 653 \\\ \hline \text { Carpenters } & 485 & 646 & 675 & 382 & 819 & 600 & 559 & 501 \\\ \hline \end{array}$$ Does this information indicate a difference (either way) in the average weekly incomes of electricians compared to those of carpenters? Use a \(5 \%\) level of significance.

Step by step solution

01

State Hypotheses

Define the null and alternative hypotheses. - Null Hypothesis (\(H_0\)): The average weekly income of electricians is equal to that of carpenters. \(H_0: \mu_E = \mu_C\) - Alternative Hypothesis (\(H_1\)): The average weekly income of electricians is not equal to that of carpenters. \(H_1: \mu_E eq \mu_C\).

02

Calculate Mean Incomes

Compute the mean income for electricians and carpenters. - Electricians' average: \( \frac{461+713+593+468+730+690+740+572+805+593+593+700+572+863+599+596+653}{17} \) - Carpenters' average: \( \frac{540+812+512+473+686+507+785+657+475+485+646+675+382+819+600+559+501}{17} \).

03

Calculate Standard Deviations

Find standard deviations for each group to understand income variability. 1. Calculate variance by finding the average of the squared differences from the mean for each group. 2. Take the square root of the variance to find the standard deviation for electricians and carpenters.

04

Conduct a t-Test for Independent Samples

We perform a t-test as follows:1. Calculate the t-statistic using the formula: \[t = \frac{\bar{x}_E - \bar{x}_C}{\sqrt{\frac{s_E^2}{n_E} + \frac{s_C^2}{n_C}}}\]where \( \bar{x}_E \) and \( \bar{x}_C \) are the sample means, \( s_E \) and \( s_C \) are the standard deviations, and \( n_E \) and \( n_C \) are the sample sizes (both 17).2. Determine the degrees of freedom and use a t-distribution table or software to find the critical t-value for a 5% significance level.

05

Make a Decision

Compare the calculated t-statistic to the critical t-value: - If the absolute value of the t-statistic is greater than the critical t-value, reject the null hypothesis. - If the absolute value of the t-statistic is less than or equal to the critical t-value, do not reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Null Hypothesis

When conducting hypothesis testing, the null hypothesis is the starting point. It represents a statement of no effect or no difference. In the context of our problem, the null hypothesis (H_0) posits that there is no difference in the average weekly income between electricians and carpenters.
The null hypothesis can be mathematically represented as \(H_0: \mu_E = \mu_C\). This means any observed differences in the sample data are due to random fluctuations, and not a genuine difference.
Why do we start with the null hypothesis? It's because the null hypothesis allows us to assess if there's enough evidence to support a contrary claim. We will either reject or fail to reject this null hypothesis based on our statistical analysis î‚• which in this case is the t-test for independent samples.

Alternative Hypothesis

The alternative hypothesis is what you might think of as the opposite of the null hypothesis. Here, it claims that there is a real difference in average weekly incomes between electricians and carpenters.
In symbols, the alternative hypothesis (H_1) is expressed as \(H_1: \mu_E eq \mu_C\). This suggests that the two groups do not have the same average income, meaning the differences observed are not by chance.
Embracing the alternative hypothesis requires compelling evidence revealed through statistical testing.

• If the data provides enough evidence against the null hypothesis, we accept the alternative hypothesis.
• Testing the alternative hypothesis involves setting a significance level, here it's 5%, which is the probability threshold of making a Type I error  rejecting the null hypothesis when it is indeed true.

t-test for Independent Samples

The t-test for independent samples is crucial for comparing the means of two distinct groups to see if there's a significant difference. In our exercise, it's used to compare electricians' and carpenters' incomes.
This test requires several calculations:

• Firstly, compute the means and standard deviations for both groups (electricians and carpenters).
• Next, the t-statistic is calculated using the formula: \[t = \frac{\bar{x}_E - \bar{x}_C}{\sqrt{\frac{s_E^2}{n_E} + \frac{s_C^2}{n_C}}}\]where \(\bar{x}_E\) and \(\bar{x}_C\) are the mean incomes, \(s_E\) and \(s_C\) are the standard deviations, and \(n_E\) and \(n_C\) are sample sizes.
• We then compare this t-statistic to a critical value determined by the degrees of freedom and the chosen significance level (5% in this case).

A high absolute value of t-statistic, exceeding the critical value, leads to the rejection of the null hypothesis, indicating that average incomes differ significantly between the two professions. Otherwise, we would not reject the null hypothesis, suggesting no statistically significant difference exists.