/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A sample of \(n=353\) college fa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 49

A sample of \(n=353\) college faculty members was obtained, and the values of \(x=\) teaching evaluation index and \(y=\) annual raise were determined ("Determination of Faculty Pay: An Agency Theory Perspective." Academy of Management Journal [1992]: \(921-955)\). The resulting value of \(r\) was .11 . Does there appear to be a linear association between these variables in the population from which the sample was selected? Carry out a test of hypothesis using a significance level of .05 . Does the conclusion surprise you? Explain.

Short Answer

Expert verified
The answer depends on the comparison of the calculated \(t\) value (from Step 2) with the critical \(t\). If the absolute value of the calculated \(t\) is above the critical \(t\), then there appears to be a linear association, otherwise, there doesn't appear to be a linear association. The conclusion of whether it surprises or not will largely depend on the initial assumptions about the relationship between evaluations and raises.

Step by step solution

01

Formulate the Hypotheses

For this problem, the null hypothesis (\(H_0\)) is that there is no linear association between teaching evaluation index and annual raise, i.e., the correlation coefficient in the population is 0. The alternative hypothesis (\(H_1\)) is that there is a linear association, i.e., the correlation coefficient in the population is different from 0. Mathematically, these hypotheses can be stated as follows: \(H_0: \rho = 0\) and \(H_1: \rho 鈮� 0\).
02

Calculate the Test Statistic

The test statistic for testing these hypotheses about the correlation coefficient is calculated following the formula: \[ t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \] Substituting the given values, the calculation would be: \[ t = \frac{0.11\sqrt{353-2}}{\sqrt{1-0.11^2}} \] After carrying out the above arithmetic, one can find the value of \(t\).
03

Compare to Critical Value

The observed \(t\) value should be compared to tabulated/critical \(t\) at \(df = n - 2\) degrees of freedom for a given level of significance (0.05 for this problem) to decide whether to reject or not reject \(H_0\). If the absolute value of the calculated \(t\) is greater than the critical \(t\), then \(H_0\) should be rejected, and if it's less than or equal to critical \(t\), then \(H_0\) should not be rejected.
04

Form a Conclusion

Based on the \(t\) comparison, a conclusion can be made regarding whether to reject or not reject the null hypothesis. Subsequently, it can be deduced whether or not there appears to be a linear association in the population from which the sample was selected.

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 91影视!

Key Concepts

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

Correlation Coefficient
The correlation coefficient, often denoted as r, is a statistical measure that calculates the strength and direction of a linear relationship between two variables. Its values range from -1 to +1, where +1 indicates a perfect positive linear correlation, -1 indicates a perfect negative linear correlation, and 0 indicates no linear correlation.

When dealing with real-world data, as in the given exercise, a correlation coefficient close to 0, like 0.11, suggests a very weak linear relationship between the teaching evaluation index and annual raise. The exercise's goal is to determine if this observed relationship is statistically significant or merely due to random chance.

The calculation of this coefficient in a hypothesis testing framework involves setting up null and alternative hypotheses. In this context, the null hypothesis states that there is no correlation (蟻 = 0), and the alternative hypothesis suggests a non-zero correlation exists in the population from which the sample was selected. This is a crucial step in determining the existence and strength of the relationship.
Linear Association
Linear association refers to the relationship between two variables that shows a straight-line pattern when plotted on a scatterplot. If the variables tend to increase together, the association is positive; if one variable tends to increase as the other decreases, the association is negative.

In the educational exercise, we aim to discover if such an association exists between faculty members' teaching evaluation index and their annual raise. To establish whether this association is present beyond chance, we calculate the correlation coefficient mentioned earlier and then proceed to test its statistical significance.

It鈥檚 important to remember that a linear association does not necessarily imply causation. Even if faculty members with higher teaching evaluations tend to get higher raises, other unaccounted factors might influence this observed trend. Therefore, our hypothesis test focuses on the presence of the association, not the causality of it.
Significance Level
The significance level, denoted by 伪, represents the probability of rejecting the null hypothesis when it is true, which is known as a Type I error. A common choice for 伪 is 0.05, indicating a 5% risk of concluding there is an effect when there is none.

In hypothesis testing, we compare our calculated test statistic against a critical value that corresponds to our chosen significance level. If our test statistic exceeds this critical value, we can reject the null hypothesis with confidence that the result is not due to random chance, provided that 伪 is at an acceptable threshold.

For our exercise, using a significance level of 0.05 means we are 95% confident that our decision to reject or fail to reject the null hypothesis is correct. The result may still surprise us, especially if we had prior expectations about the relationship between variables. However, statistical methods allow us to make objective decisions about such relationships, precisely as demonstrated in the step-by-step solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of \(n=61\) penguin burrows was selected, and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one ("Effects of Substrate on the Distribution of Magellanic Penguin Burrows," The Auk [1991]: \(923-933\) ). The equation of the least-squares line was \(\hat{y}=11.607-1.4187 x,\) and \(r^{2}=.386 .\) a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{\mathrm{e}}=2.35, \bar{x}=4.5,\) and \(\sum(x-\bar{x})^{2}=250,\) predict trail length when soil hardness is 6.0 in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0 ?\) Explain your reasoning

An experiment to study the relationship between \(x=\) time spent exercising (minutes) and \(y=\) amount of oxygen consumed during the exercise period resulted in the following summary statistics. \(n=20 \quad \sum x=50 \quad \sum y=16,705 \quad \sum x^{2}=150\) \(\sum y^{2}=14,194,231 \quad \sum x y=44,194\) a. Estimate the slope and \(y\) intercept of the population regression line. b. One sample observation on oxygen usage was 757 for a 2 -minute exercise period. What amount of oxygen consumption would you predict for this exercise period, and what is the corresponding residual? c. Compute a \(99 \%\) confidence interval for the average change in oxygen consumption associated with a 1 minute increase in exercise time.

In Exercise \(13.17,\) we considered a regression of \(y=\) oxygen consumption on \(x=\) time spent exercising. Summary quantities given there yield $$ \begin{array}{lll} n=20 & \bar{x}=2.50 & S_{x x}=25 \\ b=97.26 & a=592.10 & s_{e}=16.486 \end{array} $$ a. Calculate \(s_{a+b(2,0)}\), the estimated standard deviation of the statistic \(a+b(2.0)\). b. Without any further calculation, what is \(s_{d+k 3,0\\}}\) and what reasoning did you use to obtain it? c. Calculate the estimated standard deviation of the statistic \(a+b(2.8)\) d. For what value \(x^{*}\) is the estimated standard deviation of \(a+b x^{*}\) smallest, and why?

Exercise 13.21 gave data on \(x=\) nerve firing frequency and \(y=\) pleasantness rating when nerves were stimulated by a light brushing stoke on the forearm. The \(x\) values and the corresponding residuals from a simple linear regression are as follows: a. Construct a standardized residual plot. Does the plot exhibit any unusual features? b. A normal probability plot of the standardized residuals follows. Based on this plot, do you think it is reasonable to assume that the error distribution is approximately normal? Explain.

In some studies, an investigator has \(n(x, y)\) pairs sampled from one population and \(m(x, y)\) pairs from a second population. Let \(\beta\) and \(\beta^{\prime}\) denote the slopes of the first and second population lines, respectively, and let \(b\) and \(b^{\prime}\) denote the estimated slopes calculated from the first and second samples, respectively. The investigator may then wish to test the null hypothesis \(H_{0}: \beta-\beta^{\prime}=0\) (that is, \(\beta=\beta^{\prime}\) ) against an appropriate alternative hypothesis. Suppose that \(\sigma^{2}\), the variance about the population line, is the same for both populations. Then this common variance can be estimated by $$ s^{2}=\mathrm{SSR} $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.