Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Test the correlation, as indicated. Show all details of the test. Test for evidence of a linear association; \(r=0.28 ; n=10\)

Short Answer

Expert verified
The answer is found by testing the null hypothesis of no linear association. This process involves the calculation of a test statistic and comparison of it with a critical value from a t-distribution table. Depending on the result, a conclusion is drawn whether there is evidence for a linear association between the variables or not.

Step by step solution

01

Null and Alternative Hypotheses

The null hypothesis is \(H_0: \rho = 0\), meaning there is no linear association between the variables. The alternative hypothesis is \(H_a: \rho \neq 0\), meaning there is a linear association between the variables.
02

Calculate the Test Statistic

The test statistic \(t\) for testing the hypothesis about the population correlation coefficient can be calculated as follows: \(t = r \sqrt{\frac{n-2}{1-r^2}}\). Substituting the values given in the problem, we get \(t = 0.28 \sqrt{\frac{10-2}{1-0.28^2}}\).
03

Find the Critical Value

The critical value of \(t\) for a chosen significance level (for instance, 0.05) can be found from a t-distribution table. The degrees of freedom will be \(n-2\), equal to 8 in this case.
04

Decision Rule

Compare the calculated value \(t\) from Step 2 with the critical value from Step 3. If the absolute value of calculated \(t\) is less than the critical value, we do not reject \(H_0\). If the absolute value of calculated \(t\) is greater than the critical value, we reject \(H_0\).
05

Conclusion

If we reject \(H_0\), the conclusion is that there is evidence for a linear association between the variables (with the given level of significance). If we do not reject \(H_0\), the conclusion is that there is no evidence for a linear association between the variables (at the chosen significance level).

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 Vaia!

Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is fundamental in hypothesis testing. In the context of correlation, the null hypothesis, denoted as \( H_0 \), usually states that there is no relationship between the two variables we’re interested in; that is, the correlation coefficient \( \rho \) is zero. This is the default position that assumes no effect or association until evidence suggests otherwise.

When a researcher sets out to test the correlation, the null hypothesis serves as a starting point for statistical significance testing. If later analyses suggest that this hypothesis is unlikely, we might have grounds to consider the alternative hypothesis as more plausible.
Alternative Hypothesis
The alternative hypothesis, \( H_a \), is the counterpart to the null hypothesis and represents what the researcher aims to support. Regarding correlation hypothesis testing, the alternative hypothesis suggests that a non-zero correlation exists between the two variables – that is, there is some degree of linear association.

In our exercise, the alternative hypothesis proposed a specific type of non-zero relationship, \( \rho eq 0 \), which means that any non-zero correlation would support \( H_a \). The alternative hypothesis is what we hope to conclude given sufficient evidence from our sample data.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It’s a crucial part of the puzzle as it helps us determine how far our sample statistic lies from the null hypothesis.

In testing for correlation, we use a particular formula for our test statistic (\( t \)) that incorporates the sample correlation coefficient (\( r \)) and the sample size (\( n \)). This formula enables us to translate our sample data into a single, comparable value. The result, when compared against values in a reference distribution, can suggest whether the observed data is unusual under the assumption of the null hypothesis.
T-Distribution
The t-distribution is a probability distribution that arises when estimating the mean of a normally distributed population in situations where the sample size is small and population standard deviation is unknown. It plays a vital role when conducting tests on samples where the normal distribution can't be applied confidently.

The shape of the t-distribution is similar to a normal distribution but has heavier tails, which means it is more prone to producing values that fall far from its mean. This property makes it useful for hypothesis testing, especially for smaller sample sizes, as it gives more conservative estimates and helps mitigate the impact of outliers or small samples.
Degrees of Freedom
Degrees of freedom are a concept tied to the t-distribution and many other statistical calculations that involve some form of variability. They represent the number of independent values that can vary in an analysis without breaking any constraints.

In our correlation test, the degrees of freedom (df) are calculated as the sample size minus two (\( n - 2 \)). The rationale behind this is that two parameters (mean and standard deviation) have been estimated from our data and are used in the computation of the test statistic, thus reducing the number of 'free' pieces of information by two. Degrees of freedom are a vital component in determining the appropriate values from the t-distribution table for hypothesis testing.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold that the researcher sets to decide when to reject the null hypothesis. It is the probability of making a Type I error, which is the error of incorrectly rejecting a true null hypothesis (a false positive).

Common significance levels include 0.05, 0.01, and 0.10, which correspond to a 5%, 1%, and 10% risk of a Type I error, respectively. In the example exercise, a significance level of 0.05 could be used to compare the observed test statistic to critical values from the t-distribution, which would help us make a decision about the hypotheses.
Linear Association
Linear association is a measure of the strength and direction of a relationship that can be described with a straight line. In our context, it refers to the relationship between two variables, measured by the correlation coefficient (\( r \)).

The value of \( r \) can range from -1 to 1, where -1 indicates a perfect negative linear relationship, 1 indicates a perfect positive linear relationship, and 0 suggests no linear association. In our exercise, we use hypothesis testing to determine if the observed value of \( r = 0.28 \) for our sample size of \( n = 10 \) provides evidence of a linear association in the population from where the sample was drawn.

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

How well does a student's Verbal SAT score (on an 800 -point scale) predict future college grade point average (on a four-point scale)? Computer output for this regression analysis is shown, using the data in StudentSurvey: The regression equation is \(\mathrm{GPA}=2.03+0.00189\) VerbalSAT Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 6.8029 & 6.8029 & 48.84 & 0.000 \\ \text { Residual Error } & 343 & 47.7760 & 0.1393 & & \\ \text { Total } & 344 & 54.5788 & & & \end{array}\) (a) What is the predicted grade point average of a student who receives a 550 on the Verbal SAT exam? (b) Use the information in the ANOVA table to determine the number of students included in the dataset. (c) Use the information in the ANOVA table to compute and interpret \(R^{2}\). (d) Is the linear model effective at predicting grade point average? Use information from the computer output and state the conclusion in context.

We give computer output with two regression intervals and information about the percent of calories eaten during the day. Interpret each of the intervals in the context of this data situation. (a) The \(95 \%\) confidence interval for the mean response (b) The \(95 \%\) prediction interval for the response The intervals given are for mice that eat \(10 \%\) of their calories during the day: DayPci 1.164 \(\begin{array}{rr}95 \% \mathrm{Cl} & 95 \% \mathrm{P} 1 \\\ (-0.013,4.783) & (-2.797,7.568)\end{array}\) 85 10.0 3888 2:

Exercises 9.66 and 9.67 refer to the regression line (given in Exercise 9.46 ): Cognition \(=102.3-3.34 \cdot\) Years using years playing football to predict the score on a cognition test. In each exercise, (a) Find the predicted cognition score for that case. (b) Two intervals are shown: one is a \(95 \%\) confidence interval for the mean response and the other is a \(95 \%\) prediction interval for the response. Which is which? A person who has played football for 8 years. \(\begin{array}{ll}\text { Interval I: }(22.7,128.5) & \text { Interval II: }(63.4,87.8)\end{array}\)

Show some computer output for fitting simple linear models. State the value of the sample slope for each model and give the null and alternative hypotheses for testing if the slope in the population is different from zero. Identify the p-value and use it (and a \(5 \%\) significance level) to make a clear conclusion about the effectiveness of the model.$$ \begin{array}{lrrrr} \text { Coefficients: } & \text { Estimate } & \text { Std.Error } & \mathrm{t} \text { value } & \operatorname{Pr}(>|\mathrm{t}|) \\ \text { (Intercept) } & 807.79 & 87.78 & 9.30 & 0.000 \\ \mathrm{~A} & -3.659 & 1.199 & -3.05 & 0.006 \end{array} $$

Use the computer output (from different computer packages) to estimate the intercept \(\beta_{0},\) the slope \(\beta_{1},\) and to give the equation for the least squares line for the sample. Assume the response variable is \(Y\) in each case. $$ \begin{aligned} &\text { The regression equation is }\\\ &Y=808-3.66 \mathrm{~A}\\\ &\begin{array}{lrrrr} \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 807.79 & 87.78 & 9.20 & 0.000 \\ \text { A } & -3.659 & 1.199 & -3.05 & 0.006 \end{array} \end{aligned} $$

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free