/*! 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 53 The correlation coefficient for ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 53

The correlation coefficient for the heights and weights of ten offensive backfield football players was determined to be \(r=.8261\) a. What percentage of the variation in weights was explained by the heights of the players? b. What percentage of the variation in heights was explained by the weights of the players? c. Is there sufficient evidence at the \(\alpha=.01\) level to claim that heights and weights are positively correlated? d. What is the attained significance level associated with the test performed in part (c)?

Short Answer

Expert verified
a) 68.24%, b) 68.24%, c) Yes, there is significant evidence, d) P-value is much less than 0.01.

Step by step solution

01

Understanding R-squared for Variation Percentage

The correlation coefficient \( r \) provides information about the strength of a linear relationship between two variables. To find the percentage of the variation explained, we calculate the square of \( r \), known as \( R^2 \). In this case, \( r = 0.8261 \). The formula is \( R^2 = r^2 \).
02

Calculate R-squared

Calculate \( R^2 \) by squaring the correlation coefficient: \[ R^2 = (0.8261)^2 = 0.6824. \] This means 68.24% of the variation in weights is explained by the heights.
03

Answer for Part (b)

Since \( R^2 \) measures the proportion of the total variation in one variable explained by another, the same percentage (68.24%) of the variation in heights is explained by the weights. Hence, 68.24% of the variation in heights is explained by weights.
04

Conducting Hypothesis Test for Correlation

For part (c), we perform a hypothesis test for correlation. **Null Hypothesis**: \( H_0: \rho = 0 \) (no correlation). **Alternative Hypothesis**: \( H_a: \rho > 0 \) (positive correlation). We use a t-test to check the significance of correlation with \( \alpha = 0.01 \). The t-statistic is calculated as \[ t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}}, \] where \( n \) is the sample size (10).
05

Calculate Test Statistic

Substitute the known values into the t-statistic formula. Here, \( n = 10 \), so: \[ t = \frac{0.8261 \times \sqrt{10-2}}{\sqrt{1-0.6824}} = \frac{0.8261 \times 2.828}{0.5599} \approx 5.048. \]
06

Determine Critical Value and Decision

For a right-tailed test at \( \alpha = 0.01 \) and 8 degrees of freedom (\( n-2 \)), the critical value from the t-distribution table is approximately 2.896. Since \( t = 5.048 \) is greater than 2.896, we reject the null hypothesis.
07

Conclusion for Correlation Test

There is sufficient evidence to claim that heights and weights are positively correlated at the \( \alpha = 0.01 \) significance level.
08

Attained Significance Level (P-value)

The attained significance level is the P-value corresponding to the calculated \( t \) statistic. Using statistical software or a t-table, find the P-value for \( t = 5.048 \) with 8 degrees of freedom. It is significantly less than 0.01, indicating strong evidence against \( H_0 \).

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.

R-squared
R-squared, also known as the coefficient of determination, is a crucial statistic used in regression analysis. It quantifies how well a predictive model represents the data. In simpler terms, it tells us what percentage of the total variation in the dependent variable can be explained by the independent variable, using the given linear model.

In the given problem, the correlation coefficient was calculated as \( r = 0.8261 \). To find the R-squared value, we square the correlation coefficient: \( R^2 = (0.8261)^2 = 0.6824 \).

This means that 68.24% of the variation in weights can be explained by the variation in heights. Similarly, the same percentage (68.24%) of variation in heights is explained by weights. This dual explanation is because R-squared is symmetric regarding these two variables in a simple linear regression.

With R-squared values:
  • High values close to 1 indicate strong explanatory power.
  • Low values near 0 suggest weak model performance.
Modifying the independent variables, refining model selection, or utilizing more sophisticated models can improve R-squared values.
Hypothesis Test
Hypothesis testing is a fundamental method in statistics used to decide whether there is enough evidence to support a particular hypothesis about a population parameter. In our scenario, we are testing if there is any significant positive correlation between heights and weights of football players.

The hypothesis test follows these steps:
  • **Null Hypothesis (\( H_0 \))**: States that there is no correlation, \( \rho = 0 \).
  • **Alternative Hypothesis (\( H_a \))**: Suggests a positive correlation, \( \rho > 0 \).
By calculating the t-statistic with the formula:
\[ t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \]
where \( n \) is the sample size, we can determine whether to reject \( H_0 \).

In this exercise, the calculated t-statistic is approximately 5.048. Given the critical value of 2.896 for a 0.01 significance level and 8 degrees of freedom, we discover that \( t = 5.048 \) is higher than 2.896. Hence, we reject the null hypothesis, suggesting there is significant evidence of a positive correlation between height and weight.
P-value
The P-value is a measure that helps us determine the strength of our results in hypothesis testing. It tells us the probability of observing the results as extreme as those observed in our sample data, assuming that the null hypothesis \( H_0 \) is true.

A small P-value indicates strong evidence against the null hypothesis, leading us to reject \( H_0 \). Conversely, a large P-value suggests there is not enough evidence to reject the null hypothesis.

For this correlation test, the obtained t-value was checked against t-distribution tables or statistical software to find the P-value. In this problem, the t-statistic of 5.048 yields a P-value significantly less than 0.01. This small P-value is critical because:
  • It indicates strong evidence against the null hypothesis of no correlation.
  • The result confirms the existence of a significant positive correlation.
The P-value thus helps to solidify our decision by quantifying the strength of evidence against the null hypothesis.

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

The data in the following table give the miles per gallon obtained by a test automobile when using gasolines of varying octane levels.$$\begin{array}{cc} \hline \text { Miles per Gallon }(y) & \text { Octane }(x) \\ \hline 13.0 & 89 \\\13.2 & 93 \\\13.0 & 87 \\\13.6 & 90 \\\13.3 & 89 \\\13.8 & 95 \\\14.1 & 100 \\\14.0 & 98\\\\\hline\end{array}$$a. Calculate the value of \(r\). b. Do the data provide sufficient evidence to indicate that octane level and miles per gallon are dependent? Give the attained significance level, and indicate your conclusion if you wish to implement an \(\alpha=.05\) level test.

Television advertising would ideally be aimed at exactly the audience that observes the ads. A study was conducted to determine the amount of time that individuals spend watching TV during evening prime-time hours. Twenty individuals were observed for a 1 -week period, and the average time spent watching TV per evening, \(Y\), was recorded for each. Four other bits of information were also recorded for each individual: \(x_{1}=\) age, \(x_{2}=\) education level, \(x_{3}=\) disposable income, and \(x_{4}=\) IQ. Consider the three models given below: Model I: $$Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{3}+\beta_{4} x_{4}+\varepsilon$$ Model II: $$Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\varepsilon$$ Model III: $$Y=\beta_{0}+\beta_{1} x_{1}+\beta_{2} x_{2}+\beta_{3} x_{1} x_{2}+\varepsilon$$ Are the following statements true or false? a. If Model I is fit, the estimate for \(\sigma^{2}\) is based on 16 df. b. If Model II is fit, we can perform a \(t\) test to determine whether \(x_{2}\) contributes to a better fit of the model to the data. c. If Models I and II are both fit, then \(\mathrm{SSE}_{\mathrm{I}} \leq \mathrm{SSE}_{\mathrm{II}}\) d. If Models I and II are fit, then \(\hat{\sigma}_{1}^{2} \leq \widehat{\sigma}_{\Pi}^{2}\). e. Model II is a reduction of model I. f. Models I and III can be compared using the complete/reduced model technique presented in Section 11.14.

Is the plant density of a species related to the altitude at which data are collected? Let \(Y\) denote the species density and \(X\) denote the altitude. A fit of a simple linear regression model using 14 observations yielded \(\hat{y}=21.6-7.79 x\) and \(r^{2}=.61\) a. What is the value of the correlation coefficient \(r ?\) b. What proportion of the variation in densities is explained by the linear model using altitude as the independent variable? c. Is there sufficient evidence at the \(\alpha=.05\) to indicate that plant densities decrease with an increase in altitude?

A study was conducted to determine the effects of sleep deprivation on subjects' ability to solve simple problems. The amount of sleep deprivation varied over \(8,12,16,20,\) and 24 hours without sleep. A total of ten subjects participated in the study, two at each sleep-deprivation level. After his or her specified sleep-deprivation period, each subject was administered a set of simple addition problems, and the number of errors was recorded. The results shown in the following table were obtained. $$\begin{array}{l|ccccc}\text { Number of Errors }(y) & 8,6 & 6,10 & 8,14 & 14,12 & 16,12 \\\\\hline \text { Number of Hours without Sleep }(x) & 8 & 12 & 16 & 20 & 24\end{array}$$ a. Find the least-squares line appropriate to these data. b. Plot the points and graph the least-squares line as a check on your calculations. c. Calculate \(S^{2}\).

In the biological and physical sciences, a common model for proportional growth over time is $$ E(Y)=1-e^{-\beta t} $$ where \(Y\) denotes a proportion and \(t\) denotes time. \(Y\) might represent the proportion of eggs that hatch, the proportion of an organism filled with diseased cells, the proportion of patients reacting to a drug, or the proportion of a liquid that has passed through a porous medium. With \(n\) observations of the form \(\left(y_{i}, t_{i}\right),\) outline how you would estimate and then form a confidence interval for \(\beta\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Decision Maths

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.