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

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. \(\begin{array}{ll}\text { Interval I: }(22.7,128.5) & \text { Interval II: }(63.4,87.8)\end{array}\)

Short Answer

Expert verified
The predicted cognition score for a person who has played football for 8 years is 75.58. Interval II (63.4, 87.8) is the 95% prediction interval for the response and interval I (22.7, 128.5) is the 95% confidence interval for the mean response.

Step by step solution

01

Calculate the predicted cognition score

Substitute the value of 8 years in the given linear regression equation: \(Cognition = 102.3 - 3.34 * Years\). After calculation, the predicted cognition score will be \(Cognition = 102.3 - 3.34 * 8 = 75.58.\)
02

Identify the confidence and prediction intervals

Given the intervals I: (22.7,128.5) and II: (63.4,87.8). Confidence interval represents the range within which the average cognition scores for people who have played football for about 8 years would fall. Prediction interval represents the range within which an individual player's cognition score is likely to fall. Since the predicted score 75.58 falls within the range of interval II, interval II should be the 95% prediction interval and interval I should be the 95% confidence interval.

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.

Confidence Interval
Understanding the confidence interval (CI) is crucial in statistics, especially in regression analysis. Imagine we are trying to estimate the average cognitive ability score for individuals who have played football for a certain number of years. The CI gives us a range that we believe contains the true average of the population's cognitive scores, with a specific level of confidence, commonly 95%.

For example, in the exercise, the 95% CI for the mean response, given a person has played football for 8 years, is Interval I: (22.7, 128.5). This interval suggests that we can be 95% confident that the true average cognitive score for all individuals who played for 8 years falls between 22.7 and 128.5. Here's an important point to remember: A CI concerns the average (mean) response for a group, not an individual prediction.

CIs are generated by considering the sample data, the variability of the data, and the size of the sample. If our CI is narrow, it indicates a more precise estimate of the population mean, while a wider CI suggests less precision. In practical terms, the exercise tells us that our linear regression model, based on sampled data, allows us to make somewhat broad but statistically supported estimates about the average cognitive score.
Prediction Interval
The prediction interval (PI) is slightly different from the confidence interval and addresses a distinct question. While CI is about the mean score, PI gives us a range where we can expect to find the actual score of an individual with a certain degree of certainty, again often 95%.

In the context of our exercise, Interval II: (63.4, 87.8) is a 95% prediction interval. This suggests that a specific individual who has played football for 8 years is likely to have a cognitive score between 63.4 and 87.8, with 95% confidence. The narrower range, compared to CI, indicates a higher precision for individual predictions, but it's still important to recognize the inherent uncertainty in predicting individual outcomes.

Creating a PI includes additional variability compared to the CI, considering individual differences. It’s broader than a CI for the same data because it attempts to account for the variability between individual observations and not just the average trend. Essentially, PI helps us manage expectations about how much a specific score might differ from the predicted average.
Regression Analysis
Regression analysis is a cornerstone of statistical modeling, allowing us to understand relationships between variables. In our exercise, we use linear regression, which is one of the simplest forms of regression analysis. It models the relationship between two variables by fitting a linear equation to observed data. One variable is considered to be an explanatory variable (in this case, years playing football), and the other is the dependent variable (cognition score).

The equation presented, Cognition = 102.3 - 3.34 * Years, describes how the cognition score is expected to change with each additional year of playing football. The coefficients (102.3 and -3.34) were derived from the data and tell us the starting score (intercept) and the change in score for each year of playing (slope).

Regression analysis helps in making predictions and also in understanding the strength and nature of the relationship. For instance, the negative coefficient of years played (-3.34) suggests a negative relationship; as the years playing football increase, the model predicts that the cognition score will likely decrease. However, it's essential to bear in mind that correlation does not imply causation, and other factors could be influencing these scores.

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

Use this information to fill in all values in an analysis of variance for regression table as shown. $$ \begin{array}{|l|l|l|l|l|l|} \hline \text { Source } & \text { df } & \text { SS } & \text { MS } & \text { F-statistic } & \text { p-value } \\ \hline \text { Model } & & & & & \\ \hline \text { Error } & & & & & \\ \hline \text { Total } & & & & & \\ \hline \end{array} $$ SSModel \(=800\) with SSTotal \(=5820\) and a sample size of \(n=40\)

Predicting Re-Election Margin Data 2.9 on page 106 introduces data on the approval rating of an incumbent US president and the margin of victory or defeat in the subsequent election (where negative numbers indicate the margin by which the incumbent president lost the re-election campaign). The data are reproduced in Table 9.5 and are available in ElectionMargin. Computer output for summary statistics for the two variables and for a regression model to predict the margin of victory or defeat from the approval rating is shown: The regression equation is Margin \(=-36.76+0.839\) Approval \(\begin{array}{lrrrr}\text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \text { Constant } & -36.76 & 8.34 & -4.41 & 0.001 \\ \text { Approval } & 0.839 & 0.155 & 5.43 & 0.000 \\\ S=5.66054 & R-S q=74.64 \% & R-S q(a d j) & =72.10 \%\end{array}\) \(\begin{array}{lrrrrr}\text { Analysis of Variance } & & & & & \\ \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 943.0 & 943.04 & 29.43 & 0.000 \\ \text { Residual Error } & 10 & 320.4 & 32.04 & & \\ \text { Total } & 11 & 1263.5 & & & \end{array}\) Use values from this output to calculate and interpret the following. Show your work. (a) A \(95 \%\) confidence interval for the mean margin of victory for all presidents with an approval rating of \(50 \%\) (b) A \(95 \%\) prediction interval for the margin of victory for a president with an approval rating of \(50 \%\) (c) A \(95 \%\) confidence interval for the mean margin of victory if we have no information about the approval rating. (Hint: This is just an ordinary confidence interval for a mean based only on the single sample of Margin values.)

Test the correlation, as indicated. Show all details of the test. Test for evidence of a linear association; \(r=0.28 ; n=100\).

Exercise A .97 on page 189 , we introduce a study about mating activity of water striders. The dataset is available as WaterStriders and includes the variables FemalesHiding, which gives the proportion of time the female water striders were in hiding, and MatingActivity, which is a measure of mean mating activity with higher numbers meaning more mating. The study included 10 groups of water striders. (The study also included an examination of the effect of hyper-aggressive males and concludes that if a male wants mating success, he should not hang out with hyper-aggressive males.) Computer output for a model to predict mating activity based on the proportion of time females are in hiding is shown below, and a scatterplot of the data with the least squares line is shown in Figure 9.12 . The regression equation is MatingActivity \(=0.480-0.323\) FemalesHiding \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 0.48014 & 0.04213 & 11.40 & 0.000 \\ \text { FemalesHiding } & -0.3232 & 0.1260 & -2.56 & 0.033\end{array}\) \(\begin{array}{lll}S=0.101312 & \text { R-Sq }=45.1 \% & \text { R-Sq(adj) }=38.3 \%\end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 0.06749 & 0.06749 & 6.58 & 0.033 \\ \text { Residual Error } & 8 & 0.08211 & 0.01026 & & \\ \text { Total } & 9 & 0.14960 & & & \end{array}\) (a) While it is hard to tell with only \(n=10\) data points, determine whether we should have any serious concerns about the conditions for fitting a linear model to these data. (b) Write down the equation of the least squares line and use it to predict the mating activity of water striders in a group in which females spend \(50 \%\) of the time in hiding (FemalesHiding = 0.50) (c) Give the hypotheses, t-statistic, p-value, and conclusion of the t-test of the slope to determine whether time in hiding is an effective predictor of mating activity. (d) Give the hypotheses, F-statistic, p-value, and conclusion of the ANOVA test to determine whether the regression model is effective at predicting mating activity. (e) How do the two p-values from parts (c) and (d) compare? (f) Interpret \(R^{2}\) for this model.

A common (and hotly debated) saying among sports fans is "Defense wins championships." Is offensive scoring ability or defensive stinginess a better indicator of a team's success? To investigate this question we'll use data from the \(2015-2016\) National Basketball Association (NBA) regular season. The data \(^{6}\) stored in NBAStandings2016 include each team's record (wins, losses, and winning percentage) along with the average number of points the team scored per game (PtsFor) and average number of points scored against them ( PtsAgainst). (a) Examine scatterplots for predicting \(\operatorname{WinPct}\) using PtsFor and predicting WinPct using PtsAgainst. In each case, discuss whether conditions for fitting a linear model appear to be met. (b) Fit a model to predict winning percentage (WinPct) using offensive ability (PtsFor). Write down the prediction equation and comment on whether PtsFor is an effective predictor. (c) Repeat the process of part (b) using PtsAgainst as the predictor. (d) Compare and interpret \(R^{2}\) for both models. (e) The Golden State Warriors set an NBA record by winning 73 games in the regular season and only losing 9 (WinPct \(=0.890\) ). They scored an average of 114.9 points per game while giving up an average of 104.1 points against. Find the predicted winning percentage for the Warriors using each of the models in (b) and (c). (f) Overall, does one of the predictors, PtsFor or PtsAgainst, appear to be more effective at explaining winning percentages for NBA teams? Give some justification for your answer.

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