/*! 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 73 Assume that in a political scien... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Assume that in a political science class, the teacher gives a midterm exam and a final exam. Assume that the association between midterm and final scores is linear. The summary statistics have been simplified for clarity see Guidance on page \(209 .\) Midterm: Mean \(=75, \quad\) Standard deviation \(=10\) Final: Mean \(=75, \quad\) Standard deviation \(=10\) Also, \(r=0.7\) and \(n=20\). According to the regression equation, for a student who gets a 95 on the midterm, what is the predicted final exam grade? What phenomenon from the chapter does this demonstrate? Explain. See page 209 for guidance.

Short Answer

Expert verified
The predicted final exam grade for a student who gets a 95 on the midterm is 89.5. This demonstrates the phenomenon of regression to the mean.

Step by step solution

01

Calculating the slope m

The slope m is given by m = r*(std_dev(y)/std_dev(x)). Substituting the given values, we get m = 0.7 * (10/10) = 0.7.
02

Calculating the bias b

The bias b is given by b = mean(y) - m * mean(x). Substituting the given values, we get b = 75 - 0.7 * 75 = 22.5.
03

Predicting the final score

The final score y for a given midterm score x is given by y = mx + b. Substituting the given midterm score of 95 and the values calculated in steps 1 and 2, we get y = 0.7 * 95 + 22.5 = 89.5.
04

Understanding the phenomenon

The phenomenon being demonstrated here is called regression to the mean. Even though the student scored 95 in the midterm, which is significantly above the mean, the predicted final score is 89.5, which is closer to the mean. This is because of the correlation between the midterm and final scores. If the correlation had been perfect (i.e., r = 1), then the predicted final score would have been the same as the midterm score, but since r < 1, the predicted final score regresses towards the mean.

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Key Concepts

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

Statistical Regression
Statistical regression is a method that enables us to understand how the value of one variable, often called the dependent variable, changes when another independent variable is varied. Applied to classroom settings, regression analysis helps predict the final examination score of a student based on their midterm score.

Our example demonstrates that a student who scored highly on the midterm might not necessarily achieve an equally high score on the final exam. This tendency for extreme scores to fall back towards the average is termed 'regression to the mean'. It's a fundamental concept in statistics where the extreme values in a data set are followed by values that are closer to the mean or average. Here, even an outstanding midterm score aligns closer to the class's average performance on the final exam, epitomizing regression towards the mean.
Correlation Coefficient
The correlation coefficient, designated as \( r \), quantifies the strength and direction of the linear relationship between two variables. Values of \( r \) range from -1 to 1, where 1 implies a perfect positive linear relationship, -1 implies a perfect negative linear relationship, and 0 indicates no linear relationship at all.

In our exercise, \( r = 0.7 \) signifies a strong, positive correlation between midterm and final exam scores: as one goes up, so tends to the other. Just as the 'm' in our slope calculation reflects this correlation, it's pivotal to expect that an exceptionally high or low midterm score will result in a final score that, while still similarly high or low, is often closer to the class average because the correlation isn't perfect (\( r \) isn't 1).
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

The fact that both the midterm and final exams have standard deviations of \(10\) points suggests that students' scores on both exams vary around the mean (75) to a similar extent. This uniformity in variability assists in creating a more predictive model because we understand how far scores typically deviate from the mean, allowing us to predict final exam scores from midterms with some reliability.
Predictive Modeling
Predictive modeling encompasses a variety of techniques, including statistical models, that are used to predict future events or outcomes. By assigning numerical values to variables, we can form a predictive equation, like the one used to estimate a student's final exam score based on their midterm grade.

Here, predictive modeling uses the previous knowledge (midterm scores, along with correlation and standard deviation) to forecast the final exam score. The model's effectiveness, however, hinges on the assumptions made, like linearity and the regression to the mean demonstrated in this exercise. It's a powerful tool in many fields, from academics to finance to healthcare, to anticipate future trends or behaviors based on past data.

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Most popular questions from this chapter

Answer the questions using complete sentences. a. What is an influential point? How should influential points be treated when doing a regression analysis? b. What is the coefficient of determination and what does it measure? c. What is extrapolation? Should extrapolation ever be used?

Pitchers The table shows the Earned Run Average (ERA) and WHIP rating (walks plus hits per inning) for the top 40 Major League Baseball pitchers in the 2017 season. Top pitchers will tend to have low ERA and WHIP ratings. (Source: ESPN.com) a. Make a scatterplot of the data and state the sign of the slope from the scatterplot. Use WHIP to predict ERA. b. Use linear regression to find the equation of the best-fit line. Show the line on the scatterplot using technology or by hand. c. Interpret the slope. d. Interpret the \(y\) -intercept or explain why it would be inappropriate to do so. $$ \begin{aligned} &\begin{array}{|ll|} \hline \text { WHIP } & \text { ERA } \\ \hline 0.87 & 2.25 \\ \hline 0.95 & 2.31 \\ \hline 0.9 & 2.51 \\ \hline 1.02 & 2.52 \\ \hline 1.15 & 2.89 \\ \hline 0.97 & 2.9 \\ \hline \end{array}\\\ &\begin{array}{|c|c|} \hline \text { WHIP } & \text { ERA } \\ \hline 1.21 & 3.55 \\ \hline 1.22 & 3.64 \\ \hline 1.22 & 3.66 \\ \hline 1.27 & 3.82 \\ \hline 1.15 & 3.83 \\ \hline 1.16 & 3.86 \\ \hline \end{array} \end{aligned} $$

The table shows the heights (in inches) and weights (in pounds) of 14 college men. The scatterplot shows that the association is linear enough to proceed. $$ \begin{aligned} &\begin{array}{|c|c|} \hline \begin{array}{c} \text { Height } \\ \text { (inches) } \end{array} & \begin{array}{c} \text { Weight } \\ \text { (pounds) } \end{array} \\ \hline 68 & 205 \\ \hline 68 & 168 \\ \hline 74 & 230 \\ \hline 68 & 190 \\ \hline 67 & 185 \\ \hline 69 & 190 \\ \hline 68 & 165 \\ \hline \end{array}\\\ &\begin{array}{|c|c|} \hline \begin{array}{c} \text { Height } \\ \text { (inches) } \end{array} & \begin{array}{c} \text { Weight } \\ \text { (pounds) } \end{array} \\ \hline 70 & 200 \\ \hline 69 & 175 \\ \hline 72 & 210 \\ \hline 72 & 205 \\ \hline 72 & 185 \\ \hline 71 & 200 \\ \hline 73 & 195 \\ \hline \end{array} \end{aligned} $$ a. Find the equation for the regression line with weight (in pounds) as the response and height (in inches) as the predictor. Report the slope and the intercept of the regression line, and explain what they show. If the intercept is not appropriate to report, explain why. b. Find the correlation between weight (in pounds) and height (in inches). c. Find the coefficient of determination and interpret it. d. If you changed each height to centimeters by multiplying heights in inches by \(2.54\), what would the new correlation be? Explain. e. Find the equation with weight (in pounds) as the response and height (in inches) as the predictor, and interpret the slope. f. Summarize what you found: Does changing units change the correlation? Does changing units change the regression equation?

The following table shows the fat content (in grams) and calories for a sample of granola bars. (Source: calorielab. com ) $$ \begin{array}{|c|l|} \hline \text { Fat (in grams) } & \text { Calories } \\ \hline 7.6 & 370 \\ \hline 3.3 & 106.1 \\ \hline 18.7 & 312.4 \\ \hline \end{array} $$ $$ \begin{array}{|c|c|} \hline \text { Fat (in grams) } & \text { Calories } \\ \hline 3.8 & 113.1 \\ \hline 5 & 117.8 \\ \hline 5.5 & 131.9 \\ \hline 7.2 & 140.6 \\ \hline 6.1 & 118.8 \\ \hline 4.6 & 124.4 \\ \hline 3.9 & 105.1 \\ \hline 6.1 & 136 \\ \hline 4.8 & 124 \\ \hline 4.4 & 119.3 \\ \hline 7.7 & 192.6 \\ \hline \end{array} $$ a. Use technology to make a scatterplot of the data. Use fat as the independent variable \((x)\) and calories as the dependent variable \((y)\). Does there seem to be a linear trend to the data? b. Compute the correlation coefficient and the regression equation, using fat as the independent variable and calories as the dependent variable. c. What is the slope of the regression equation? Interpret the slope in the context of this problem. d. What is the \(y\) -intercept of the regression equation? Interpret the \(y\) -intercept in the context of this problem or explain why it would be inappropriate to do so. e. Find and interpret the coefficient of determination. f. Use the regression equation to predict the calories in a granola bar containing 7 grams of fat. g. Would it be appropriate to use the regression equation to predict the calories in a granola bar containing 25 grams of fat? If so, predict the number of calories in such a bar. If not, explain why it would be inappropriate to do so. h. Looking at the scatterplot there is a granola bar in the sample that has an extremely high number of calories given the moderate amount of fat it contains. Remove its data from the sample and recalculate the correlation coefficient and regression equation. How did removing this unusual point change the value of \(r\) and the regression equation?

Data on the 3-point percentage, field-goal percentage, and free-throw percentage for a sample of 50 professional basketball players were obtained. Regression analyses were conducted on the relationships between 3-point percentage and field-goal percentage and between 3-point percentage and freethrow percentage. The StatCrunch results are shown below. (Source: nba.com) Simple linear regression results: Dependent Variable: 3 Point \% Independent Variable: Field Goal \% 3 Point \(\%=40.090108-0.091032596\) Field Goal \% Sample size: 50 \(\mathrm{R}\) (correlation coefficient) \(=-0.048875984\) \(\mathrm{R}-\mathrm{sq}=0.0023888618\) Estimate of error standard deviation: \(7.7329785\) Simple linear regression results: Dependent Variable: 3 Point \% Independent Variable: Free Throw \% 3 Point \(\%=-8.2347225+0.54224127\) Free Throw \(\%\) Sample size: 50 \(\mathrm{R}\) (correlation coefficient) \(=0.57040364\) \(\mathrm{R}-\mathrm{sq}=0.32536031\) Estimate of error standard deviation: \(6.3591944\) Based on this sample, is there a stronger association between 3 -point percentage and field-goal percentage or 3 -point percentage and freethrow percentage? Provide a reason for your choice based on the StatCrunch results provided.

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