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Tennis racquets vary in their physical characteristics. The data in the accompanying table give measures of bending stiffness and twisting stiffness as measured by engineering tests for 12 tennis racquets: $$\begin{array}{ccc}\hline & \begin{array}{l}\text { Bending } \\\\\text { Racquet }\end{array} & \begin{array}{l}\text { Twisting } \\\\\text { Stiffness, } x\end{array} & \text { Stiffness, } y \\\\\hline 1 & 419 & 227 \\\2 & 407 & 231 \\\3 & 363 & 200 \\\4 & 360 & 211 \\\5 & 257 & 182 \\\6 & 622 & 304 \\\7 & 424 & 384 \\\8 & 359 & 194 \\\9 & 346 & 158 \\\10 & 556 & 225 \\\11 & 474 & 305 \\\12 & 441 & 235 \\\\\hline\end{array}$$ a. If a racquet has bending stiffness, is it also likely to have twisting stiffness? Do the data provide evidence that \(x\) and \(y\) are positively correlated? b. Calculate the coefficient of determination \(r^{2}\) and interpret its value.

Short Answer

Expert verified
If so, what is the coefficient of determination and what does it mean? Answer: Yes, there is a positive correlation between bending stiffness and twisting stiffness for the tennis racquets, with a correlation coefficient of approximately 0.603. The coefficient of determination is approximately 0.363, meaning that 36.3% of the variation in twisting stiffness can be explained by the linear relationship with bending stiffness.

Step by step solution

01

Calculate the correlation coefficient (r)

To calculate the correlation coefficient (r), we will make use of the following formula: $$r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n(\sum x^2) - (\sum x)^2][n(\sum y^2) - (\sum y)^2]}}$$ Where \(n\) is the number of pairs, \(x\) is the bending stiffness, and \(y\) is the twisting stiffness. Based on the given data, we calculate the following values: \(\sum x = 4608\) \(\sum y = 2772\) \(\sum x^2 = 2029080\) \(\sum y^2 = 904364\) \(\sum xy = 1261142\) \(n = 12\) Now we can plug these values into the formula to calculate the correlation coefficient r: $$r = \frac{12(1261142) - (4608)(2772)}{\sqrt{[12(2029080) - (4608)^2][12(904364) - (2772)^2]}}$$ $$r \approx 0.603$$
02

Determine if there is a positive correlation

Now that we have calculated the correlation coefficient (r), we can determine if there is a positive correlation between bending stiffness (x) and twisting stiffness (y). Since \(r \approx 0.603\), which is positive and between 0 and 1, there is a positive correlation between bending stiffness and twisting stiffness. As a racquet's bending stiffness increases, its twisting stiffness tends to increase as well.
03

Calculate the coefficient of determination \(r^2\)

To calculate the coefficient of determination (\(r^2\)), we square the correlation coefficient (r). $$r^2 = (0.603)^2$$ $$r^2 \approx 0.363$$
04

Interpret the value of \(r^2\)

The coefficient of determination (\(r^2\)) is approximately 0.363. This means that about 36.3% of the variation in twisting stiffness (y) can be explained by the linear relationship with bending stiffness (x). In conclusion, there is a positive correlation between bending stiffness and twisting stiffness for these tennis racquets, with about 36.3% of the variation in twisting stiffness being explained by their bending stiffness.

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Key Concepts

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

Correlation Coefficient
The correlation coefficient, denoted by the symbol 'r', is a statistical measure that calculates the strength and direction of a linear relationship between two variables. It's a value that ranges between -1 and 1 where 1 indicates a perfect positive linear correlation, -1 indicates a perfect negative linear correlation, and 0 implies no correlation at all.

It is calculated using the formula: \[r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n(\sum x^2) - (\sum x)^2][n(\sum y^2) - (\sum y)^2]}}\]In the context of the tennis racquet data, 'x' represents the bending stiffness and 'y' represents the twisting stiffness. A correlation coefficient of approximately 0.603 suggests a moderate strength of the positive linear relationship between the two variables. To truly understand their relationship, we also need to look at the coefficient of determination, which we'll discuss next.

When creating educational content that aims to explain the correlation coefficient, it's crucial to present real-world examples, like the tennis racquet case, that make the abstract idea tangible. Providing visual aids, such as scatter plots with a line of best fit, can also be extremely helpful to students in visualizing the relationship between variables.
Positively Correlated
Two variables are considered to be positively correlated when an increase in one variable tends to be associated with an increase in the other variable. In the example of tennis racquets, the bending stiffness and twisting stiffness are positively correlated because as one measure increases, the other tends to increase as well.

This concept reflects a direct relationship, where variables move together in the same direction. It's essential for students to recognize that correlation does not imply causation; just because two variables move together does not mean that one variable causes the change in the other. This distinction is necessary for a complete understanding of correlation in a statistical context.

In educational resources, using data visualization and employing consistent, real-world scenarios can support the student's comprehension of positive correlation. Encouraging students to think about their daily life observations can create a bridge to understanding the concept deeply. Examples such as the relationship between study time and exam scores or temperature and ice cream sales can solidify this understanding.
Variation in Statistics
Variation in statistics refers to the degree to which data points in a statistical distribution or dataset differ from each other and the mean (average) of the dataset. It is a core concept since it gives insight into the spread or dispersion of the data. The various measures to quantify variation include range, variance, and standard deviation.

In the context of the racquets' study, the coefficient of determination (\(r^2\)) which is approximately 0.363, tells us that around 36.3% of the variation in twisting stiffness can be explained by its linear relationship with bending stiffness. The rest of the variation is due to other factors or inherent variability.

Understanding variation is essential in making predictions and decisions based on data. When explaining variation to students, it's helpful to use graphical representations, like bar charts or histograms, to illustrate how data can vary around the mean. It's also important to stress that while some variation can be predicted by one variable, there's often a portion that cannot be explained by the model at hand.

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Most popular questions from this chapter

The following data (Exercise 16, Section 12.2) were obtained in an experiment relating the dependent variable \(y\) (texture of strawberries) with \(x\) (coded storage temperature). $$ \begin{array}{l|rrrrr} x & -2 & -2 & 0 & 2 & 2 \\ \hline y & 4.0 & 3.5 & 2.0 & 0.5 & 0.0 \end{array} $$ a. Estimate the expected strawberry texture for a coded storage temperature of \(x=-1\). Use a \(99 \%\) confidence interval. b. Predict the particular value of \(y\) when \(x=1\) with a \(99 \%\) prediction interval. c. At what value of \(x\) will the width of the prediction interval for a particular value of \(y\) be a minimum, assuming \(n\) remains fixed?

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An informal experiment was conducted at McNair Academic High School in Jersey City, New Jersey. Twenty freshman algebra students were given a survey at the beginning of the semester, measuring his or her skill level. They were then allowed to use laptop computers both at school and at home. At the end of the semester, their scores on the same survey were recorded \((x)\) along with their score on the final examination \((y) .^{9}\) The data and the MINITAB printout are shown here. $$ \begin{array}{ccc} \hline \text { Student } & \text { End-of-Semester Survey } & \text { Final Exam } \\ \hline 1 & 100 & 98 \\ 2 & 96 & 97 \\ 3 & 88 & 88 \\ 4 & 100 & 100 \\ 5 & 100 & 100 \\ 6 & 96 & 78 \\ 7 & 80 & 68 \\ 8 & 68 & 47 \\ 9 & 92 & 90 \\ 10 & 96 & 94 \\ 11 & 88 & 84 \\ 12 & 92 & 93 \\ 13 & 68 & 57 \\ 14 & 84 & 84 \\ 15 & 84 & 81 \\ 16 & 88 & 83 \\ 17 & 72 & 84 \\ 18 & 88 & 93 \\ 19 & 72 & 57 \\ 20 & 88 & 83 \\ \hline \end{array} $$ $$ \begin{aligned} &\text { Analysis of Variance }\\\ &\begin{array}{lrrrrr} \text { Source } & \text { DF } & \text { Adj SS } & \text { AdjMS } & \text { F-Value } & \text { P-Value } \\ \hline \text { Regression } & 1 & 3254.03 & 3254.03 & 56.05 & 0.000 \\ \text { Error } & 18 & 1044.92 & 58.05 & & \\ \text { Total } & 19 & 4298.95 & & & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Model Summary }\\\ &\begin{array}{ccc} \mathrm{S} & \mathrm{R}-\mathrm{sq} & \mathrm{R}-\mathrm{sq}(\mathrm{adj}) \\ \hline 7.61912 & 75.69 \% & 74.34 \% \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Coefficients }\\\ &\begin{array}{lrrrr} \text { Term } & \text { Coef } & \text { SE Coef } & \text { T-Value } & \text { P-Value } \\ \hline \text { Constant } & -26.8 & 14.8 & -1.82 & 0.086 \\ \mathrm{x} & 1.262 & 0.169 & 7.49 & 0.000 \end{array} \end{aligned} $$ Regression Equation $$ y=-26.8+1.262 x $$ a. Construct a scatterplot for the data. Does the assumption of linearity appear to be reasonable? b. What is the equation of the regression line used for predicting final exam score as a function of the endof-semester survey score? c. Do the data present sufficient evidence to indicate that final exam score is linearly related to the end-ofsemester survey score? Use \(\alpha=.01\). d. Find a \(99 \%\) confidence interval for the slope of the regression line. e. Use the MINITAB printout to find the value of the coefficient of determination, \(r^{2}\). Show that \(r^{2}=\) SSR/Total SS. f. What percentage reduction in the total variation is achieved by using the linear regression model?

What diagnostic plot can you use to determine whether the data satisfy the normality assumption? What should the plot look like for normal residuals?

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