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Chapter 3: Problem 72

Each year, students in an elementary school take a standardized math test at the end of the school year. For a class of fourth-graders, the average score was 55.1 with a standard deviation of \(12.3 .\) In the third grade, these same students had an average score of 61.7 with a standard deviation of \(14.0 .\) The correlation between the two sets of scores is \(r=0.95\). Calculate the equation of the least-squares regression line for predicting a fourth-grade score from a third-grade score. (a) \(\hat{y}=3.60+0.835 x\) (b) \(\hat{y}=15.69+0.835 x\) (c) \(\hat{y}=2.19+1.08 x\) (d) \(\hat{y}=-11.54+1.08 x\) (e) Cannot be calculated without the data.

Short Answer

Expert verified
The correct answer is (a) \(\hat{y} = 3.60 + 0.835 x\).

Step by step solution

01

Understand What is Asked

The problem requires determining the equation of the least-squares regression line used to predict the fourth-grade score based on the third-grade score.
02

Recall the Regression Line Equation

The regression line equation is given by: \[\hat{y} = a + b x\]where \(\hat{y}\) is the predicted fourth-grade score, \(a\) is the y-intercept, \(b\) is the slope, and \(x\) is the third-grade score.
03

Calculate the Slope (b)

Use the formula for the slope of the regression line, \[b = r \cdot \frac{s_y}{s_x}\]where \(r = 0.95\), \(s_y = 12.3\) (standard deviation of fourth-grade scores), and \(s_x = 14.0\) (standard deviation of third-grade scores). Substitute these values to calculate \(b\):\[b = 0.95 \cdot \frac{12.3}{14.0} = 0.835\]
04

Calculate the Y-Intercept (a)

The formula for the y-intercept is:\[a = \bar{y} - b \bar{x}\]where \(\bar{y} = 55.1\) (mean of the fourth-grade scores) and \(\bar{x} = 61.7\) (mean of the third-grade scores). Substitute these values to find \(a\):\[a = 55.1 - 0.835 \times 61.7 = 55.1 - 51.5145 = 3.5855 \a \approx 3.60\]
05

Formulate the Regression Equation

Substitute the values of \(a\) and \(b\) into the regression equation:\[\hat{y} = 3.60 + 0.835x\]This matches option (a) from the given choices.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure that describes the amount of variation or dispersion in a set of values. In simpler terms, it tells us how spread out the numbers in a data set are from the mean (average). A low standard deviation means the values are close to the mean, whereas a high standard deviation indicates that the values are spread out over a larger range.
For example, in the exercise given, the standard deviation of the fourth-grade scores is 12.3, while for the third-grade scores, it is 14.0. This means that the scores of the third graders have a slightly wider spread around the mean compared to the fourth graders.
Understanding standard deviation helps in comparing the variability between two data sets and provides insight into how reliable the mean is as a measure of central tendency in each set.
  • Lower standard deviation: values are close to the mean
  • Higher standard deviation: values are spread out
Hence, when you see a standard deviation number, it helps you quickly gauge how consistent the data points are and how much they deviate from the average.
Correlation Coefficient
The correlation coefficient, often represented by the letter 'r', measures the strength and direction of a linear relationship between two variables. The value of the correlation coefficient ranges from -1 to 1. A correlation of 1 indicates a perfect positive linear relationship, while -1 indicates a perfect negative linear relationship. A correlation of 0 indicates no linear relationship whatsoever.
In the given problem, the correlation coefficient between the third-grade scores and the fourth-grade scores is 0.95. This is a strong positive correlation, implying that the scores in the two grades are closely related. As the third-grade scores increase, we can expect the fourth-grade scores to increase as well, in a nearly linear fashion.
A high correlation coefficient helps us reliably use the data from one grade to predict scores in the other grade, simply because their changes tend to follow the same pattern. Here are a few things to remember:
  • Strong correlation (close to 1 or -1): strong relationship
  • Weak correlation (close to 0): weak or no relationship
  • The sign of the correlation coefficient (+/-) signifies the direction of the relationship
Slope of a Line
The slope of a line in a linear regression equation represents the rate at which the dependent variable changes with respect to the independent variable. In the context of regression analysis, the slope tells us how much the predicted value (dependent variable) is expected to increase (or decrease) when the independent variable increases by one unit.
From the problem, we calculated the slope, represented as 'b', using the formula:\[b = r \cdot \frac{s_y}{s_x}\]Substituting the given values, the slope was found to be 0.835. This means that for each additional point in the third-grade score, the predicted fourth-grade score increases by approximately 0.835 points.
This relationship provides a predictable and quantifiable measure of how changes in third-grade scores influence fourth-grade scores. Some key takeaways about the slope:
  • A positive slope (e.g., 0.835) indicates that as the third-grade score increases, the predicted fourth-grade score also increases.
  • The steeper the slope, the greater the change in the predicted value for a unit change in the independent variable.
  • Understanding the slope helps in predicting outcomes and understanding the relationship between variables.

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Most popular questions from this chapter

Data on the IQ test scores and reading test scores for a group of fifth-grade children give the following regression line: predicted reading score \(=-33.4+0.882(\mathrm{IQ}\) score \()\) (a) What's the slope of this line? Interpret this value in context. (b) What's the \(y\) intercept? Explain why the value of the intercept is not statistically meaningful. (c) Find the predicted reading score for a child with an IQ score of 90 .

The figure below plots the average brain weight in grams versus average body weight in kilograms for 96 species of mammals. \({ }^{12}\) There are many small mammals whose points overlap at the lower left. (a) The correlation between body weight and brain weight is \(r=0.86 .\) Explain what this value means. (b) What effect does the elephant have on the correlation? Justify your answer.

The gas mileage of an automobile first increases and then decreases as the speed increases. Suppose that this relationship is very regular, as shown by the following data on speed (miles per hour) and mileage (miles per gallon). $$\begin{array}{lccccc}\hline \text { Speed: } & 20 & 30 & 40 & 50 & 60 \\\\\text { Mileage: } & 24 & 28 & 30 & 28 & 24 \\\\\hline \end{array}$$ (a) Make a scatterplot to show the relationship between speed and mileage. (b) Calculate the correlation for these data by hand or using technology. (c) Explain why the correlation has the value found in part (b) even though there is a strong relationship between speed and mileage.

The percent of an animal species in the wild that survives to breed again is often lower following a successful breeding season. A study of merlins (small falcons) in northern Sweden observed the number of breeding pairs in an isolated area and the percent of males (banded for identification) that returned the next breeding season. Here are data for seven years: $$\begin{array}{llllllll}\hline \text { Breeding pairs: } & 28 & 29 & 29 & 29 & 30 & 32 & 33 \\\\\text { Percent return: } & 82 & 83 & 70 & 61 & 69 & 58 & 43 \\\\\hline \end{array}$$ Make a scatterplot to display the relationship between breeding pairs and percent return. Describe what you see.

Measurements on young children in Mumbai, India, found this least-squares line for predicting height \(y\) from \(\operatorname{arm} \operatorname{span} x:\) $$\hat{y}=6.4+0.93 x$$ Measurements are in centimeters \((\mathrm{cm})\). In addition to the regression line, the report on the Mumbai measurements says that \(r^{2}=0.95\). This suggests that (a) although arm span and height are correlated, arm span does not predict height very accurately. (b) height increases by \(\sqrt{0.95}=0.97 \mathrm{~cm}\) for each additional centimeter of arm span. (c) \(95 \%\) of the relationship between height and arm span is accounted for by the regression line. (d) \(95 \%\) of the variation in height is accounted for by the regression line. (e) \(95 \%\) of the height measurements are accounted for by the regression line.
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