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Chapter 9: Problem 13

Let \(x\) be per capita income in thousands of dollars. Let \(y\) be the number of medical doctors per 10,000 residents. Six small cities in Oregon gave the following information about \(x\) and \(y\) (based on information from Life in America's Small Cities by G. S. Thomas, Prometheus Books). $$ \begin{array}{l|lrlrrr} \hline x & 8.6 & 9.3 & 10.1 & 8.0 & 8.3 & 8.7 \\ \hline y & 9.6 & 18.5 & 20.9 & 10.2 & 11.4 & 13.1 \\ \hline \end{array} $$ Complete parts (a) through (e), given \(\Sigma x=53, \Sigma y=83.7, \Sigma x^{2}=471.04\) \(\Sigma y^{2}=1276.83, \Sigma x y=755.89\), and \(r \approx 0.934 .\) (f) Suppose a small city in Oregon has a per capita income of 10 thousand dollars. What is the predicted number of M.D.s per 10,000 residents?

Short Answer

Expert verified
Approximately 20 medical doctors per 10,000 residents.

Step by step solution

01

Identify the need for Linear Regression

We need to find a predictive relationship between per capita income \(x\) and the number of medical doctors \(y\). Given \(r \approx 0.934\), there's a strong linear correlation between these variables, suggesting linear regression is appropriate.
02

Calculate the slope (b)

The slope \(b\) for the regression line is calculated using the formula: \[ b = \frac{n \Sigma xy - \Sigma x \Sigma y}{n \Sigma x^2 - (\Sigma x)^2} \] where \(n=6\), \(\Sigma xy = 755.89\), \(\Sigma x = 53\), \(\Sigma y = 83.7\), \(\Sigma x^2 = 471.04\). Substituting these values, we find: \[ b = \frac{6 \cdot 755.89 - 53 \cdot 83.7}{6 \cdot 471.04 - 53^2} = \frac{4525.34 - 4436.1}{2826.24 - 2809} = \frac{89.24}{17.24} \approx 5.178 \]
03

Calculate the intercept (a)

The intercept \(a\) for the regression line is calculated using the formula: \[ a = \frac{\Sigma y - b \Sigma x}{n} \] Using \(b = 5.178\), \(\Sigma y = 83.7\), and \(\Sigma x = 53\), the calculation gives: \[ a = \frac{83.7 - 5.178 \times 53}{6} = \frac{83.7 - 274.434}{6} = \frac{-190.734}{6} \approx -31.789 \]
04

Write the equation of the line

Substitute \(a\) and \(b\) into the linear equation \(y = a + bx\): \[ y = -31.789 + 5.178x \] This equation can be used to predict \(y\), the number of doctors per 10,000 residents, given \(x\), the per capita income.
05

Predict the value of y when x = 10

Substitute \(x = 10\) into the equation \(y = -31.789 + 5.178 \times x\): \[ y = -31.789 + 5.178 \times 10 = -31.789 + 51.78 = 19.991 \]

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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 is a statistical measure that describes the strength and direction of a relationship between two variables. In this exercise, it is represented by \( r \), which is approximately 0.934. A correlation coefficient close to 1 or -1 indicates a strong relationship. In this case, it is close to 1, signifying a strong positive linear relationship between per capita income \( x \) and the number of medical doctors \( y \). This means that as income increases, the number of doctors generally increases as well.Understanding the correlation coefficient can guide decisions in linear regression analysis, which involves predicting one variable based on another. Since the correlation here is high, it confirms that using linear regression to model the relationship between income and doctors is reasonable. This strong relationship also inspires confidence in the predictions made from the regression analysis.
Predictive Relationship
A predictive relationship in statistics refers to using known values of one variable to forecast the unknown values of another. In this case, the goal is to predict the number of medical doctors based on per capita income using a linear regression model. The correlation coefficient previously calculated indicates a strong predictive ability. The process of establishing a predictive relationship involves:
  • Calculating the correlation coefficient to determine the strength of the relationship.
  • Using regression analysis to develop an equation that describes the relationship.
  • Applying this equation to predict future or unobserved values.
In practice, this means that given a per capita income figure, it's possible to estimate the number of medical doctors that might be expected in a particular city or region. This kind of analysis is invaluable in planning and resource allocation in various fields.
Slope Calculation
The slope calculation is a critical step in building the regression equation. The slope \( b \) measures how much \( y \) (doctors per 10,000 residents) changes for a one-unit change in \( x \) (per capita income). The slope is computed using the formula: \[ b = \frac{n \Sigma xy - \Sigma x \Sigma y}{n \Sigma x^2 - (\Sigma x)^2} \]Substituting the given values, we find \( b = \frac{89.24}{17.24} \approx 5.178 \).This slope indicates that for each additional thousand dollars in per capita income, the number of doctors per 10,000 residents increases, on average, by about 5.178 doctors. Hence, the slope gives insight into the rate of change and helps quantify the predictive relationship further.Understanding the slope is vital, as it tells us how sensitive the dependent variable is to changes in the independent variable, making it a crucial aspect of predictive analytics.
Regression Equation
The regression equation summarizes the predictive relationship between the independent variable \( x \) (income) and the dependent variable \( y \) (doctors). The standard form of the regression equation is given by: \[ y = a + bx \]Here, \( a \) is the y-intercept and \( b \) is the slope. From the solution, \( a \) is approximately -31.789 and \( b \) is approximately 5.178. Therefore, the regression equation becomes:\[ y = -31.789 + 5.178x \]This equation means that without accounting for any income (\( x = 0 \)), there are almost no doctors per 10,000 residents in this model, which is not realistic but helps in creating a linear shift with the constant changes. The number 5.178 represents how each unit increase in income changes the number of doctors. Such equations are used for prediction. In this problem, if per capita income is 10, then substituting \( x = 10 \) leads to a prediction of around 19.991 doctors per 10,000 residents. The regression equation is a powerful tool used widely in data-driven forecasting and decision-making.

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Most popular questions from this chapter

In Section 5.1, we studied linear combinations of independent random variables. What happens if the variables are not independent? A lot of mathematics can be used to prove the following: Let \(x\) and \(y\) be random variables with means \(\mu_{x}\) and \(\mu_{y}\), variances \(\sigma_{x}^{2}\) and \(\sigma_{y}^{2}\), and population correlation coefficient \(\rho\) (the Greek letter rho). Let \(a\) and \(b\) be any constants and let \(w=a x+b y .\) Then, In this formula, \(\rho\) is the population correlation coefficient, theoretically computed using the population of all \((x, y)\) data pairs. The expression \(\sigma_{x} \sigma_{y} \rho\) is called the covariance of \(x\) and \(y .\) If \(x\) and \(y\) are independent, then \(\rho=0\) and the formula for \(\sigma_{w}^{2}\) reduces to the appropriate formula for independent variables (see Section 5.1). In most real-world applications, the population parameters are not known, so we use sample estimates with the understanding that our conclusions are also estimates. Do you have to be rich to invest in bonds and real estate? No, mutual fund shares are available to you even if you aren't rich. Let \(x\) represent annual percentage return (after expenses) on the Vanguard Total Bond Index Fund, and let \(y\) represent annual percentage return on the Fidelity Real Estate Investment Fund. Over a long period of time, we have the following population estimates (based on Morningstar Mutual Fund Report). $$ \mu_{x} \approx 7.32 \quad \sigma_{x} \approx 6.59 \quad \mu_{y} \approx 13.19 \quad \sigma_{y} \approx 18.56 \quad \rho \approx 0.424 $$ (a) Do you think the variables \(x\) and \(y\) are independent? Explain. (b) Suppose you decide to put \(60 \%\) of your investment in bonds and \(40 \%\) in real estate. This means you will use a weighted average \(w=0.6 x+0.4 y\). Estimate your expected percentage return \(\mu_{w}\) and risk \(\sigma_{w}\) (c) Repeat part (b) if \(w=0.4 x+0.6 y\). (d) Compare your results in parts (b) and (c). Which investment has the higher expected return? Which has the greater risk as measured by \(\sigma_{w} ?\)

For a fixed confidence level, how does the length of the confidence interval for predicted values of \(y\) change as the corresponding \(x\) values become further away from \(\bar{x}\) ?

Given the linear regression equation \(x_{3}=-16.5+4.0 x_{1}+9.2 x_{4}-1.1 x_{7}\) (a) Which variable is the response variable? Which variables are the explanatory variables? (b) Which number is the constant term? List the coefficients with their corresponding explanatory variables. (c) If \(x_{1}=10, x_{4}=-1\), and \(x_{7}=2\), what is the predicted value for \(x_{3} ?\) (d) Explain how each coefficient can be thought of as a "slope." Suppose \(x_{1}\) and \(x_{7}\) were held as fixed but arbitrary values. If \(x_{4}\) increased by 1 unit, what would we expect the corresponding change in \(x_{3}\) to be? If \(x_{4}\) increased by 3 units, what would be the corresponding expected change in \(x_{3}\) ? If \(x_{4}\) decreased by 2 units, what would we expect for the corresponding change in \(x_{3}\) ? (e) Suppose that \(n=15\) data points were used to construct the given regression equation and that the standard error for the coefficient of \(x_{4}\) is \(0.921\). Construct a \(90 \%\) confidence interval for the coefficient of \(x_{4}\). (f) Using the information of part (e) and level of significance \(1 \%\), test the claim that the coefficient of \(x_{4}\) is different from zero. Explain how the conclusion has a bearing on the regression equation.

Fuming because you are stuck in traffic? Roadway congestion is a costly item, in both time wasted and fuel wasted. Let \(x\) represent the average annual hours per person spent in traffic delays and let \(y\) represent the average annual gallons of fuel wasted per person in traffic delays. A random sample of eight cities showed the following data (Reference: Statistical Abstract of the United States, 122 nd Edition). $$ \begin{array}{l|llllllll} \hline x(\mathrm{hr}) & 28 & 5 & 20 & 35 & 20 & 23 & 18 & 5 \\ \hline y(\mathrm{gal}) & 48 & 3 & 34 & 55 & 34 & 38 & 28 & 9 \\ \hline \end{array} $$ (a) Draw a scatter diagram for the data. Verify that \(\Sigma x=154, \Sigma x^{2}=3712\), \(\Sigma y=249, \Sigma y^{2}=9959\), and \(\Sigma x y=6067\). Compute \(r\) The data in part (a) represent average annual hours lost per person and average annual gallons of fuel wasted per person in traffic delays. Suppose that instead of using average data for different cities, you selected one person at random from each city and measured the annual number of hours lost \(x\) for that person and the annual gallons of fuel wasted \(y\) for the same person. $$ \begin{array}{l|cccccccc} \hline x(\mathrm{hr}) & 20 & 4 & 18 & 42 & 15 & 25 & 2 & 35 \\ \hline y(\mathrm{gal}) & 60 & 8 & 12 & 50 & 21 & 30 & 4 & 70 \\ \hline \end{array} $$ (b) Compute \(\bar{x}\) and \(\bar{y}\) for both sets of data pairs and compare the averages. Compute the sample standard deviations \(s_{x}\) and \(s_{y}\) for both sets of data pairs and compare the standard deviations. In which set are the standard deviations for \(x\) and \(y\) larger? Look at the defining formula for \(r\), Equation \(1 .\) Why do smaller standard deviations \(s_{x}\) and \(s_{y}\) tend to increase the value of \(r\) ? (c) Make a scatter diagram for the second set of data pairs. Verify that \(\Sigma x=161, \quad \Sigma x^{2}=4583, \quad \Sigma y=255, \quad \Sigma y^{2}=12,565\), and \(\Sigma x y=7071 .\) Compute \(r\). (d) Compare \(r\) from part (a) with \(r\) from part (c). Do the data for averages have a higher correlation coefficient than the data for individual measurements? List some reasons why you think hours lost per individual and fuel wasted per individual might vary more than the same quantities averaged over all the people in a city.

The following data are based on information from the Harvard Business Review (Vol. 72, No. 1). Let \(x\) be the number of different research programs, and let \(y\) be the mean number of patents per program. As in any business, a company can spread itself too thin. For example, too many research programs might lead to a decline in overall research productivity. The following data are for a collection of pharmaceutical companies and their research programs: $$ \begin{array}{l|rrrrrr} \hline x & 10 & 12 & 14 & 16 & 18 & 20 \\ \hline y & 1.8 & 1.7 & 1.5 & 1.4 & 1.0 & 0.7 \\ \hline \end{array} $$ Complete parts (a) through (e), given \(\Sigma x=90, \Sigma y=8.1, \Sigma x^{2}=1420\), \(\Sigma y^{2}=11.83, \Sigma x y=113.8\), and \(r \approx-0.973 .\) (f) Suppose a pharmaceutical company has 15 different research programs. What does the least-squares equation forecast for \(y=\) mean number of patents per program?
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