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Chapter 13: Problem 5

Suppose that a simple linear regression model is appropriate for describing the relationship between \(y=\) house price and \(x=\) house size (sq ft) for houses in a large city. The true regression line is \(y=23,000+47 x\) and \(\sigma=5000\). a. What is the average change in price associated with one extra sq \(\mathrm{ft}\) of space? With an additional 100 sq \(\mathrm{ft}\) of space? b. What proportion of \(1800-\) sq-ft homes would be priced over \(\$ 110,000 ?\) Under \(\$ 100,000\) ?

Short Answer

Expert verified
The average increase in the price of a house for one extra sq ft is $47, and for an additional 100 sq ft, the average price increase is $4700. The proportion of 1800 sq-ft homes priced over $110,000 or under $100,000 can be found by looking up z-table values corresponding to the z-scores -0.12 and -2.12, respectively.

Step by step solution

01

Calculate the average change in price for one extra sq ft

To calculate the average change in price for one extra sq ft, look at the coefficient of 'x' in the regression equation. The coefficient of 'x' in our equation is 47. Thus, for each additional sq ft, the price of the house increases, on average, by $47.
02

Calculate the average change in price for an additional 100 sq ft

If you want to find out how much the price changes for an additional 100 sq ft, multiply the average change in price for one extra sq ft (which is $47) by 100. Hence, for an additional 100 sq ft, the price will increase by 47 * 100 = $4700.
03

Calculate the price for 1800 sq-ft homes and find z-scores

If we want to find the proportion of homes that cost over $110,000 or under $100,000 for 1800 sq ft homes, first we need to find the price for 1800 sq-ft homes using the regression equation \(y=23000 + 47x\), where x = 1800. Substituting the value of 'x', we obtain a price of \(y=23000 + 47*1800\$ = \$110600\) . Given the standard deviation (\(\sigma = 5000\)), we are able to calculate the z-score for $110,000 and for $100,000. The formula to find the z-value is \(z= (X-碌)/蟽\), where 'X' is the value for which we find the z-score, '碌' is the mean, and '蟽' is the standard deviation. For $110,000, the z-score is \(( 110000 - 110600 ) / 5000 = -0.12\) . Similarly, for $100,000, the z-score is \(( 100000 - 110600 ) / 5000 = -2.12\)
04

Find the proportion of houses with given price using z-scores

Given a z-score, we can find the proportion of that value in a normal distribution using a standard z-table or calculator. For an area to the right of z (which corresponds to the proportion of homes that cost more than $110,000), one would subtract the z-value from 1 since the total area under the distribution curve is 1. So, the proportion of homes with a cost over $110,000 is \(1 - P(Z < -0.12 )\) . For an area to the left of z (which corresponds to the proportion of homes that cost less than $100,000), directly use the z-value. So, the proportion of homes with a cost less than $100,000 is \(P(Z < -2.12 )\). Thus, we need to check the standard normal distribution table to find the values of these proportions.

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Key Concepts

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

House Price Prediction
House price prediction is a common and important application of linear regression. By using a linear regression equation, we can predict the price of a house based on its features, like its size. In this particular example, the equation given is \[ y = 23,000 + 47x \]where \(y\) represents the house price, and \(x\) corresponds to the size of the house in square feet.
  • The intercept, 23,000, refers to the starting price of a house with 0 square feet, although this is purely hypothetical.
  • The coefficient 47 tells us how much the house price increases, on average, for every additional square foot.

In practical applications, these models help homeowners, sellers, and real estate professionals make informed decisions by understanding how factors such as house size impact pricing.
Regression Coefficient
The regression coefficient is a crucial part of any linear regression model. It quantifies the relationship between variables. Here, the regression coefficient is 47, which can be interpreted as the average increase in house price per additional square foot. Essentially, it serves as the slope of the regression line. An intuitive way to comprehend the regression coefficient is to consider its role:
  • It specifies the amount of change in the dependent variable (house price) for a one-unit increase in the independent variable (house size).
  • In our equation, for every square foot added to a house, the price increases by \(47\, \text{USD}\).
  • This number is derived from historical data, making it a powerful predictor in regressions.

Understanding the regression coefficient helps in predicting future outcomes and analyzing the effect of changing one or more input variables on the output.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of our problem, it illustrates how much individual house prices deviate, on average, from the predicted price given by the regression equation. A smaller standard deviation means that the prices are closer to the predicted values, while a larger one signifies more variation.
  • For our linear regression model, the standard deviation \(\sigma\) is \(5,000\, \text{USD}\).
  • This means that most house prices will fall within \(5,000\, \text{USD}\) of the predicted price.
  • In practical terms, it helps calculate how reliable the prediction is and to what extent one might expect actual prices to fluctuate around predicted values.

Standard deviation is vital for understanding prediction intervals and making informed predictions on varied data sets.
Z-score Calculation
The z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It is expressed as the number of standard deviations a particular value is from the mean. Z-scores are instrumental when deciding the proportion of data points within certain thresholds.
  • To calculate a z-score, the formula used is \( z = \frac{X - \mu}{\sigma} \), where \(X\) is the value being evaluated, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • For example, a house priced at \(110,000\, \text{USD}\) results in a z-score of \(-0.12\), meaning it is 0.12 standard deviations below the mean.
  • Z-scores help assess what proportion of homes are priced above or below specified price thresholds by referring to standard normal distribution tables.

They are invaluable for determining probabilities in normally distributed data and understanding how unusual a data point is in relation to the overall distribution.

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Most popular questions from this chapter

A sample of \(n=61\) penguin burrows was selected. and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one ("Effects of Substrate on the Distribution of Magellanic Penguin Burrows," The Auk [1991]: 923-933). The equation of the least-squares line was \(\hat{y}=11.607-\) \(1.4187 x\), and \(r^{2}=.386\). a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{e}=2.35, \bar{x}=4.5\), and \(\sum(x-\bar{x})^{2}=250\), predict trail length when soil hardness is \(6.0\) in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0 ?\) Explain your reasoning.

An experiment to study the relationship between \(x=\) time spent exercising (min) and \(y=\) amount of oxygen consumed during the exercise period resulted in the following summary statistics. $$ \begin{aligned} &n=20 \quad \sum x=50 \quad \sum y=16,705 \quad \sum x^{2}=150 \\ &\sum y^{2}=14,194,231 \quad \sum x y=44,194 \end{aligned} $$ a. Estimate the slope and \(y\) intercept of the population regression line. b. One sample observation on oxygen usage was 757 for a 2 -min exercise period. What amount of oxygen consumption would you predict for this exercise period, and what is the corresponding residual? c. Compute a \(99 \%\) confidence interval for the true average change in oxygen consumption associated with a 1 -min increase in exercise time.

If the sample correlation coefficient is equal to 1 , is it necessarily true that \(\rho=1 ?\) If \(\rho=1\), is it necessarily true that \(r=1 ?\)

The employee relations manager of a large company was concerned that raises given to employees during a recent period might not have been based strictly on objective performance criteria. A sample of \(n=20\) employees was selected, and the values of \(x\), a quantitative measure of productivity, and \(y\), the percentage salary increase, were determined for each one. A computer package was used to fit the simple linear regression model, and the resulting output gave the \(P\) -value \(=.0076\) for the model utility test. Does the percentage raise appear to be linearly related to productivity? Explain.

Give a brief answer, comment, or explanation for each of the following. a. What is the difference between \(e_{1}, e_{2}, \ldots, e_{n}\) and the \(n\) residuals? b. The simple linear regression model states that \(y=\alpha+\beta x\) c. Does it make sense to test hypotheses about \(b\) ? d. SSResid is always positive. e. A student reported that a data set consisting of \(n=6\) observations yielded residuals \(2,0,5,3,0\), and 1 from the least-squares line. f. A research report included the following summary quantities obtained from a simple linear regression analysis: $$ \sum(y-\bar{y})^{2}=615 \quad \sum(y-\hat{y})^{2}=731 $$
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