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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 ft of space? With an additional \(100 \mathrm{sq} \mathrm{ft}\) 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 change in price associated with one extra sq ft of space is \$47. For an additional 100 sq ft, the average increase in price is \$4700. For 1800 sq ft homes, the average price is \$110,600, hence approximately 50% of such homes would be priced over \$110,000 and 100% would be priced over \$100,000.

Step by step solution

01

Interpretation of Coefficients

In a simple linear regression model \(y = a + b x\), where \(y\) is the dependent variable and \(x\) is the independent variable, \(a\) is the y-intercept and \(b\) is the slope of the line. The slope \(b\) tells us the average change in \(y\) for one unit increase in \(x\). For this given model, we have \(y = 23,000 + 47 x\), where \(y\) represents house price and \(x\) is the house size. Thus, the average change in price associated with one extra sq ft of space is the slope, which is \$47.
02

Compute Change for Additional 100 sq ft

To compute the change in price with an additional hundred sq ft of space, we simply multiply the slope by 100. Hence, the average change in price associated with an additional 100 sq ft of space is \(47 * 100 = \$4700\).
03

Compute Price for 1800 sq ft

To compute the price for a house of size 1800 sq ft, we substitute \(x = 1800\) into the equation \(y = 23,000 + 47 x\). This gives \(y = 23,000 + 47 * 1800 = \$110,600.\) This is the average price of a 1800 sq ft house.
04

Proportion Calculation

Since \(\sigma=5000\), which represents the standard deviation, it is assumed that the prices have a normal distribution. A rule of thumb (68-95-99.7 rule) for normal distribution states that 95% of data falls within 2 standard deviations of the mean. Given that the mean price for a 1800 sq ft home is \$110,600, it can be said that 95% of such homes would be priced between \$100,600 and \$120,600. For homes priced over \$110,000, the proportion would be approximately 50% (as \$110,600 is the mean). Additionally, all homes would be priced over \$100,000 as this lower limit falls below our mean minus 2 standard deviations, and the normal distribution curve implies all values are likely within this range.

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Key Concepts

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

Regression Analysis
Regression analysis is a statistical method used to study the relationship between variables. Specifically, in simple linear regression, the focus is on two variables, with the goal to quantify or predict the impact of changes in the independent variable on the dependent variable.

In our exercise, the relationship between house size (independent variable) and house price (dependent variable) is explored. By analyzing real estate data, the simple linear regression model aims to create a formula that accurately predicts house prices based on their size. The linear formula you saw in the problem, \(y=23,000+47x\), is the result of such an analysis.

Moving on, understanding the significance of the coefficients in this formula is part of regression analysis. For example, the coefficient of \(x\), which is 47, informs us that for every one square foot increase in house size, we can expect, on average, a \$47 increase in the price of the house. This understanding is fundamental to utilizing regression analysis in decision-making, be it for pricing strategies, investment considerations, or market analysis.
Dependent and Independent Variables
In any analytical context, distinguishing between dependent and independent variables is critical to understanding the dynamic of the functions or equations at hand.

The independent variable is the factor that we hypothesize to influence, or predict, the outcome - it鈥檚 the cause in a cause-and-effect relationship. In contrast, the dependent variable is the outcome of interest - the effect.

In the provided exercise, house size, measured in square feet \(x\), is the independent variable that predicts or influences our outcome of interest, which is house price \(y\). When we say that house prices depend on their square footage, we are effectively saying that square footage is the independent variable, while the price is the dependent variable. This relationship is core to the study of regression analysis, as it lays the foundation for our predictions and interpretations.
Normal Distribution
The normal distribution is a continuous probability distribution that is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In the context of our exercise, the term \(\sigma=5000\) represents the standard deviation, a measure of the spread of house prices around the mean house price. It's assumed that the house prices follow a normal distribution, meaning most houses will be priced close to the average, with fewer homes at extremely high or low prices.

This concept is crucial when we're calculating proportions of homes in specific price ranges. For instance, we can predict that approximately 68% of the data will be within one standard deviation of the mean, and about 95% will be within two standard deviations, according to the empirical rule. Knowing this, you can make informed estimates about the proportion of homes exceeding or falling below certain prices, like in the solutions provided for the problem regarding the 1800 sq-ft homes and their associated prices of over or under \$110,000 and \$100,000 respectively.

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Most popular questions from this chapter

Exercise \(13.8\) gave data on \(x=\) treadmill run time to exhaustion and \(y=20-\mathrm{km}\) ski time for a sample of 11 biathletes. Use the accompanying MINITAB output to answer the following questions. The regression equation is ski \(=-88.8-2.33\) tread \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & p \\\ \text { Constant } & 88.796 & 5.750 & 15.44 & 0.000 \\ \text { tread } & 2.3335 & 0.5911 & 3.95 & 0.003 \\ s=2.188 & \text { R-sq }=63.4 \% & \text { R-sq }(a d j)=59.3 \% & \end{array}\) Analysis of Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 74.630 & 74.630 & 15.58 & 0.003 \\ \text { Error } & 9 & 43.097 & 4.789 & & \\ \text { Total } & 10 & 117.727 & & & \end{array}\) a. Carry out a test at significance level \(.01\) to decide whether the simple linear regression model is useful. b. Estimate the average change in ski time associated with a 1 -minute increase in treadmill time, and do so in a way that conveys information about the precision of estimation. c. MINITAB reported that \(s_{a+b(10)}=.689\). Predict ski time for a single biathlete whose treadmill time is \(10 \mathrm{~min}\), and do so in a way that conveys information about the precision of prediction. d. MINITAB also reported that \(s_{a+b(11)}=1.029 .\) Why is this larger than \(s_{a+b(10) ?}\)

The authors of the article "Age, Spacing and Growth Rate of Tamarix as an Indication of Lake Boundary Fluctuations at Sebkhet Kelbia, Tunisia" (Journal of Arid Environments [1982]: 43-51) used a simple linear regression model to describe the relationship between \(y=\) vigor (average width in centimeters of the last two annual rings) and \(x=\) stem density (stems/m \(^{2}\) ). The estimated model was based on the following data. Also given are the standardized residuals. \(\begin{array}{lrrrrr}x & 4 & 5 & 6 & 9 & 14 \\ y & 0.75 & 1.20 & 0.55 & 0.60 & 0.65 \\ \text { St. resid. } & -0.28 & 1.92 & -0.90 & -0.28 & 0.54\end{array}\) $$ \begin{array}{lrrrrr} x & 15 & 15 & 19 & 21 & 22 \\ y & 0.55 & 0.00 & 0.35 & 0.45 & 0.40 \\ \text { St. resid. } & 0.24 & -2.05 & -0.12 & 0.60 & 0.52 \end{array} $$ a. What assumptions are required for the simple linear regression model to be appropriate? b. Construct a normal probability plot of the standardized residuals. Does the assumption that the random deviation distribution is normal appear to be reasonable? Explain. c. Construct a standardized residual plot. Are there any unusually large residuals? d. Is there anything about the standardized residual plot that would cause you to question the use of the simple linear regression model to describe the relationship between \(x\) and \(y ?\)

The employee relations manager of a large company was concemed 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.

The article "Effect of Temperature on the pH of Skim Milk" (Journal of Dairy Research [1988]: 277- 280) reported on a study involving \(x=\) temperature \(\left({ }^{\circ} \mathrm{C}\right)\) under specified experimental conditions and \(y=\) milk \(\mathrm{pH}\). The accompanying data (read from a graph) are a representative subset of that which appeared in the article: \(\begin{array}{rrrrrrrrr}x & 4 & 4 & 24 & 24 & 25 & 38 & 38 & 40 \\ y & 6.85 & 6.79 & 6.63 & 6.65 & 6.72 & 6.62 & 6.57 & 6.52\end{array}\) $$ \begin{array}{lrrrrrrrr} x & 45 & 50 & 55 & 56 & 60 & 67 & 70 & 78 \\ y & 6.50 & 6.48 & 6.42 & 6.41 & 6.38 & 6.34 & 6.32 & 6.34 \\ \sum x=678 & \sum y=104.54 & \sum x^{2}=36,056 & \\ \sum y^{2}=683.4470 & & \sum x y=4376.36 & & \end{array} $$ Do these data strongly suggest that there is a negative linear relationship between temperature and \(\mathrm{pH}\) ? State and test the relevant hypotheses using a significance level of \(.01\).

According to "Reproductive Biology of the Aquatic Salamander Amphiuma tridactylum in Louisiana" (Journal of Herpetology [1999]: \(100-105\) ), the size of a female salamander's snout is correlated with the number of eggs in her clutch. The following data are consistent with summary quantities reported in the article. MINITAB output is also included. \(\begin{array}{lrrrrr}\text { Snout-Vent Length } & 32 & 53 & 53 & 53 & 54 \\ \text { Clutch Size } & 45 & 215 & 160 & 170 & 190 \\ \text { Snout-Vent Length } & 57 & 57 & 58 & 58 & 59 \\\ \text { Clutch Size } & 200 & 270 & 175 & 245 & 215 \\ \text { Snout-Vent Length } & 63 & 63 & 64 & 67 & \\ \text { Clutch Size } & 170 & 240 & 245 & 280 & \end{array}\) The regression equation is \(\begin{array}{lrrrr}Y=-133+5.92 x & & & & \\\ \text { Predictor } & \text { Coef } & \text { StDev } & T & P \\ \text { Constant } & 133.02 & 64.30 & 2.07 & 0.061 \\ x & 5.919 & 1.127 & 5.25 & 0.000 \\ s=33.90 & \text { R-Sq }=69.7 \% & \quad R-S q(a d j)=67.2 \% & \end{array}\) Additional summary statistics are $$ \begin{aligned} &n=14 \quad \bar{x}=56.5 \quad \bar{y}=201.4 \\ &\sum x^{2}=45,958 \quad \sum y^{2}=613,550 \quad \sum x y=164,969 \end{aligned} $$ a. What is the equation of the regression line for predicting clutch size based on snout-vent length? b. Calculate the standard deviation of \(b\). c. Is there sufficient evidence to conclude that the slope of the population line is positive. d. Predict the clutch size for a salamander with a snoutvent length of 65 using a \(95 \%\) interval. e. Predict the clutch size for a salamander with snout-vent length of 105 .
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