Question

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

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

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.

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.

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 \(z= (X-µ)/σ\), where 'X' is the value for which we find the z-score, 'µ' is the mean, and 'σ' is the standard deviation. For $110,000, the z-score is \(( 110000 - 110600 ) / 5000 = -0.12\) . Similarly, for $100,000, the z-score is \(( 100000 - 110600 ) / 5000 = -2.12\)

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.