.69. Suppose that the linear regression model E{y) =

a + fix with normality and constantstandard deviation a is truly appropriate. Then the interval of numbers Roughly, the correlation is the average crossproduct of the z-score for x times the z-score for y.

Using this formula, explain why

(a) the correlation has the same value when x predictsy as when y predicts x,

(b) the correlation does not depend on the units of measurement. (Note: For the population, the correlation is often defined as Covariance ofx and y

(Standard deviation of x)(Standard deviation of y)

The covariance between x and y is the average of the cross-products (x - ixx){y - /Xy) about the population means.)

y ± t.oz5s

(A- -

- *)2 predicts where a new observation on y will fall at that value of x. This interval, which for large n is roughly y ± 2s, is a 95% prediction interval for y.

To make an inference about the mean ofy (rather than a single value of y) at that value of x, one can use the confidence interval y ± tsnssJ -

(-^ -

SCx - xpFor large n near x this is roughly y ± 2syjl/n. The r-value in these intervals is based on df = n — 2.

Most software has options for calculating these formulas. Refer to the "house selling price" data file at the text Web site.

(a) Using software, find a 95% prediction interval at house size x = 2000.

(b) Using software, find a 95% confidence interval for the mean selling price at house size x = 2000.

(c) Explain intuitively why a prediction interval for a single observation is much wider than a confidence interval for the mean.

(d) Explain how prediction intervals would likely be in error if, in fact, (i) the variability in housing prices tends to increase as house size increases, (ii) the response variable is highly discrete, such as y = number of children in Exercise 9.30.