Economics 202
Spring 2007
Homework 10
Book problems: 12.10, 12.22, 12.32, 12.34, 12.48, 12.62
12.10
a. r = -0.23 states that there is a weak inverse linear relationship between account balance
and number of withdrawals during the month.
b. H0: ρ = 0
H1: ρ≠ 0
Decision rule: Reject H0 if |t| > 1.6759
t = (-.23)/√(1-(-.23)2/(50-2) = -1.63
Since t is less than the critical value (in absolute value), conclude that the population
correlation between account balance and number of withdrawals is not significantly
different from zero (or that the variables are not linearly related).
c. Type II error… not rejecting the null hypothesis when it is false.
12.22
a. The slope coefficient says that a 1 mile increase in the distance from headquarters
means an average reduction of $10.12 in dollar volume of business.
b. z = -10.12 – 0 / 3.12 = -3.24 (this is a large sample; a t value is ok, too)
The critical value of z is 2.33, so reject the null hypothesis and conclude that the slope
coefficient is significantly different from zero, which means that the farther a customer is
from corporate headquarters, the lower the dollar volume of sales.
12.32
a. 40,000 ± (2.4469)(√145.40/6)√1/8 + (0)2/134,679
40.000 ± 4.258
39,995.74-40,004.25
b. 36,320 ± (2.4469)(√145.40/6)√1 + 1/8 + (0)2/134,679
36,320 ± (2.4469)(4.9227)(1.06)
36,320 ± 12.77
36,315.74-36,324.25
c. 36,320 ± (2.4469)(√145.40/6)√1 + 1/8 + (43,000-40,000)2/134,679
36,320 ± 99.29
36,307.22-36332.77
The precision of the estimate decreases as the value of xp moves away from the mean (in
either direction. The most precise estimate (of either the average y or a particular y) is at
the mean of the x’s.
12.34
a. 0.124 ± (1.9840)(2.88)
-5.58992-5.837
b. Since zero is within the confidence interval, the value of β1 could be zero. It is thus
possible that there is no relationship between age and speed.
12.48
a. H0: ρ = 0
H1: ρ≠ 0
Decision rule: Reject H0 if |t| > 2.0484 (I used a 5% level of significance.)
t = (.67)/√(1-(.67)2/(30-2) = 4.7757
Reject the null hypothesis and conclude that sales and advertising dollars are positively
linearly related.
b. Since this is a simple (one variable) model, the coefficient of determination would be
(.67)2 = .4489. Thus variation in advertising would explain about 45% of the variation in
sales.
12.62
a. t = (0.028)/[(0.20)/√148,885.73] = 54.022
At any significance level, you would reject the null hypothesis and conclude that as test
score increases, GPA increases.
b. First, figure out the predicted GPA: y = 1 + 0.028(80) = 3.24
3.24 ± (1.6479)(.20)√1 + 1/400 + (80-68)2/148,885.73
3.24 ± (1.6479)(.20)(1.005)
3.24 ± .33
2.91-3.57
We are 90% confident that a specific student who scores 80% on the test will have a GPA
between 2.91 and 3.57.
c. It depends on whether you view 2.90 as being in the prediction interval or not. If so, I
would conclude that the student’s GPA reflects an average grading standard. If it is
below prediction interval, I would conclude (marginally) that the student’s GPA reflects a
relatively higher grading standard than students at other schools with an equivalent GPA.
(Technically, this score is below any value in the 90% confidence interval, but if you had
used a 95% interval, it would be within it.)
d. First, figure out the predicted GPA: y = 1 + 0.028(65) = 2.82
2.82 ± (1.6479)(.20)√1 + 1/400 + (65-68)2/148,885.73
2.82 ± (1.6479)(.20)(1.005)
2.82 ± .33
2.49-3.15
The student’s GPA again falls below the prediction interval. You could again conclude
that the student’s GPA reflects a relatively higher grading standard than students at other
schools with an equivalent GPA.