Oct. 28, 2003
ECON 240A-1
Midterm
L. Phillips
1. (15) According to TNS International, 69% of wireless Web users use it primarily for
receiving and sending e-mail. Suppose that three wireless Web users are selected at
random.
a. What is the probability that all of them use it primarily for e-mail?
b. What is the probability that none of them use it primarily for e-mail?
c. What is the purpose of selecting them at random?
This is a binomial problem and could be solved with a branching diagram or using the
binomial probability distribution:, p(k succeses) = [n!/(k!n-k!)] pk (1-p)n-k
0.69
0.69
0.69
Primarily e-mail
Primarily e-mail
Primarily e-mail
Etc.
a. by independence, p =(0.69)(0.69)(0.69) = 0.328 = [3!/(3!0!)] (0.69)3 (0.31)0
b. ditto, p = (0.31)(0.31)(0.31)=0.03 = [3!/(3!0!)] = (0.69)0 (0.31)3
c. Then the outcome for each user is independent of the others and the result is
the product of individual probabilities.
2. (15) The natural remedy echinacea is reputed to boost the immune system, which will
reduce flu and colds.. A 6-month study was undertaken to determine whether the remedy
works. From this study, the following probability distribution of the number of
respiratory infections per year, x, for echinacea users was produced.
X, infections/year
Frequency or probability
0
0.45
1
0.31
2
0.17
3
0.06
4
0.01
a. Are the numbers in this table consistent with a probability distribution?
Explain.
b. What is the probability an echinacea user has more than one infection per
year?
c. What is the probability an echinacea user has no infections per year?
a. Yes the frequencies are non-negative and sum to one.
b. P(x>1) = 0.17+0.06+0.01=0.24
c. P(x=0) = 0.45
3. (15) According to the American Academy of Cosmetic Dentistry, 75% of adults
believe that an unattractive smile hurts career success. Suppose that 25 adults are
randomly selected. What is the probability that 15 or more of them would agree with
the claim?
Oct. 28, 2003
ECON 240A-2
Midterm
L. Phillips
This problem can be solved using the Tables for the cumulative binomial
probability distribution in the Appendix, or by using the normal approximation. From
Table 1, p. B-5, for n = 25, p(success) = 0.75: p(k<=14) = 0.03, so p(k>14) = 0.97 since
these probabilities must sum to one. Alternatively, using the normal approximation, with
mean = n*p = 25*0.75 = 18.75. and variance = n*p*(1-p) = 25* 0.75 *0.25 = 4.6875, so
the standard deviation is the square root of this number, i.e. 2.165. So the standardized
normal variate, z, for k=15 is: z = (15 – 18.75) /2.165 = - 1.73. From Appendix B, Table
3, p. B-8, for the normal distribution, z = 1.73 has an area (probability) above the mean of
0.4582 so 0.5 minus 0.4582 = 0.04 is in the upper tail implying that p(z<-1.73) = 0.04 and
so p(z>-1.73) = 0.96 = p(k>=15), close to the answer from Table 1.
4. (15) Suppose the amount of time teenagers spend weekly at part-time jobs is normally
distributed with a standard deviation of 20 minutes. A random sample of 100
observations is drawn and the sample mean is computed as 125 minutes. Determine the
95% confidence interval estimate of the population mean. p(-1.96<z<1.96} = 0.95, where
z = (125 – u)/(std.dev./10) = (125-u)/(20/10) = (125-u)/2, so
p(-1.96<[125-u]/2<1.96)=0.95, or p(-3.92<125-u<3.92)=0.95, and subtracting 125 from
all three parts of the inequality, p(-3.92-125<-u<3.92-125) or p(-128.92<-u<-121.08=.95
and multiplying all three parts of the inequality by minus one, changing the signs of the
inequality, p(128.92>u>121.08) = 0.95.
5. (15) The University of California’s share, in percent, of California General Fund
expenditure is plotted against a time index (t=0 is 1968-69, t=35 is 2003-04) in Figure 5.1
below. A linear trend line was used to fit this data, with parameter estimates and
goodness of fit appearing on the chart. The Excel regression output appears in Figure 5.1.
A plot of actual, fitted, and residual from Eviews is displayed in Figure 5.2
a. Do you think this trend-line is very useful for predicting UC’s budget share
for the next fiscal year, 2004-05? Explain.
b. How would you interpret the fitted parameter 0.0682?
c. How much do you think UC’s share of the pie (budget share) might shrink in
ten years, i.e. by 2013-14?
d. Is this linear trend model of UC’s Budget Share statistically significant?
Explain.
e. Are any of the assumptions underlying ordinary least squares violated?
a. No. This model captures the long run trend but not the shorter run cycle:
conceptual general time series model: time series = trend + cycle + seasonal +
random. There is no seasonal since this is annual data, and the error term in
the regression model is consequently capturing cycle plus random since only
trend is explicitly modeled.
b. It is the fitted UC budget share in 1968-69, i,e. at t=0, y(0) =0.0682.
Oct. 28, 2003
ECON 240A-3
Midterm
L. Phillips
c. Relying on the regression estimate of the slope of –0.0008, ten times this
would mean a decline of about 0.8 of one percent in budget share, down to
about 3%.
d. Yes, the F-statistic of 115.8 is large and highly significant. The t-statistic for
the estimate of the slope is also large, -10.76, and highly significant. For this
bivariate regression, the square of this t-statistic is the F-statistic.
e. From Figure 5.2, the estimated residual is autocorrelated, so the assumption
that the errors are independent, i.e. E[e(i)e(j)] = 0 is violated.
Figure 5.1: Linear Trend Fit to UC’s Share of California General Fund Expenditures
University of California's Share of State General Fund Expenditure
8.00%
7.00%
6.00%
Percent
5.00%
4.00%
3.00%
y = -0.0008x + 0.0682
R2 = 0.773
2.00%
1.00%
0.00%
0
5
10
15
20
Year( 0 = 1968-69)
25
30
35
40
Oct. 28, 2003
ECON 240A-4
Midterm
L. Phillips
Table 5.1: Regression Results from the Linear Trend Model of UC’s Budget Share
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.879206
R Square
0.773003
Adjusted R Square 0.766326
Standard Error
0.004834
Observations
36
ANOVA
df
Regression
Residual
Total
Intercept
X Variable 1
SS
MS
F
Significance F
1 0.002706 0.002706 115.781431
1.729E-12
34 0.000795 2.34E-05
35 0.003501
Coefficients
Standard Error t Stat
0.068168 0.001578 43.1858
-0.00083 7.76E-05 -10.7602
P-value
Lower 95% Upper 95% Lower 95.0% Upper 95.0%
2.7345E-31 0.064959709 0.071375 0.06495971 0.07137538
1.729E-12 -0.000992206 -0.00068 -0.00099221
-0.000677
F igure 5. 2: Actual, F itted and Residual-Linear T rend Model of UC's Budget Share
8
7
6
5
4
1.0
3
0.5
0.0
-0.5
-1.0
-1.5
70
75
80
Residual
85
90
Actual
95
00
Fitted