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Homework on chapter 3
Introduction to Econometrics
Please, box your answers
1. 15 points
Suppose a distribution’s mean is 100 and variance 43. In a simple random sample of size …
(a) …n=100, find Pr( Y < 101)
The sample mean’s SE = (43/100)½ = 0.656, and Pr( Y < 101) = Pr(( Y - 100)/SE <
1/.656) ~ Φ(1.52) ~ .936
(b) …n=64, find Pr(101 < Y < 103)
The sample mean’s SE = (43/64)½ = 0.82, and Pr(101 < Y < 103) = Pr(1/.82 <Z<
3/.82) ~ Φ(3.66) - Φ(1.22) ~ 1 - .8888 = .1112
(c) …n=100, find Pr( Y > 98)
The sample mean’s SE = (43/100)½ = 0.656, and Pr( Y > 98) ~ Pr(Z > -2/.656) =
Pr(Z < 2/.656 = 3.05) = Φ(3.05) ~ .9988 ~ 1, to 4 decimals
2. 25 points, 5 each
In a survey of 400 prospective voters, 215 would vote for incumbent and 185 for challenger. Let
p be the fraction of all prospective voters who preferred the incumbent at the time of the survey,
and let p^ be the survey’s fraction for the incumbent. (Hint: Bernoulli)
(a) Use the survey’s results to estimate p.
p
^ = 215/400 = .5375
(b) Following slide 17, ch3 use this formula for the standard error: SE=(p^ (1-p^ )/n)½ .
SE = (.5375*.4625/400)½ = 0.02493
(c) What is the p-value for the test H0:p=.5, H1:p≠.5 , given this sample? Interpret this
number.
The p-value is defined in terms of the t-statistic t = (.5375-.5)/.02493 = 1.504 as
p-value = 2Φ(-|t|) = 2Φ(-1.504) ~ 2·.0662 = .1324
There is not enough statistical evidence to reject the null at the 5% significance level
(d) What is the p-value for the test H0:p=.5, H1:p>.5 , given this sample?
It is p-value = 1 - Φ(t) = 1 - Φ(1.504) ~ 1 - .934 = .066
(e) Did the survey contain statistically significant evidence (at 5% level) that the incumbent
was ahead of the challenger at the time of the survey? Explain.
No, the p-value of .066 exceeds .05, the significance level. (Or t-statistic fails to
exceed critical value 1.64.)
3. 15 points
Using the data in the above exercise,
(a) Construct a 95% confidence interval for p.
.5375 ± 1.96*SE with SE = 0.02493 gives interval [.489,.586]
(b) Construct a 99% confidence interval for p.
.5375 ± 2.58*SE with SE = 0.02493 gives interval [.473,.602]
(c) Without any additional calculations, test the hypothesis H0:p=.5, H1:p≠.5 at the 5%
significance level.
Not rejected since .5 belongs in (a)’s interval.
4. 10 points
Data on fifth-grade test scores for 420 school districts in CA yield Y = 646.2 and sY = 19.5.
(a) Construct a 95% confidence interval for the mean test score in the population. (Slide 14)
646.2 ± 1.96*SE with SE = 19.5/√420 = .952 gives interval [644.33,648.07]
(b) When the districts were divided into those with small classes (<20 students/teacher) and
those with large classes (>20 students/teacher), the following was found:
Class size
Mean score ( Y )
St. dev. (sY)
N
Small
Large
657.4
650
19.4
17.9
238
182
Is there statistically significant evidence that districts with smaller classes have higher mean
test scores? Use a one-sided test and explain.
The critical value of the t-statistic for the null of no difference versus the alternative
that difference is positive is t > t*=1.64, at the 5% significance level. To compute it,
t≡
( 657.4 − 650 ) − 0 ≈ 4.048...SE ≡
1.8281
19.42 17.9 2
+
= 1.8281
238
182
Since 4.048 > 1.64, we reject the null in favor of the alternative that districts with small
class sizes have a mean test score that exceeds that of those with large class sizes.
5. 20 points
Grades on a standardized test are known to have a mean of 1,000 for students in the US. 453
randomly selected FL students take the test, yielding sample mean of 1,013 and sample standard
deviation (s) of 108.
(a) Construct a 95% confidence interval for the mean test score for FL students.
(b) 1,013 ± 1.96*SE with SE = 108/√453 = 5.07 gives interval [1003.06,1022.94]
(c) Is there statistically significant evidence that FL students perform differently than US
students?
Yes, since the null of no difference is rejected at the 5% significance level (interval
excludes US sample mean of 1,000)
(d) Another 503 FL students are randomly selected to take a 3-hour prep course and then the
test. Their average score is 1,019 with a standard deviation of 95.
(i)
Construct a 95% confidence interval for the change in average test score
associated with the prep course.
Here the SE is
952 1082
+
= 6.61
503 453
The interval is (1,019-1,013) ± 1.96*SE = [-7,+19]
. SE =
(ii)
Is there statistically significant evidence that the prep course helped?
No, the interval includes 0, the null difference between the two populations
(FL with no prep, FL with prep).