ECO220Y1 Midterm: ECO220Y1Y UTSG Term Test 3 220 JAN15 Solution
Term Test #3, ECO220Y1Y, January 30, 2015, Prof. Murdock: SUGGESTED SOLUTIONS
(1) (a) As the sample size rises from 20 to 100 the sampling error (aka sampling variability or sampling noise) on the
sample median, a sample statistic, will decrease. Hence the sampling distribution, which the Monte Carlo results
simulate, should be less spread out. Hence the 5th percentile will increase (be less extreme in the left tail), which means
(A) is the correct choice. We would not expect any change in the 50th percentile (the center of the sampling distribution)
as the change in sample size affects the spread not the central tendency, which is why (B) is incorrect. We would expect
the 95th percentile, the standard deviation and the interquartile range in the STATA summary to all DECREASE as the
sampling distribution would be less spread out, which is why (C) – (E) are incorrect answers to the question.
(b) (<4)=?
~[0,10]
[]==+
2=0+10
2=5
[]==(−)
12 =(10−0)
12 =8.3333
[]==5
[]=
=8.3333
21 =0.3968
[]=
√=√8.3333
√21 =0.6299
Given that the population is Uniformly distributed, which is not
too far from Normal, a sample size of 21 is sufficiently large to
apply the Central Limit Theorem (CLT): the sampling
distribution of will be Normal.
(<4)=< 4−5
√0.3968=(<−1.59)
=0.5−0.4441=0.0559
(2) (a)
:=0.10
:>0.10
(b) The significance level is the maximum chance of a Type I error I will tolerate, which in this case would be wrongfully
terminating a grader whose error rate is not above 10% (i.e. inferring that the grader has an error rate above 10% when
in fact s/he does not). This would suggest a low . However, this must be weighed against a Type II, which would mean
4.000
0
Density
3 4 5 6 7
X-bar (minutes)
P(X-bar < 4) = 0.056
Term Test #3, ECO220Y1Y, January 30, 2015, Prof. Murdock: SUGGESTED SOLUTIONS
failing to terminate an error-prone grader. The chance of a Type II error increases as is decreased. The choice of the
significance level reflects a personal decision in light of these competing concerns. Personally, I would probably choose a
significance level of =0.05 as I am concerned about both Type I errors and Type II errors as I would want to use the
most accurate graders to ensure fairness to the students (which means not firing good graders and not failing to fire bad
graders).
(c) Find the rejection region:
(>)=
(>1.645)=0.05
= −
(1−)
1.645= −0.10
0.10(1−0.10)
120
..=0.145
:=0.10(1−0.10)
120 =0.027)
Hence, the unstandardized rejection region is (0.145,∞) (which could also be written as (0.145,1)).
Next, find the probability of making a Type II error (failing to
reject the false null hypothesis). In other words, what is the
chance the sample proportion of marking errors for a grader
with an overall error rate of 15% does not end up in the
rejection region?
<0.145=0.15,=120)=?
<0.145=
< 0.145−0.15
0.15(1−0.15)
120
=(<−0.1534)=0.5−0.06=0.44=
:=0.15=0.15(1−0.15)
120 =0.033)
Hence, the chance that the coordinator will fail to dismiss the incompetent grader is 0.44, which is a pretty high chance
of a Type II error. In other words, the power (0.56=1−) is not particularly high. (Note: If =0.01 is chosen, then
the c.v. 0.164 is, is 0.67 and power is 0.33. If =0.10 is chosen, then the c.v. is 0.135, is 0.32 and power is 0.68.)
0.145
0
Density
0.05 .1 .15 .2 .25 .3
P-hat
alpha = 0.05, c.v. = 0.145
0.145
0
Density
0.05 .1 .15 .2 .25 .3
P-hat
beta = 0.44
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