Economics 203 Midterm 1 Solutions — Spring 2010 — Form A
1. A. Mean, median, and mode are the three common measures of central tendency, or “where the middle of
the data lies”.
2. B. The empirical rule says that approximately 68% of observations fall within one standard deviation of the
mean, and approximately 95% within two standard deviations, and virtually all of the observations will be
within three standard deviations of the mean. Using symmetry we infer that 34% of the stores will be between
1 and 0 standard deviations, or between 19 and 23 employees. Similarly, approximately 47.5% of the stores
will be between 0 and +2 standard deviations, or between 23 and 31 employees. Adding these gives 81.5%.
3. E.
H
0
:
σ
2
≥
7
.
4 vs.
H
1
:
σ
2
<
7
.
4 would also be correct.
4. C. The test statistic is
χ
2
=
(
n

1)
s
2
σ
2
=
(50

1)(5
.
1)
7
.
4
= 33
.
770
5. B. The test statistic is
z
=
(71

64
.
5)

5
√
4
2
/
14+3
2
/
17
= 1
.
1599
6. B.
α
is the probability of rejecting the null hypothesis when it is true. A higher
α
means we reject the null
hypothesis more often, but we make the mistake of rejecting it when it is true more often also.
7. A. The onetailed test referred to in the output shows a signiﬁcant diﬀerence; this refers to 1. The twotailed
test is not signiﬁcant, as the pvalue given is greater than 5%. So we do not reject the null hypothesis in 2.
¯
x
1
>
¯
x
2
, so we would not reject the null in 3 at any sensible level of
α
.
8. D. First compute the grand mean:
¯
¯
x
=
n
1
¯
x
1
+
n
2
¯
x
2
+
n
3
¯
x
3
+
n
4
¯
x
4
n
1
+
n
2
+
n
3
+
n
4
=
30(3
.
52)+16(3
.
4)+20(2
.
8)+24(3)
30+16+20+24
=
288
90
= 3
.
2. Then
the sum of squares for treatments is
SST
=
n
1
(¯
x
1

¯
¯
x
)
2
+
n
2
(¯
x
2

¯
¯
x
)
2
+
n
3
(¯
x
3

¯
¯
x
)
2
+
n
4
(¯
x
4

¯
¯
x
)
2
=
30(3
.
52

3
.
2)
2
+ 16(3
.
4

3
.
2)
2
+ 20(2
.
8

3
.
2)
2
+ 24(3

3
.
2)
2
=
7.872
9. C. The ANOVA test is always a onesided greater than test, so we use
α
, not
α/
2. The degrees of freedom
are
k

1 and
n

k
. Hence the correct choice is
F
0
.
05
,
3
,
86
10. B. Because this is an unknown but equal variance ttest, we ﬁrst calculate the pooled variance:
s
2
p
=
(20

1)
s
2
1
+(10

1)
s
2
2
20+10

2
= 1660. Then the test statistic is
t
=
123

141
√
1660(1
/
20+1
/
10)
=

1
.
1407
11. C. For the pooled variance ttest, the degrees of freedom are simply
n
1
+
n
2

2, or 28.
12. D. This is a onesided test, with a rejection region to the right. The test statistic is negative, however, so the
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