Exam 1 Key
MA 223
1. Recall IID is “Independent, Identically Distributed”.
Data Set 1: Plausibly IID..
Data Set 2: Obvious increase in variance—fails the “identical” part.
Data Set 3: Obvious dependence—repeated plus/minus oscillation.
Data Set 4: Obvious trend—fails the “identical” part.
2. (a) Note that
(
)
100
100
100
1 ∑
1 ∑
1 ∑
ȳ =
yi =
(xi − 10) =
xi − 20 = x̄ − 10.
100 i=1
100 i=1
100 i=1
So ȳ = 73.
(b) Note that
z̄ =
100
100
100
1 ∑
1 ∑
1 1 ∑
1
zi =
xi /10 =
xi = x̄.
100 i=1
100 i=1
10 100 i=1
10
So z̄ = 8.3. Then
σz2
100
100
100
1 ∑
1 ∑
1 1 ∑
1 2
2
2
=
(zi − z̄) =
(xi /10 − x̄/10) =
(xi − x̄)2 =
σ .
99 i=1
99 i=1
100 99 i=1
100 x
That is, σz2 = 20.5/100 = 0.205.
3. (a) Eyeball it—about 620.
(b) About 470 to 740.
(c) Again, eyeball. Quartile one is at around 550 or so, quartile three at 670, IQ is
around 120.
(d) The fraction of data below 550 is about 0.25 or 25 percent, way over five percent.
(e) The plot is quite linear, so normality is supported. In fact, the y intercept at 612
is a good measure of the approximate median.
4. (a) The odds of being put on hold all 5 times is (0.8)5 = 0.32768.
(
)
5
(b) The odds of being put on hold k times is
(0.8)k (0.2)5−k . The odds of being
k
put on hold twice in five calls is obtained by using k = 2 and yields 0.0512.
5. (a) Given that r = 0.98 we find that r2 = 0.982 = 0.9604. That is, 96.04 percent of
the output variation is explained by the regression.
(b) An obvious pattern to the residuals—positive, zero to negative, then positive
again. The IID assumption is violated, and the underlying relationship between
x and y is clearly not linear.
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