1. Which is wider (all else being equal), a 95% or a 90% confidence interval?
A 95 percent CI is wider.
2. Students were tested in their statistics knowledge, then given a review sheet, after
which they were tested again. The results are below:
ID number before review after review
3472
52
58
4381
58
49
44
51
7724
3737
67
66
39
48
3465
5472
51
61
increase
6
-9
7
-1
9
10
(a) If you were hoping that the review made the test scores go up, what would the
null and alternative hypotheses be for a hypothesis test?
Let µ be the average increase.
H0 : µ ≤ 0
H1 : µ > 0
(b) Test this hypothesis.
The mean difference is d = 3.667 and the standard deviation is s = 7.312. Of
course the sample size is n = 6. Our t statistic is
d−0
3.667 − 0
√ = 1.23.
=
s/sqrtn
7.312/ 6
Looking this up on the T -table with n − 1 = 5 degrees of freedom we get an
upper tail area of .1367, which since this is a one sided test is the same as the
p-value. Thus we fail to reject H0 with a p-value of 0.1367 and find no evidence
of improvement.
(c) Give a 95% confidence interval for how much scores went up.
Note this should be a 1-sided confidence interval since we want the score increase.
s
7.312
µ > d − 2.02 √ = 3.667 − 2.02 √ = −2.36
n
6
Thus we are 95% confident that scores increase at least −2.36 points.
3. Students in a MATH 1343 class can be sorted into two groups, “Fools,” and “Uncools.”
The Uncools think they’re smarter than the Fools, while the Fools think they’re hipper
than the Uncools. There are 17 Fools and 14 Uncools in the class.
(a) The class is given a statistics practice test. Because the math department gives
this test every semester they know the standard deviation of the scores is 3.2
points no matter who takes the test. The average score for the Fools is 63.2, while
the average for the Uncools was 71.5. If you wanted to set up a hypothesis test to
see whether the Uncools are really smarter, what would your null and alternative
hypotheses be?
H0 : µuncools ≤ µf ools
H1 : µuncools > µf ools
(b) Test your hypothesis above.
(71.5 − 63.2) − 0
q
= 7.19
3.22
3.22
+ 14
17
We look this up on the normal table, and note that it is off the end of the table.
Thus our p-value will be less than 0.0001, so we reject the null hypothesis and
conclude that Uncools are smarter than Fools (i.e. their average test scores are
higher).
(c) Give a 90% confidence interval for the difference in average scores.
Note that this is again a one-sided CI.
r
3.22 3.22
+
= 6.39
(71.5 − 62.2) − 1.65
17
14
We conclude that Uncools score at least 6.39 points higher than fools on this test.
(d) A CS 101 class, for a final class project, comes up with a website “coolornot.com”
that allows users to rate one another’s coolness on a 0-10 scale, kind of like “hotornot.com” purports to do for looks. In spite of obvious design flaws (no 0 score
for CS majors, no 11 score for Fonzie) the Fools in the MATH 1343 class decide
to use the site to embarrass the Uncools. If you wanted to help the Fools set up a
hypothesis test to test their belief, what would the null and alternative hypotheses
be?
We let µ denote the average “coolornot.com” score, with a subscript indicating
the group.
H0 : µf ools − µuncools ≤ 0
H1 : µf ools − µuncools > 0
(e) The average coolornot score for the Fools was 7.4, with a standard deviation of
1.3, while the average coolornot score for the Uncools was 6.2, with a standard
deviation of 1.8. Test the hypothesis above.
We need
16(1.3)2 + 13(1.8)2
s2p =
= 2.38
17 + 14 − 2
Now we have
(7.4 − 6.2) − 0
q
= 2.16.
2.38
2.38
+ 14
17
We look this up on a T -table with 17 + 14 − 2 = 29 degrees of freedom and find
a 0.9804 lower tail area, and hence a 0.0196 upper tail area. Our p-value is thus
0.0196 and we reject H0 and conclude that the average coolornot score is higher
for Fools than it is for Uncools.
(f) The divide in the class has now deepened, despite the efforts of the foolish and
uncool professor to mediate, and both sides go looking for support in the larger
campus community. The trouble is, the typical student is not very self-aware.
A sample of 83 students resulted in 56 of them self-reporting their Fool/Uncool
status incorrectly. Give a 90% confidence interval for the percentage of all students
who will self-report their Fool/Uncool status incorrectly.
We have p̂ = 56/83 = .675 as the proportion of the sample incorrectly selfreporting. To compute a 90% confidence interval we do:
r
.675(1 − .675)
= (.590, .760).
.675 ± 1.65
83
Thus somewhere between 59% and 76% of the campus community will self-report
incorrectly.
(g) Fools accuse Uncools of improperly labeling themselves “cooler than the other
side of the pillow,” and hence logically not in the Uncool group, with a different
frequency than Fools rate themselves “inteligynt,” and hence logically not in the
“Fool” group. Never able to quite grasp that negative number thing they are
unwilling to risk looking silly making a more precise accusation than that. If you
were to set up null and alternative hypotheses to test this accusation what would
the null and alternative hypotheses be?
Let ρ denote the population proportion of incorrect self-reporters, with a subscript
denoting the group.
H0 : ρuncool − ρf ools = 0
H1 : ρuncool − ρf ools 6= 0
(h) A sample of 48 Fools found 17 of them claimed they were “inteligynt” while a
sample of 53 Uncools found 22 of them claimed to be “cooler than the other side
of the pillow.” Test the hypotheses above.
17+22
Jointly in the sample we have p̂pooled = 48+53
= .386. Separately we have p̂f ool =
17/48 = .354 and p̂uncool = 22/53 = .415. We compute our test statistic
(.415 − .354) − 0
q
.386(1−.386)
48
+
= .63.
.386(1−.386)
53
We look this up on the normal table to find a lower tail area of 0.7357. Thus we
have an upper tail area of 0.2643. Since this is a two-tailed test, we double this
for the p-value of 0.5286. Hence we fail to reject H0 and are unable to detect any
difference in incorrect self-reporting between the two populations.
(i) Compute a 95% confidence interval for the difference.
r
(.415 − .354) ± 1.96
.415(1 − .415) .354(1 − .354)
+
= (−.1288, .2508)
48
53
Not surprisingly, given the outcome of the hypothesis test above, we find a confidence interval that contains zero.
4. A 2009 health and fitness study classified Americans into 4 groups according to their
weekly level of exercise. The groups and study results are below.
ˆ I) Total Couch Potatoes
ˆ II) Lazy Bums
ˆ III) People who exercise less than they say they do
ˆ IV ) People with an unhealthy obsession with exercise
I
II III
118 142 133
IV
122
Test at the 5% level whether you can find unequal proportions of the population falling
into the four groups.
Obs
Exp
O−E
(O − E)2
(O − E)2 /E
I
II
III
IV
118
142
133
122
128.75
128.75 128.75 128.75
−10.75
13.25
4.25
−6.75
15.5625 175.5625 18.0625 45.5625
0.8976
1.3636 0.1403 −.3539
Summing the last row we get 2.7554. We look this up on a χ2 table with 3 degrees
of freedom and find 0.4301. Thus our p-value is 0.4301 and we fail to reject the null
hypothesis that the population is evenly split among the four groups.
5. A more detailed analyis in the study elicited information about people’s honesty in
reporting the number of days per week they exercised, determining whether they A)
exercised less than they claimed to, B) exercised exactly what they claimed to, or C)
exercised more than they claimed to. Results are below. Note the totals for the groups
are the same as above.
I
II III IV Total
A)
22 34
72
5
133
B)
77 66
31 101 275
C)
19 42
30
16
107
Total 118 142 133 122 515
Is there a difference in honesty between the four groups (test at the 1% level).
Expected:
I
II
A) 30.4738 36.6718
B) 63.0097 75.8252
C) 24.5165 29.5029
Observed-Expected:
I
II
A) -8.4738 -2.6718
B) 13.9903 -9.8252
C) -5.5165 12.4971
III
IV
34.3476 31.5068
71.0194 64.1456
27.6330 25.3476
III
IV
37.6524 -26.5068
-40.0194 35.8544
2.3670
-9.3476
(O − E)2 :
I
II
III
IV
A) 71.805
7.139 1417.703 702.610
B) 195.728 96.535 1601.552 1285.538
5.603
87.378
C) 30.432 156.178
(O − E)2 /E:
I
A) 2.356
B) 3.106
C) 1.241
II
0.195
1.273
5.294
III
41.275
21.637
0.203
IV
22.300
20.041
3.316
If we sum these we get 122.237. We will “look this up” on a χ2 table with (R − 1) ×
(C − 1) = 2 × 3 = 6 degrees of freedom. Obviously this number is way of the charts, so
we conclude that the p-value is less than 0.0001. Thus we reject the null hypothesis of
no relationship between the groups and their honesty about their exercise habits and
conclude that there indeed is a difference.
6. The folow-up study also measured the LDL cholesterol levels of people in the study.
Results are summarized below. Note the n for each group is the same as in the above
problems.
group mean LDL sd LDL
I
171
18.1
II
152
15.2
III
133
11.5
IV
102
16.4
Test whether there is a difference in the average LDL level for the four groups.
We need the overall mean. Recall that the group totals are 118, 142, 133, and 122.
x=
118 · 171 + 142 · 152 + 133 · 133 + 122 · 102
= 145.2424
118 + 142 + 133 + 102
We can now compute the group sum of squares:
118·(171−142.2424)2 +142·(152−142.2424)2 +133·(133−142.2424)2 +102·(102−142.2424)2 = 287651.
The error sum of squares will be:
117 · 18.12 + 141 · 15.22 + 132 · 11.52 + 101 · 16.42 = 115529.
We can now construct our anova table:
Source Sum Sq DF Mean Sq
F
Group 287651
3 95883.67 424.1139
Error
115529 511
226.08
Total
403180 514
We “look this number up” on the F -table with 3 numerator and 511 denominator
degrees of freedom. Obviously the p-value is less than 0.01, so we reject H0 and find
that there is indeed a difference in LDL level among the groups.
7. Consider the data
x y
4 -5
3 4
7 21
5 14
1 11
The line y = 2x + 1 is the least squares regression line.
(a) Compute the fitted values and the residuals.
x y ŷ = 2x + 1 = (y − ŷ)
4 -5
9
-14
3 4
7
-3
7 21
15
6
5 14
11
3
1 11
3
8
(b) Compute a 95% confidence interval for m.
We’ll need
sP
r
2i
196 + 9 + 36 + 9 + 64
s=
=
= 10.231.
n−2
3
P
The standard error for the slope requires (xi − x)2 so we compute this. Note
that x = 4.
x (xi − 4) (xi − 4)2
4
0
0
3
-1
1
3
9
7
5
1
1
-3
9
1
sum:
20
We can now compute the standard error for the slope to be
s
pP
= 2.288.
(xi − x)2
Thus a 90% confidence interval for the slope will be (using the T -table with 4
degrees of freedom):
2 ± 2.78 · 2.288 = (−4.36, 8.36).
(c) Based on your interval above, would you reject or fail to reject the null hypothesis
H0 : m = 0?
Since 0 is in the confidence interval, we would not reject H0 .
(d) Predict y for an x-value of 6.
2 · 6 + 1 = 13
(e) Compute a 90% confidence interval for your prediction.
r
13 ± 2.14 · 10.231 1 +
1 (6 − 4)2
+
= (−12.91, 38.91)
5
20