Exam 3 Key
MA 223
1. Note we have large (> 30) sample sizes, so the relevant statistics will be normal rather
than t.
The test statistic is
Z=
√
x̄1 − x̄2 − (µ1 − µ2 )
s
where s = s21 /n1 + s22 /n2 (the “pooled” variance). This statistic will be standard
normal—note sample sizes exceed 30. Under H0 the Z statistic is just Z = (x̄1 − x̄2 )/s.
We will fail to reject if −1.96 < Z < 1.96. In this case we obtain Z = −3.83, so we
REJECT H0 . The p-value is off the table, essential ZERO.
The assumptions underlying this test are minimal—iid data, but because the samples
sizes are large we do not need normality of the underlying pdf’s; this also allows us to
approximate the population variances σ1 and σ2 involved with the sample variances s1
and s2 .
2. THIS IS VITAL: You need to run a PAIRED comparison here. An unpaired comparison gives away all of the power of the experiment (and involves slightly different
assumptions)!
(a) The null hypothesis is H0 : µd = 0, where di denotes the difference sample i of
Brand 1 minus sample i of Brand 2 and µd is the difference in the mean life of the
tires (under such identical or “paired” conditions). The alternative is Ha : µd ̸= 0
(a two-sided test).
(b) The differences of the eight tires car-by-car are 4, 6, 0, 2, 7, 0, 4, 1. The appropriate
statistic is
d¯
t= √
s/ 8
which under H0 will be t-distributed with 7 df if the underlying quantity di is IID
normal. From the table we will fail to reject H0 when −2.365 < t < 2.365. In
this case we obtain d¯ = 3.0, s = 2.673, and t = 3.175, so we REJECT H0 ; p value
is about 0.016.
If you run an unpaired comparison, you won’t reject, and can’t see any differences
in the tire lifetimes!
3. (a) Here’s the completed table.
Source df
Factor
2
Error
27
Total
29
SS
MS
F
774.88 387.44 23.2
450.90 16.70
1225.78
p-value
0.000
(b) The p-value for this f -ratio is less than 0.05, practically zero, indicating the null
hypothesis that all treatments are the same should be REJECTED. You don’t
need to fill any of the table out to see this.
(c) The statistic t = √ȳ2 −µ2
M SE /n
should be t-distributed with 27 df (note µ2 = µ + τ2 ).
We should thus have −2.052 < t < 2.052 for 95 percent of such experiments. In
this case we obtain confidence interval
√
√
ȳ2 − 2.052 M SE /n < µ2 < ȳ2 + 2.052 M SE /n
which becomes, after we fill in M SE = 16.7, n = 10, and ȳ2 = 92.5,
89.85 < µ2 < 95.15.
(d) The essential assumptions are that the data can be modelled as yij = µ + τi + eij
where the eij are N (0, σ 2 ) and independent.
4. (a) Let πk for 1 ≤ k ≤ 6 be the probability of face k appearing in a random role.
Then
H0 : πk = 1/6, 1 ≤ k ≤ 6
is the null hypothesis versus
Ha : at least one πk ̸= 1/6.
(b) In each case the expected number under H0 is 20. The appropriate statistic is
∑
χ2 = ((expected − actual)2 /expected) with 5 df. The critical value is χ2 > 11.0
to reject. For our data
χ2 = (82 + 102 + 52 + 32 + 22 + 82 )/20 = 13.3
We REJECT H0 . The p-value is about 0.021.