Stat 571 Second Midterm Exam November 20, 2007
1(a) df= 10+ (13−1) −2 = 20 since one zero is dropped in
the second sample. With a 2-sided test .02 < p < .05
and we conclude that the two groups have different
variances (reject “σ1=σ2”).
(b) Using the t-test that does not require equal variance
we get s2
¯y1−¯y2=8.062
10 +2.052
13 = 6.82 and t= (90.30 −
86.65)/√6.82 = 1.398. Using df= 9 and a 2-sided test
we get .10 <p<.20. We fail to reject “µ1=µ2”:
there is no significant difference between the average
size of the two Lyrebird species.
2(a) the Mann-Whitney test p-value is 0.028 because the
ranks do not change. The t-test p-value is smaller
than 0.006 because only ¯y1−¯y2changes (increases).
Sample sizes (4) and variability (s1and s2) stay the
same. So tincreases and the p-value decreases.
(b) Data set 3 (top). Since the ranks are the same in
data set 3 and in data set 1, the p-value from Mann-
Whitney test is 0.028, which is < .05. With data set
4 the rank sums would be 23 and 13, and 13 is too
big to get significance at the .05 level (the table says
we need 10 or less). Note: data set 3 returns p > .05
with a t-test because the variability in sample “b” is
large. Data set 4 also returns p>.05 with a t-test
because the difference in sample means is small.
(c) For data set 3, Mann-Whitney test would be most
appropriate, because we are not sure the data are
normally distributed (presence of an ’outlier’ in sam-
ple “b”) and the sample size in each sample is small
(4). For data set 4, a t-test would be most appro-
priate, because the normal distribution seems appro-
priate in each sample and the t-test is more powerful
than Mann-Whitney test when both can be used.
3(a) Smaller with a one-sided test in the correct direction.
The formula for nis σ2(zα/2+zβ)2/(µ0−µa)2for a
2-sided test, while we replace zα/2=z.015 = 2.17
by zα=z.03 = 1.88 for a one-sided test. zαbeing
smaller, nis also smaller for a one-sided test.
(b) Using n=σ2(zα/2+zβ)2/(µ0−µa)2we get 25 =
104 ∗(2.17 + zβ)2/(0 −5)2i.e. zβ=√6.01 −2.17 =
0.281 and with Table A β=.3897 so the power is
61%. One can get the same result by determining the
rejection region: outside 0±2.17∗p104/25 = ±4.426.
Then the power is P{Z < −9.426/2.04}+P{Z >
−0.574/2.04}= 0 + P{Z > −0.281}= 1 −0.3897 =
0.61.
4(a) ˆpA= 63/220 = .2864, ˆpC= 59/80 = .7375 and
ˆp= (63 + 59)/300 = .4067. We can approximate the
binomial with a normal distribution because 220∗ˆp=
41.3>5 and 220 ∗ˆq= 178.7>5 and similarly
80 ∗ˆp= 37.4>5 and 80 ∗ˆq > 5. Then z= (.7375 −
.2864)/p.4067 ∗.5933 ∗(1/220 + 1/80) = 7.034. Us-
ing the normal distribution we get p < 2∗2.9∗10−7is
way smaller than .0001 and way smaller than α=.10.
We strongly reject pA=pC: there is strong evidence
that the locus is linked to some genetic region affect-
ing flowering time, the C allele being linked to early
flowering. Assumptions include random sampling of
and independence among plants.
(b) Observations are clearly paired: each leaf observa-
tion is paired to the root observation made on the
same plant. Therefore, one would use a two-sided
paired-sample t-test, provided that the distribution
of difference (leaf expression -root expression) is not
too far from a normal distribution. Note: there is no
need to assume that σ1=σ2and no need to assume
normality of gene expression. Only the normality of
the expression difference is needed.
5(a) false: we mostly use a t-distribution. The normal
distribution is only used when the variance is known.
true
true because z-quantiles are smaller than t-quantiles.
false: the confidence interval is centered at the sam-
ple mean (¯y, which is random) while the not-rejection
region is centered at the null hypothesis µ0(not ran-
dom - known before collecting the data)
true
true: they both have the same length, which is twice
zα/2∗σ/√n.
Summary of grades:
Frequency
20 40 60 80 100
0 10 20 30 40
●● ●● ●● ●
76 83 89
1