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Statistical Methods - Math 425/525 (CRN 23436/23445), Winter 2010
Practice Final Answers: March 14, 2010
SHORT CALCULATIONS:
(1) For 10 degrees of freedom, find A so that P (−A < t < A) = .99.
SOLUTION:From table 4, t.005 = 3.169. So this is the value of A.
(2) For 15 degrees of freedom, find A so that P (t < A) = .95.
SOLUTION:From table 4, t.05 = 1.753 so A = 1.753
(3) For 5 degrees of freedom find A so that P (t < A) = .05.
SOLUTION:From table 4, t.05 = 2.015 so A = −2.015.
(4) Let X be normal with mean 20 and standard deviation 4. Find the
interval around the mean of X that contains 85% of the distribution.
SOLUTION:From table 3, P (Z < −1.44) = .075. So P (−1.44 <
Z < 1.44) = .85. X and Z are related by Z = X−20
4 , or equivalently,
X = 4Z + 20. 4 × 1.44 = 5.76. So
P (−5.76 + 20 < X < 5.76 + 20) = P (14.24 < X < 25.76).
(5) If you are doing a hypothesis test at significance level .05, what is
the probability of falsely rejecting H0 ? What is the name of this
type of error?
SOLUTION:The probability of falsely rejecting H0 is .05. This is
called a type I error.
(6) Suppose X is a random variable with mean 19 and standard deviation 8. Let X be the r.v. obtained by taking means of samples of
size 400. What is the mean and standard deviation of X?
SOLUTION:The mean is the same (19). The sample standard
deviation is the population standard deviation divided by the square
root of the sample size, so is 8/20 = .4.
EXPLANATIONS:
(7) Suppose your supervisor asks you if H0 is true. To answer this,
you conduct a hypothesis test and you discover that your p-value is
greater than your significance level α. How should you now answer
your supervisor?
SOLUTION:Our test didn’t give us enough information to show
H0 is false. So it may or may not be true.
(8) Suppose you do a confidence interval estimate with margin of error
.01. You take a sample of 100 and use confidence level 98%, and you
get a margin of error of .03. You wish for a margin of error of .01,
but you actually get an estimate that has a margin of error of .03.
Explain two changes you can make to your procedure to decrease
your margin of error.
SOLUTION:We can either increase our sample size, or we can
decrease our confidence level. Either will decrease the margin of
error.
LONG ANSWER:
(9) Scientists counted the number of sea whelks on the ocean floor at
the coast in .25 square meter samples. They did this in 7 locations
in California and in 6 locations in Oregon. The average number they
found in California were 11.9 with sample standard deviation 2.68,
and in Oregon 26.9 with sample standard deviation 1.56.
Test the hypothesis that there are more sea whelks on the coast in
Oregon than on the coast of California at the .01 significance level.
SOLUTION: We have small sample sizes, and samples from two
populations. So we use the small sample procedure for the difference
of two population means.
Let µ1 be the average population of sea whelks in a .25 square meter
space in Oregon, and µ2 be the same for California.
(a) α = .01. H0 : µ1 = µ2 , Ha : µ1 < µ2 .
(b) We calculate our test statistic. The pooled variance is
5 · 1.562 + 6 · 2.682
= 5.02.
11
Our t-statistic is given by the formula
26.9 − 11.9
t= q
= 12.03
5.02( 16 + 17 )
s2 =
with 11 degrees of freedom.
(c) Consulting table 4, we get a critical value of t.01 = 2.718.
(d) Since 12.03 is greater than the critical value, we reject H0 and
conclude that the density of sea whelks is greater off the coast
of Oregon than off the coast of California if the locations were
chosen randomly.
(10) A school district in Milwaukee assigns children randomly to a Montessori kindergarten and a traditional kindergarten. They measured a
score on a mathematics problem solving test. There was a sample of
30 five year olds from the Montessori school, and their average score
was 19 with a standard deviation of 3.11. There was a sample of 25
five year olds from traditional school, and their average score was 17
with a standard deviation of 4.19.
(a) Give a 95% confidence interval for the average score of the
Montessori students minus the average score of the traditional
students.
SOLUTION: Let µ1 be the mean score of Montessori students,
and µ2 be the mean score of the traditional students.
We first calculate the pooled variance
29 × 3.112 + 24 × 4.192
= 13.242
53
Next we find the appropriate critical value for a 95% confidence
interval with 53 degrees of freedom. This is close to a normal
variable (since 51 > 30) so we use 1.96 as our critical value.
Now we need the standard error
r
1
1
13.228( + ) = .985.
30 25
So µ1 − µ2 is 2 ± 1.96 × .985 = 2 ± 1.93 with 95% confidence.
(b) If the school each student went to was determined by the parents, instead of randomly, how would that affect the validity of
this information?
SOLUTION: You might then be learning about the difference
between children whose parents choose traditional schools vs.
Montessori schools rather than the difference produced by the
different educational approaches.
(11) A U.S. survey in 2004 estimated that 20% of all Americans of ages 1620 drove under the influence of drugs or alcohol in the previous year.
New Zealand plans a similar survey and wants a 95% confidence
interval with a margin of error of .04.
(a) If they expect a similar answer, what sample size do they need?
SOLUTION: If the expectation
p is that p̂ will be about .2, then
the standard error will be .2 × .8/n where n is the sample
size.
So if we want
p
1.96 .16/n = .04
s2 =
this is equivalent to (1.96)2 .16/n = .0016 or n = 384.16. Since
n has to be an integer, we take n = 385.
(b) If they have no idea what the answer will be, what sample size
do they need?
p
p(1 − p)/n so the maxiSOLUTION: The standard error is p
√
mum value is when p = .5, which is .25/n = .5/ n. If we
have
√
1.96 · .5/ n = .04
we get .9604/.0016 = n = 600.25. So we need n ≥ 601.
(12) According to an exit poll in the 2000 New York senate election,
55.7% of a sample of size 2232 voted for Hillary Clinton.
(a) Test the hypotheses (based on that sample) that Clinton got
over 50% of the vote at the 5% significance level.
SOLUTION:
(i) p is the proportion who voted for Clinton. H0 : p = .5,
Ha : p > .5, α = .05.
(ii) Our test statistic is
.557 − .5
z=p
= 5.39.
.5(1 − .5)/2232
(iii) The p-value is P (z > 5.39) which is not in our table, but
which is very, very small.
(iv) The p-value is smaller than our significance level so we
reject the null hypothesis that Clinton’s vote is 50%, and
conclude her level of the vote is greater than 50% (at
significance level 5%).
(b) If Clinton actually had gotten 40% of the vote, how likely is a
type II error at the 5% significance level, with a sample size of
2232?
SOLUTION: We are asking with p = .4 what the probability is
of p̂ > .5. Normalizing,
.5 − .4
= 9.64.
z=p
.4 × .6/2232
So we are asking for P (z > 9.64), which is essentially zero
(smaller than .0003 - the lowest value in our table).
(13) A test of the breaking strengths of two types of cables is conducted
using sample sizes of n1 = n2 = 100. We get x1 = 1925, s1 = 40,
x2 = 1905, s2 = 30. Does the data show at significance level α = .05
that there is a difference between the mean breaking strengths of the
two types of cables?
SOLUTION: Since n1 , n2 > 30 we can use the large sample method.
Let µ1 be the mean breaking strength of the first type of cable, µ2
be the mean breaking strength of the second type of cable.
(a) α = .05, H0 : µ1 = µ2 . Ha : µ1 6= µ2 .
(b) The test statistic is
20
1925 − 1905
=
= 4.
z=p
2
2
5
40 /100 + 30 /100
(c) Our p-value is P (|Z| > 4), which is less than .0003.
(d) The p-value is less than α, so we reject H0 , and conclude that
the breaking strengths are different at the .05 significance level.
(14) You wish to test the hypothesis that more Republicans favor the
death penalty than Democrats. Out of 200 republicans you find
136 of them support the death penalty. Out of 200 democrats, you
find 124 of them support the death penalty. Do the appropriate
hypothesis test.
SOLUTION: Let p1 be the proportion of Republicans who support the death penalty and p2 be the proportion of Democrats who
support the death penalty. p̂1 = .68 and p̂2 = .62. n1 = n2 = 200.
Since p̂1 n1 , p̂2 n2 , (1 − p̂1 )n1 , (1 − p̂2 )n2 are all > 5, we can use the
large sample procedure.
(a) We’ll take α = .05. H0 : p1 = p2 . Ha : p1 > p2 .
(b) Our pooled sample proportion is
136 + 124
260
p̂ =
=
= .65.
400
400
Our test statistic is
.06
.06
.68 − .62
z=p
=p
=
= 1.25.
.048
p̂(1 − p̂)(2/200)
.65 × .35/100
(c) The p-value is P (z > 1.25) = .1056.
(d) Since the p-value is greater than the significance level, we don’t
have enough evidence to reject H0 . We don’t conclude that
more Republicans support the death penalty than Democrats.