University of Toronto Scarborough
STAB22 Final Examination
April 2009
For this examination, you are allowed two handwritten letter-sized
sheets of notes (both sides) prepared by you, a non-programmable,
non-communicating calculator, and writing implements.
This question paper has 24 numbered pages. Before you start,
check to see that you have all the pages. You should also
have a Scantron sheet on which to enter your answers, and a set
of statistical tables. If any of this is missing, speak to an invigilator.
This examination is multiple choice. Each question has equal
weight, and there is no penalty for guessing. To ensure that
you receive credit for your work on the exam, fill in the bubbles
on the Scantron sheet for your correct student number (under
“Identification”), your last name, and as much of your first name
as fits.
Mark in each case the best answer out of the alternatives given
(which means the numerically closest answer if the answer is a
number and the answer you obtained is not given.)
If you need paper for rough work, use the back of the sheets of this
question paper. The question paper will be collected at the end of
the examination, but any writing on it will not be read or marked.
Before you begin, two more things:
• Check that the colour printed on your Scantron sheet matches
the colour of your question paper. If it does not, get a new
Scantron from an invigilator.
• Complete the signature sheet, but sign it only when the in-
vigilator collects it. The signature sheet shows that you were
present at the exam.
1
,The correct alternative has an asterisk next to it.
1. A simple random sample of 70 coffee drinkers was taken. Each sampled coffee drinker was asked to
taste two unmarked cups of coffee, one of which is actually Brand A and the other is Brand B, and
was asked which one they preferred. 44 coffee drinkers preferred Brand A, and the other 26 preferred
Brand B. Use this information to answer this question and the next one.
The people who commissioned the survey are trying to find out whether a majority of all coffee-drinkers
(that is, more than 50% of them) prefer brand A. Carry out a suitable test of significance to assess the
evidence. What is the P-value of your test?
(a) between 0.05 and 0.10
(b) between 0.025 and 0.05
(c) greater than 0.10
(d) * between 0.01 and 0.025
(e) less than 0.01
Null p = 0.50, alternative
p p > 0.50; p̂ = 44/70 = 0.629; standard error, using the null
value of p, 0.50, is (0.50)(1 − 0.50)/70 = 0.060, so z = (0.629 − 0.5)/0.060 = 2.15.
One-sided P-value is 0.0158.
2. In the situation of Question 1, it turned out that the coffee drinkers had always been given an unmarked
cup containing Brand A coffee first, and Brand B coffee second. How do you react to this knowledge?
(a) The P-value was small, so there will still be good evidence that Brand A is preferred.
(b) The coffee drinkers received their coffee in unmarked cups, so it doesn’t matter which brand is
actually tasted first.
(c) Brand A must have an advantage by being tasted first, so there cannot be a significant difference
between brands A and B.
(d) The P-value was large, so there will still be no evidence that Brand A is preferred.
(e) * A better approach would have been to toss a coin to decide whether each drinker gets Brand A
first or Brand B first.
If the test gives a small P-value (as it did), this could be because brand A really is preferred,
or because the brand given first, whichever it was, is preferred, and we have no way of telling
which is the reason for the small P-value. So (a) and (c) could be true, or could be false. (d)
was actually false anyway, but if you thought the P-value from the previous question was
large, (d) here could also be either true or false. (b) is a red herring: the fact that the cup
is unmarked doesn’t stop Brand A from being tasted first, so if being first is the reason for
the difference, whether the cup is unmarked or not will not matter. That leaves (e), and we
already know that randomization is a good thing.
3. A university financial aid office took a simple random sample of students to see how many of them were
employed the previous summer. Of the 750 men sampled, 703 had been employed the previous summer;
of the 650 women sampled, 592 had been employed the previous summer. Use this information for this
question and the next one.
Test whether the proportion of all male students employed last summer is different from the proportion
of all female students employed last summer. What is the P-value of your test?
(a) * between 0.05 and 0.10
(b) between 0.01 and 0.025
(c) between 0.025 and 0.05
(d) less than 0.01
2
, (e) greater than 0.05
Two-proportion test from §8.2, so go through the procedure, using a two-sided alternative.
p̂1 = 0.9373, p̂2 = 0.9108; p̂ = (703 + 592)/(750 + 650) = 0.9250;
p
SEDp = (0.9250)(0.0750)(1/750 + 1/650) = 0.0141;
z = (0.9373 − 0.9108)/0.0141 = 1.88; P-value 2 × (1 − 0.9699) = 0.0602. (e) was supposed to
say “greater than 0.10”; with the question as written, (e) is also correct.
4. From the information given in Question 3, what is the upper limit of a 95% confidence interval for the
difference between the proportion of men employed the previous summer and the proportion of women
employed the previous summer? (Take the difference as men minus women.)
(a) * 0.05
(b) -0.10
(c) 0.10
(d) 0.00
(e) -0.05
Section 8.2 again, only this time using SED because it’s a CI:
p
SED = (0.9373)(0.0627)/750 + (0.9108)(0.0892)/650 = 0.0143,
so the upper limit is 0.9373 − 0.9108 + 1.96(0.0143) = 0.0545. (The lower limit is close to 0.)
5. A report states that 3% of all births are “multiple births” (that is, twins, triplets, etc.). A random
sample of 5000 births is taken. Assuming the report to be correct, what is the probability that 2.5%
or less of the births in the sample are multiple births?
(a) * less than 0.02
(b) between 0.05 and 0.10
(c) greater than 0.30
(d) between 0.02 and 0.05
(e) between 0.10 and 0.30
p
Normal approximation to binomial: for p̂, mean is p = 0.03 and SD is (0.03)(0.97)/5000 =
0.0024, so z = (0.025 − 0.03)/0.0024 = −2.07, and the probability of less is 0.0191.
6. A population has a mildly non-normal shape with mean 35 and standard deviation 7.5. A sample of
size 225 is taken from this population. What is the probability that the sample mean is less than 33.5?
(a) * about 0.001
(b) 0.05
(c) 0.42
(d) more than 0.90
(e) 0.12
√
Asking about sample mean, so use method of §5.2: z =√(33.5 − 35)/(7.5/ 225) = −3, so the
probability of less is 0.0013. Don’t forget to divide by 225!
The mild nonnormality doesn’t affect the calculation. The sample size is so large that the
approximation from the central limit theorem will be very good.
3
STAB22 Final Examination
April 2009
For this examination, you are allowed two handwritten letter-sized
sheets of notes (both sides) prepared by you, a non-programmable,
non-communicating calculator, and writing implements.
This question paper has 24 numbered pages. Before you start,
check to see that you have all the pages. You should also
have a Scantron sheet on which to enter your answers, and a set
of statistical tables. If any of this is missing, speak to an invigilator.
This examination is multiple choice. Each question has equal
weight, and there is no penalty for guessing. To ensure that
you receive credit for your work on the exam, fill in the bubbles
on the Scantron sheet for your correct student number (under
“Identification”), your last name, and as much of your first name
as fits.
Mark in each case the best answer out of the alternatives given
(which means the numerically closest answer if the answer is a
number and the answer you obtained is not given.)
If you need paper for rough work, use the back of the sheets of this
question paper. The question paper will be collected at the end of
the examination, but any writing on it will not be read or marked.
Before you begin, two more things:
• Check that the colour printed on your Scantron sheet matches
the colour of your question paper. If it does not, get a new
Scantron from an invigilator.
• Complete the signature sheet, but sign it only when the in-
vigilator collects it. The signature sheet shows that you were
present at the exam.
1
,The correct alternative has an asterisk next to it.
1. A simple random sample of 70 coffee drinkers was taken. Each sampled coffee drinker was asked to
taste two unmarked cups of coffee, one of which is actually Brand A and the other is Brand B, and
was asked which one they preferred. 44 coffee drinkers preferred Brand A, and the other 26 preferred
Brand B. Use this information to answer this question and the next one.
The people who commissioned the survey are trying to find out whether a majority of all coffee-drinkers
(that is, more than 50% of them) prefer brand A. Carry out a suitable test of significance to assess the
evidence. What is the P-value of your test?
(a) between 0.05 and 0.10
(b) between 0.025 and 0.05
(c) greater than 0.10
(d) * between 0.01 and 0.025
(e) less than 0.01
Null p = 0.50, alternative
p p > 0.50; p̂ = 44/70 = 0.629; standard error, using the null
value of p, 0.50, is (0.50)(1 − 0.50)/70 = 0.060, so z = (0.629 − 0.5)/0.060 = 2.15.
One-sided P-value is 0.0158.
2. In the situation of Question 1, it turned out that the coffee drinkers had always been given an unmarked
cup containing Brand A coffee first, and Brand B coffee second. How do you react to this knowledge?
(a) The P-value was small, so there will still be good evidence that Brand A is preferred.
(b) The coffee drinkers received their coffee in unmarked cups, so it doesn’t matter which brand is
actually tasted first.
(c) Brand A must have an advantage by being tasted first, so there cannot be a significant difference
between brands A and B.
(d) The P-value was large, so there will still be no evidence that Brand A is preferred.
(e) * A better approach would have been to toss a coin to decide whether each drinker gets Brand A
first or Brand B first.
If the test gives a small P-value (as it did), this could be because brand A really is preferred,
or because the brand given first, whichever it was, is preferred, and we have no way of telling
which is the reason for the small P-value. So (a) and (c) could be true, or could be false. (d)
was actually false anyway, but if you thought the P-value from the previous question was
large, (d) here could also be either true or false. (b) is a red herring: the fact that the cup
is unmarked doesn’t stop Brand A from being tasted first, so if being first is the reason for
the difference, whether the cup is unmarked or not will not matter. That leaves (e), and we
already know that randomization is a good thing.
3. A university financial aid office took a simple random sample of students to see how many of them were
employed the previous summer. Of the 750 men sampled, 703 had been employed the previous summer;
of the 650 women sampled, 592 had been employed the previous summer. Use this information for this
question and the next one.
Test whether the proportion of all male students employed last summer is different from the proportion
of all female students employed last summer. What is the P-value of your test?
(a) * between 0.05 and 0.10
(b) between 0.01 and 0.025
(c) between 0.025 and 0.05
(d) less than 0.01
2
, (e) greater than 0.05
Two-proportion test from §8.2, so go through the procedure, using a two-sided alternative.
p̂1 = 0.9373, p̂2 = 0.9108; p̂ = (703 + 592)/(750 + 650) = 0.9250;
p
SEDp = (0.9250)(0.0750)(1/750 + 1/650) = 0.0141;
z = (0.9373 − 0.9108)/0.0141 = 1.88; P-value 2 × (1 − 0.9699) = 0.0602. (e) was supposed to
say “greater than 0.10”; with the question as written, (e) is also correct.
4. From the information given in Question 3, what is the upper limit of a 95% confidence interval for the
difference between the proportion of men employed the previous summer and the proportion of women
employed the previous summer? (Take the difference as men minus women.)
(a) * 0.05
(b) -0.10
(c) 0.10
(d) 0.00
(e) -0.05
Section 8.2 again, only this time using SED because it’s a CI:
p
SED = (0.9373)(0.0627)/750 + (0.9108)(0.0892)/650 = 0.0143,
so the upper limit is 0.9373 − 0.9108 + 1.96(0.0143) = 0.0545. (The lower limit is close to 0.)
5. A report states that 3% of all births are “multiple births” (that is, twins, triplets, etc.). A random
sample of 5000 births is taken. Assuming the report to be correct, what is the probability that 2.5%
or less of the births in the sample are multiple births?
(a) * less than 0.02
(b) between 0.05 and 0.10
(c) greater than 0.30
(d) between 0.02 and 0.05
(e) between 0.10 and 0.30
p
Normal approximation to binomial: for p̂, mean is p = 0.03 and SD is (0.03)(0.97)/5000 =
0.0024, so z = (0.025 − 0.03)/0.0024 = −2.07, and the probability of less is 0.0191.
6. A population has a mildly non-normal shape with mean 35 and standard deviation 7.5. A sample of
size 225 is taken from this population. What is the probability that the sample mean is less than 33.5?
(a) * about 0.001
(b) 0.05
(c) 0.42
(d) more than 0.90
(e) 0.12
√
Asking about sample mean, so use method of §5.2: z =√(33.5 − 35)/(7.5/ 225) = −3, so the
probability of less is 0.0013. Don’t forget to divide by 225!
The mild nonnormality doesn’t affect the calculation. The sample size is so large that the
approximation from the central limit theorem will be very good.
3
