Statistics – Final exam
Spring 2017
Solutions
Question 1
a) The probability of rolling a 4 is:
𝑃 4 = 𝑃 𝑝𝑖𝑐𝑘 𝑑𝑖𝑒 1 ∙ 𝑃 𝑟𝑜𝑙𝑙 4 𝑤𝑖𝑡ℎ 𝑑𝑖𝑒 1 + ⋯ + 𝑃 𝑝𝑖𝑐𝑘 𝑑𝑖𝑒 5 ∙ 𝑃 𝑟𝑜𝑙𝑙 4 𝑤𝑖𝑡ℎ 𝑑𝑖𝑒 5
Therefore, we need to determine the probability of rolling a 4 with each of the 5 dice
and the probability of picking each of the 5 dice. We can easily calculate these
probabilities for the three types of dice:
•
!
With a die with 6 faces, the probability of rolling a 4 is . There are two dice with
!
!
6 faces, so the probability of getting a 6-sided die is . Therefore, the probability
•
!
! !
!
! !
!"
of rolling a 4 with a 6-sided die is ∙ =
!
With a die with 8 faces, the probability of rolling a 4 is . There are two dice with
!
!
8 faces, so the probability of getting an 8-sided die is . Therefore, the probability
! !
!
! !
!"
of rolling a 4 with an 8-sided die is ∙ =
•
!
With a die with 10 faces, the probability of rolling a 4 is
!
!"
. There is only one die
!
with 10 faces, so the probability of getting a 10-sided die is . Therefore, the
!
!
!
!
probability of rolling a 4 with a 10-sided die is ∙ =
! !"
!"
Finally, the probability of rolling a 4 with a die chosen randomly from among the 5 dice
is:
1
1
1
𝑃 4 =
+
+
= 0.1367
15 20 50
b) We can apply the same strategy as above:
• With a die with 6 faces, the probability of rolling an 8 is 0. Therefore, the
!
probability of rolling an 8 with a 6-sided die is ∙ 0 = 0
!
•
!
With a die with 8 faces, the probability of rolling an 8 is . There are two dice
!
!
with 8 faces, so the probability of getting an 8-sided die is . Therefore, the
! !
!
! !
!"
probability of rolling an 8 with an 8-sided die is ∙ =
•
With a die with 10 faces, the probability of rolling an 8 is
!
!
!"
. There is only one die
!
with 10 faces, so the probability of getting a 10-sided die is . Therefore, the
!
!
!
!
probability of rolling an 8 with a 10-sided die is ∙ =
! !"
!"
Finally, the probability of rolling an 8 with a die chosen randomly from among the 5 dice
is:
1
1
𝑃 8 =
+
= 0.07
20 50
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c) We can again use the same strategy:
• With a die with 6 faces, the probability of rolling a 9 is 0. Therefore, the
!
probability of rolling a 9 with a 6-sided die is ∙ 0 = 0
!
•
•
With a die with 8 faces, the probability of rolling a 9 is 0. Therefore, the
probability of rolling a 9 with an 8-sided die is 0
!
With a die with 10 faces, the probability of rolling a 9 is . There is only one die
!"
!
with 10 faces, so the probability of getting a 10-sided die is . Therefore, the
!
!
!
!
probability of rolling a 9 with a 10-sided die is ∙ =
! !"
!"
Finally, the probability of rolling a 9 with a die chosen randomly from among the 5 dice
is:
1
𝑃 9 =
= 0.02
50
d) In order to decide which bet to take, you need to compare the probability of rolling a
4 in a) to the probability of rolling an 8 or a 9, which is the sum of the probabilities in b)
and c):
𝑃 4 = 0.1367 > 𝑃 8 or 9 = 𝑃 8 + 𝑃 9 = 0.09
Therefore, it is best to bet on a 4.
Question 2
a) Although 𝑋 can take an infinity of values, we can count these values: value number 1
is 3! , value number 2 is 3! , value number 3 is 3! , and so on. Therefore, 𝑋 is a discrete
random variable.
b) We use the formula for the expected value of a discrete random variable:
!
𝐸 𝑋 =
!
𝑥𝑓(𝑥) =
!!!
2
3 ∙ !=
3
!
!
!!!
2=∞
!!!
Therefore, 𝑋 does not have an expected value.
c) We now use the formula for an expected value of a function of 𝑋, which in this case is
ℎ 𝑋 = 𝑋 !:
!
𝐸 𝑋
!
=𝐸 ℎ 𝑋
=
!
ℎ(𝑥)𝑓(𝑥) =
!!!
3
!!!
! !
2
∙ !=
3
!
2 ∙ 3! = ∞
!!!
Therefore, 𝐸(𝑋 ! ) also doesn’t exist.
d) Recall that the variance of 𝑋 is calculated as:
𝑉𝑎𝑟 𝑋 = 𝐸 𝑋 ! − 𝐸 𝑋 !
However, neither 𝐸 𝑋 nor 𝐸(𝑋 ! ) exist (i.e., they cannot be calculated). Therefore, we
cannot calculate the variance of 𝑋.
Question 3
a) 𝑋 follows a Bernoulli distribution because it can take only two values: success (the
individual would vote for FD) or failure (the individuals would not vote for FD). We
typically indicate success with the value 1 and failure with the value 0.
b) 𝑌 counts the number of successes from a Bernoulli distribution in a given number of
tries (in this case, 100 tries). Therefore, 𝑌 follows a binomial distribution. The possible
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values of 𝑌 are all integers between 0 and 100 (the number of people who would vote
for FD lies between 0 and 100).
c) The question tells us that 𝑋 follows a Bernouli distribution with parameter 𝑝 = 0.1. In
this case, the probability function of 𝑋 is:
𝑓 0 = 𝑃 𝑋 = 0 = 1 − 𝑝 = 0.9
𝑓 1 = 𝑃 𝑋 = 1 = 𝑝 = 0.1
d) Now we know that 𝑌 follows a binomial distribution with parameters 𝑛 = 100 and
𝑝 = 0.1. Therefore, its probability function is:
𝑛 !
100
𝑓 𝑘 =
𝑝 1 − 𝑝 !!! =
∙ 0.1! ∙ 0.9!""!!
𝑘
𝑘
The probability that 15 respondents answer that they would vote for FD is:
100!
100
𝑓 15 =
∙ 0.1!" ∙ 0.9!""!!" =
∙ 0.1!" ∙ 0.9!" = 0.033
15
15! ∙ 100 − 15 !
In other words, there is a 3.3% probability that 15 respondents answer that they would
vote for FD. Note that you could calculate this probability in Excel using the formula:
= BINOM. DIST(15, 100, 0.1, FALSE)
Question 4
a) In order to construct a 95% confidence interval, we first need to find the critical value
we need to use. Note that 𝛼 = 1 − 0.95 = 0.05. Therefore, the critical value is:
𝑧!!!/! = 𝑧!!!.!"/! = 𝑧!.!"# = 1.96
Note also that the text of the question gives us the sample average 𝑋! = 0.10 for a
sample with 𝑛! = 50 elements, as well as the sample standard deviation 𝑆! = 0.15. The
95% confidence interval can then be constructed as:
𝑋 ± 𝑧!!!/! ∙
𝑆!
= 0.10 ± 1.96 ∙
𝑛
0.15!
= [0.058, 0.142]
100
b) We want to conduct a hypothesis test of the hypothesis that stock A outperforms
investments in riskless government bonds (in other words, that the return on stock A is
higher than the return on government bonds). The null and the alternative hypotheses
are:
𝐻! : 𝜇! ≥ 𝜇!
𝐻! : 𝜇! ≥ 0.03
, or
𝐻! : 𝜇! < 𝜇!
𝐻! : 𝜇! < 0.03
To conduct this test, we first need to calculate the test statistic:
𝑋! − 𝜇! 0.10 − 0.03
𝑍=
=
= 3.300
!
0.15
!
𝑆!
50
𝑛!
This is a one-sided test with a “less than” alternative hypothesis, so we would reject the
null hypothesis if the value of the test statistic is lower than the critical value 𝑧! =
𝑧!.!" = −1.645 (you can find this critical value in Excel using the formula
= NORM. S. INV(0.05)). Given that 3.300 > −1.645, we cannot reject the null hypothesis
that stock A outperforms investments in riskless government bonds. The p-value of this
hypothesis test is:
𝑝 = Φ 𝑍 = Φ 3.300 = 0.9995
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This value is much higher that the significance level of 5%, which again tells us that we
cannot reject the null hypothesis. This can also be calculated in Excel using the formula:
= NORM. S. DIST(3.300, TRUE)
c) Now we are given information about a second random variable. The sample mean of
this new variable is average 𝑋! = 0.07 for a sample with 𝑛! = 30 elements, with a
sample standard deviation 𝑆! = 0.1.
We want to test the hypothesis that the population rate of return on stock A is higher
than the population rate of return on stock B. This is a test of two population means:
𝐻! : 𝜇! ≥ 𝜇!
𝐻! : 𝜇! < 𝜇!
The test statistic for this test is:
𝑋! − 𝑋!
0.10 − 0.07
𝑍=
=
= 1.072
!
!
0.15
0.10
!
!
𝑆! 𝑆!
+
30
+
50
𝑛! 𝑛!
This is again a one-sided test with a “less than” alternative hypothesis, so we would
reject the null hypothesis if the value of the test statistic is lower than the critical value
𝑧! = 𝑧!.!" = −2.326 (again, you can find this critical value in Excel using the formula
= NORM. S. INV(0.01)). Given that 1.072 > −2.326, we cannot reject the null hypothesis
that stock A has a higher rate of return than stock B. Although not required, we can also
calculate the p-value of this hypothesis test:
𝑝 = Φ 𝑍 = Φ 1.072 = 0.8581
As before, this value is much higher than the significance level of 1%, telling us again
that we cannot reject the null. As before, the p-value can be calculated in Excel using the
formula:
= NORM. S. DIST 1.072, TRUE
d) This time we have a two-sided hypothesis test:
𝐻! : 𝜇! = 𝜇!
𝐻! : 𝜇! ≠ 𝜇!
The test statistic for this hypothesis test is:
𝑋! − 𝜇! 0.07 − 0.03
𝑍=
=
= 2.191
!
0.10
!
𝑆!
30
𝑛!
We would reject the null hypothesis if the absolute value of the test statistic is higher
than the critical value 𝑧!!!/! = 𝑧!!!.!"/! = 𝑧!.!" = 1.645 (you can find this critical value
in Excel using the formula = NORM. S. INV(0.95)). Given that 2.191 > 1.645, we reject
the null hypothesis that stock B has the same rate of return as government bonds. The pvalue of this hypothesis test:
𝑝 = 2Φ −|𝑍| = 2Φ −2.191 = 0.028
which is lower than the significance level of 10%, again indicating that we reject the null
hypothesis at a 10% significance level (note that we would not have been able to reject
the null at a 1% significance level). Similar to the previous parts, the p-value is
calculated in Excel as follows:
= 2 ∗ NORM. S. DIST −ABS(2.191), TRUE
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