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QA 233 PRACTICE PROBLEMS
PROBABILITY, SAMPLING
DISTRIBUTIONS CONFIDENCE
INTERVALS & HYPOTHESIS TESTING
These problems will give you an opportunity to
practice applying the material on Probability
Distributions (from Chapters 5 & 6 of the ASW
textbook and Slide Sets 5 & 6), Sampling
Distributions (from Chapter 7 of the ASW textbook
and Slide Set 7), Confidence Intervals (from Chapter 8
of the ASW textbook and Slide Set 8), and Hypothesis
Testing (from Chapter 9 of the ASW textbook and
Slide Set 9).
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
- What is the random variable in this problem?
Gender of a customer
- On what level is this rv measured?
Nominal level
membership
–
we
are
only
measuring
group
- What is the statistic of interest?
The number or proportion of the sample (today’s
customers) who are male
- What is the question we are being asked?
P(p  0.70)=?
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
- How is this statistic distributed?
np = 100(0.75) 5 and np = 100(1 - 0.75) 5, so by the
Central Limit Theorem (CLT) the random variable is
normally distributed with an expected value of p = 0.75
and a standard deviation (also called the standard error)
of
σp 
p1 - p 
0.751 - 0.75

n
100
 0.001875  0.043301
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
- What does this question look like?
Area of
Probability = ?
0.70
{
}
f(p)
Area of
Probability =
0.50
0.75
p
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
- How can we use this information to answer the question
we are being asked?
Since p (the sample proportion) is approximately
normally distributed with an expected value of p = 0.75
and a standard deviation (also called the standard error)
of sp = 0.043301, we can use the z-transformation to
rewrite the question being asked
P(p  0.70)=?
in terms of the standard normal random variable z
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
{
}
- What does the z-transformation look like?
Area of
f(p)
Probability =
Area of
0.50
Probability = ?
0.70
0.75
p
????
0.00
z
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
- How do we do the z-transformation?
Pp  0.70  P(0.70  p  0.75) P(0.75  p  )
 0.70- 0.75 p - p 0.75- 0.75
  0.50
= P


 0.043301

s
0.043301
p


 P(-1.15  z  0.00) 0.50
= 0.3749  0.50  0.8749
1. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. If one
hundred customers come into your store today, what is
the probability that at least 70% are male?
{
}
- What does the z-transformation look like?
Area of
Area of
f(p)
Probability =
Probability =
0.3749
0.50
0.70
0.75
p
-1.15
0.00
z
2. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. What is
the probability that no more than one of the next six
customers that come into your store today is male?
- What is the random variable in this problem?
Gender of a customer
- On what level is this rv measured?
Nominal level
membership
–
we
are
only
measuring
group
- What is the value of interest?
The number of the sample (the next six customers) who are
male
- What is the question we are being asked?
P(x  1)=?
2. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. What is
the probability that no more than one of the next six
customers that come into your store today is male?
- How is this statistic distributed?
np = 6(0.75)< 5 and n(1-p) = 6(1 - 0.75)< 5 so the random
variable is not normally distributed – we have six trials,
each with two potential outcomes, so it is binomially
distributed with p = 0.75 and n = 6.
2. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. What is
the probability that no more than one of the next six
customers that come into your store today is male?
}
- What does this question look like?
Area of
f(x)
1
0.9 Probability
= ?
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
x
4
5
6
2. Suppose that exactly three-quarters of the customers
who come into your sheet music store are male. What is
the probability that no more than one of the next six
customers that come into your store today is male?
- How can we use this information to answer the question
we are being asked?
Since x (the number of males in the next six customers) is
binomially distributed with p = 0.75 and n = 6, we can
calculate this probability directly as follows
P(x  1) = P(x = 0) + P(x = 1) = f(0) + f(1), and
6
f0   0.750 1  0.756-0
 0
6!
=
(1.00)(0.000244)
0!(6 - 0)!
= 0.000244
and
6
f1   0.751 1  0.756-1
 1
6!
=
(0.75)(0.000977)
1!(6 - 1)!
= 0.004395
So P(x  1) = P(x = 0) + P(x = 1) = f(0) + f(1) = 0.000244+
0.004395 = 0.004639
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
for at least $160.00?
- What is the random variable in this problem?
Annual amount contributed by a mid-level manager to the
United Way fund drive
- On what level is this rv measured?
Ratio level – the differences between values have
meaning and there is a natural, well-defined 0 ($0.00)
- What is the value of interest?
The contribution made by the next mid-level manager
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
for at least $160.00?
- What is the question we are being asked?
P(x  $160.00) = ?
- How is this statistic distributed?
We are told that the random variable is normally
distributed with an expected value of m = $135.00 and a
standard deviation of 55.00.
{
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
Area of
for at least $160.00?
- What does this question look like? Probability =
0.50
f(x)
$135.00
$160.00
}
}
Area of
Area of
Probability Probability
= ?
= ?
x
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
for at least $160.00?
- How can we use this information to answer the question
we are being asked?
Since x (the contribution made by an individual mid-level
manager) is normally distributed with an expected value
of m = $135.00 and a standard deviation of s = 55.0, we
can use the z-transformation to rewrite the question being
asked
P(x  $160.00)=?
in terms of the standard normal random variable z
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
for at least $160.00?
Area of
- What does the z-transformation
look like?
f(x)
Probability =
0.50
{
$135.00
0.00
$160.00
????
}
}
Area of
Area of
Probability Probability
= ?
= ?
x
z
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
for at least $160.00?
- How do we do the z-transformation?
Px  $160.00
 P($135.00 x  )- P($135.00 x  $160.00)
 $135.00- $135.00 x - m $160.00- $135.00
= 0.50- P



55.0
s
55.0


 0.50- P(0.00  z  0.45)
= 0.50- 0.1736  0.3264
3. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the next
annual contribution from a mid-level manager will be
for at least $160.00?
Area of
- What does the z-transformation
look like?
f(x)
Probability =
0.50
{
$135.00
0.00
$160.00
0.45
}
}
Area of
Area of
Probability Probability
= 0.1736
= 0.3264
x
z
4. Suppose we take a sample of thirty-four mid-level
managers and record the amounts they gave to the
United Way fund drive last year. If the sample has a
mean of $150.00 and a standard deviation of 50.0, what
would be the appropriate 95% confidence interval
estimate of the mean annual contribution made by midlevel managers to the United Way fund drive?
- What is the random variable in this problem?
Annual amount contributed by a mid-level manager to the
United Way fund drive
- On what level is this rv measured?
Ratio level – the differences between values have
meaning and there is a natural, well-defined 0 ($0.00)
- What is the value of interest?
The mean contribution made by mid-level managers
4. Suppose we take a sample of thirty-four mid-level
managers and record the amounts they gave to the
United Way fund drive last year. If the sample has a
mean of $150.00 and a standard deviation of 50.0, what
would be the appropriate 95% confidence interval
estimate of the mean annual contribution made by midlevel managers to the United Way fund drive?
- What is the question we are being asked?
What symmetric interval, when calculated over infinitely
many repeated samples, will contain the true mean
contribution made by mid-level managers 95% of the
time?
- How is this statistic distributed?
Since n = 34  30, the random variable x is normally
distributed with an expected value of m = ?? and a
s
s

standard error of x
.
n
4. Suppose we take a sample of thirty-four mid-level
managers and record the amounts they gave to the
United Way fund drive last year. If the sample has a
mean of $150.00 and a standard deviation of 50.0, what
would be the appropriate 95% confidence interval
estimate of the mean annual contribution made by midlevel managers to the United Way fund drive?
- What is the appropriate approach to building the
confidence interval?
We are trying to estimate m (the population mean). Since
we have a large sample (n = 34  30) the Central Limit
Theorem ensures us that the distribution of sample means
( x ) will be approximately normal. Also, although we
don’t know the population standard deviation s, the
large sample size ensures that the sample standard
deviation s will be a very good estimate of s. Thus we
s
will use
x z s  x z
2 x
2
n
4. Suppose we take a sample of thirty-four mid-level
managers and record the amounts they gave to the
United Way fund drive last year. If the sample has a
mean of $150.00 and a standard deviation of 50.0, what
would be the appropriate 95% confidence interval
estimate of the mean annual contribution made by midlevel managers to the United Way fund drive?
- What is the appropriate 95% confidence interval?
We have a confidence level of 95%, so z = 1.96
f(x)
}
Area of
Probability
= 0.475
x
$150.00
-1.96
0
1.96
z
4. Suppose we take a sample of thirty-four mid-level
managers and record the amounts they gave to the
United Way fund drive last year. If the sample has a
mean of $150.00 and a standard deviation of 50.0, what
would be the appropriate 95% confidence interval
estimate of the mean annual contribution made by midlevel managers to the United Way fund drive?
- We have that x = $150.00, s = 50, n = 34, and z = 1.96, so
the appropriate 95% confidence interval is
x  z/2
s
50.0
 150.0  1.96
n
34
=150.0  16.807
 133.193,166.807
- Interpretation: If we repeated this sampling process an
infinite number of times, we would expect 95% of the
resulting intervals to contain the mean contribution of all
mid-level managers in the population.
5. Suppose that we believe that the mean annual
contribution made by mid-level managers to the United
Way fund drive is at least $165.00. Based on the sample
to which we referred in the previous problem, what
does a hypothesis test suggest about our conjecture at a
0.05 level of significance? At a 0.01 level of significance?
1 State the Null and Alternative Hypotheses
Ho: m  165.00
Ha: m < 165.00
2 Select the Appropriate Test Statistic – n = 34  30, so use
x - μ 0 x - μ0
z

s
sx
n
3 State the Desired Level of Significance , Find the Critical
Value(s) and State the Decision Rule
First consider =0.05. We have a one-tailed test, so z = 1.645
0.5-=0.45
0.500
f(x)
=0.05
Reject Region
μ0  zαsx
 zα  1.645
95% Do Not
Reject Region
m0  165.00
x
0
z
Decision rule – do not reject Ho if –1.645  z
otherwise reject Ho
4 Calculate the Test Statistic
x - m 0 150.00 - 165.00
z

 1.74929
50
sx
34
5 Use the Decision Rule to Evaluate the Test Statistic and
Decide Whether to Reject or Not Reject the Null
Hypothesis
–z  z i.e., –1.645  -1.74929
so reject Ho. The sample evidence does not support the
claim that the mean contribution by mid-level managers
is at least $165.00.
We now consider =0.01 (and repeat the hypothesis
testing procedure)
1 State the Null and Alternative Hypotheses
Ho: m  165.00
Ha: m < 165.00
2 Select the Appropriate Test Statistic – n = 34  30, so use
x - μ 0 x - μ0
z

s
sx
n
4 State the Desired Level of Significance , Find the Critical
Value(s) and State the Decision Rule
Since =0.01 and we have a one-tailed test, z = -2.33
0.5-=0.49
0.500
f(x)
=0.01
Reject Region
99% Do Not
Reject Region
μ0  zαsx
m0  165.00
x
 z  2.33
0
z
Decision rule – do not reject Ho if –2.33  z
otherwise reject Ho
5 Calculate the Test Statistic
x - m 0 150.00 - 165.00
z

 1.74929
50
sx
34
6 Use the Decision Rule to Evaluate the Test Statistic and
Decide Whether to Reject or Not Reject the Null
Hypothesis
–z  z i.e., –2.33  -1.74929
so do not reject Ho. The sample evidence supports the
claim that the mean contribution by mid-level managers
is at least $165.00.
Notice two things about what happened when we
changed the level of significance () from .05 to .01:
- Only steps 3 and 5 of the hypothesis testing procedure
were impacted
- The final decision changed from reject to do not reject!
This is why it is so important to perform steps 1 – 3
BEFORE you collect or analyze sample data!
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
- What is the random variable in this problem?
Annual amount contributed by a mid-level manager to the
United Way fund drive
- On what level is this rv measured?
Ratio level – the differences between values have
meaning and there is a natural, well-defined 0 ($0.00)
- What is the statistic of interest?
The mean contribution made by sixteen SafetyCo midlevel managers
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
- What is the question we are being asked?
P( x  $150.00) = ?
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
- How is this statistic distributed?
Since our parent population from which we have drawn
our sample (the contribution made by an individual midlevel manager) is normally distributed with an expected
value of m = $135.00 and a standard deviation of 55.00, we
have that x is also normally distributed with an expected
value of m x  m  $135.00and a standard deviation
(a.k.a. standard error) of
sx  s
n
 $55.0
16
 13.75
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
}
{
- What does this question look like?
f(x) Area of Probability
= 0.50
Area of
Probability = ?
$135.00
$150.00
x
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
- How can we use this information to answer the question
we are being asked?
Since x (the sample mean) is normally distributed with an
expected value of m = $135.00 and a standard error of s x =
13.75, we can use the z-transformation to rewrite the
question being asked
P( x  $150.00) = ?
in terms of the standard normal random variable z
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
}
{
- What does this question look like?
f(x) Area of Probability
= 0.50
Area of
Probability = ?
$135.00
$150.00
x
0.00
????
z
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
- How do we do the z-transformation?
Px  $150.00
 P(-  x  $135.00) P($135.00 x  $150.00)
 $135.00- $135.00 x - m $150.00- $135.00

= 0.50  P


13.75
sx
13.75


 0.50  P(0.00  z  1.09)
= 0.50  0.3621  0.8621
6. Suppose that the annual amount given by mid-level
managers to the United Way fund drive is normally
distributed with a mean of $135.00 and a standard
deviation of 55.0. What is the probability that the mean
annual contribution of sixteen mid-level managers at
SafetyCo will be no more than $150.00?
}
{
- What does this question look like?
f(x) Area of Probability
Area of
= 0.50
Probability
= 0.1736
$135.00
$150.00
x
0.00
1.09
z
Notice that we did not need to use the notion of the
Central Limit Theorem for Problem #4 – WHY NOT?
7. Suppose that of our sample of thirty-four mid-level
managers, nineteen contributed at least $165.00 to the
United Way fund drive last year. What would be the
appropriate 87% confidence interval estimate of the
proportion of mid-level managers that contributed at
least $165.00 to the United Way fund drive last year?
- What is the random variable in this problem?
Whether individual mid-level managers contributed at
least $165.00 to the United Way fund drive last year
- On what level is this rv measured?
Nominal level – the rv represents group membership
(contributed at least $165.00 vs. contributed less than
$165.00 to the United Way fund drive last year)
- What is the value of interest?
The proportion of mid-level managers who contributed at
least $165.00 to the United Way fund drive last year
7. Suppose that of our sample of thirty-four mid-level
managers, nineteen contributed at least $165.00 to the
United Way fund drive last year. What would be the
appropriate 87% confidence interval estimate of the
proportion of mid-level managers that contributed at
least $165.00 to the United Way fund drive last year?
- What is the question we are being asked?
What symmetric interval, when calculated over infinitely
many repeated samples, will contain the true proportion
who contributed at least $165.00 to the United Way fund
drive last year?
- How is this statistic distributed?
Since n = 34  30, the random variable p is normally
distributed with an expected value of p = ? and a standard
p1 - p 
error of
sp 
n
7. Suppose that of our sample of thirty-four mid-level
managers, nineteen contributed at least $165.00 to the
United Way fund drive last year. What would be the
appropriate 87% confidence interval estimate of the
proportion of mid-level managers that contributed at
least $165.00 to the United Way fund drive last year?
- What is the appropriate approach to building the
confidence interval?
We are trying to estimate p (the population proportion).
Since we have a large sample (np = 34(.5)  30 and n(1 –
p) = 34(.5)  30) the Central Limit Theorem ensures us that
the distribution of sample proportions (p) will be
approximately normal. Thus we will use
p1 - p 
p  zsp  p  z
n
7. Suppose that of our sample of thirty-four mid-level
managers, nineteen contributed at least $165.00 to the
United Way fund drive last year. What would be the
appropriate 87% confidence interval estimate of the
proportion of mid-level managers that contributed at
least $165.00 to the United Way fund drive last year?
- What is the appropriate 87% confidence interval?
We have a confidence level of 87%, so z = 1.51
f(p)
}
Area of
Probability
= 0.435
p
19
 0.5588
34
-1.51
0
1.51
z
7. Suppose that of our sample of thirty-four mid-level
managers, nineteen contributed at least $165.00 to the
United Way fund drive last year. What would be the
appropriate 87% confidence interval estimate of the
proportion of mid-level managers that contributed at
least $165.00 to the United Way fund drive last year?
- We have that p = 0.5588 and n = 34 so the standard error
is
sp 
p1 - p
0.55881 - 0.5588

 0.085154
n
34
7. Suppose that of our sample of thirty-four mid-level
managers, nineteen contributed at least $165.00 to the
United Way fund drive last year. What would be the
appropriate 87% confidence interval estimate of the
proportion of mid-level managers that contributed at
least $165.00 to the United Way fund drive last year?
- Thus the appropriate 95% confidence interval is
p1 - p 
p  zsp  p  z
n
 0.5588 1.510.085154
 0.4302,0.6874
- Interpretation: If we repeated this sampling process an
infinite number of times, we would expect 87% of the
resulting intervals to contain the proportion of all midlevel managers in the population who contributed at least
$165.00 to the United Way fund drive last year.
8. Suppose that we believe that no more than 35% of midlevel managers contribution at least $165.00 annually to
the United Way fund drive. From our sample of thirtyfour mid-level managers, nineteen actually did
contribute at least $165.00 to the United Way fund drive
last year. Based on the sample to which we referred in
the previous problem, what hat does a hypothesis test
suggest about our conjecture at a 0.05 level of
significance? At a 0.01 level of significance?
1 State the Null and Alternative Hypotheses
Ho: p  0.35
Ha: p > 0.35
2 Select the Appropriate Test Statistic – np0 = 11.9  5 and
n(1 - p0) = 22.1  5, so use
p - p0
p - p0
z

σp
p0 1  p0 
n
3 State the Desired Level of Significance , Find the
Critical Value(s) and State the Decision Rule
=0.05 and we have a one (upper)-tailed test, so z = 1.65
f(z)
0.5
0.5-=0.45
=0.05
99% Do Not
Reject Region
p0  0.35
0
Reject Region
p
p0  zσ p
 z  1.65
Decision rule – do not reject Ho if z  1.65
otherwise reject Ho
z
4 Calculate the Test Statistic
p - p0
p - p0
z

σp
p0 1  p0 
n
19
- 0.35
0.208
34


 2.552867
0.0818
0.351  0.35
34
5 Use the Decision Rule to Evaluate the Test Statistic and
Decide Whether to Reject or Not Reject the Null
Hypothesis
z  z i.e., 2.552867  1.65
so reject Ho. The sample evidence does not support the
claim that no more than 40% of mid-level managers
contribution at least $165.00 annually to the United Way
fund drive.
We now consider =0.01 (and repeat the hypothesis
testing procedure)
1 State the Null and Alternative Hypotheses
Ho: p  0.35
Ha: p > 0.35
2 Select the Appropriate Test Statistic – np0 = 11.9  5 and
n(1 - p0) = 22.1  5, so use
z
p - p0

σp
p - p0
p0 1  p0 
n
4 State the Desired Level of Significance , Find the
Critical Value(s) and State the Decision Rule
=0.01 and we have a one (upper)-tailed test, so z = 2.33
f(z)
0.5
0.5-=0.49
=0.01
99% Do Not
Reject Region
p0  0.35
0
Reject Region
p
po  zασ p
 zα  2.33
Decision rule – do not reject Ho if z  2.33
otherwise reject Ho
z
5 Calculate the Test Statistic
p - p0
p - p0
z

sp
p0 1  p0 
n
19
- 0.35
0.208
34


 2.552867
0.0818
0.351  0.35
34
6 Use the Decision Rule to Evaluate the Test Statistic and
Decide Whether to Reject or Not Reject the Null
Hypothesis
z  z i.e., 2.552867  2.33
so reject Ho. The sample evidence does not support the
claim that no more than 40% of mid-level managers
contribution at least $165.00 annually to the United Way
fund drive.
Notice what happened when we changed the level of
significance () from .05 to .01:
- Again only steps 4 and 6 of the hypothesis testing
procedure were impacted
- The final decision did not change from reject to do not
reject In this case! This is because the evidence against
this null hypothesis was relatively strong!