Practice Test 3 –Bus 2023
Directions: For each question find the answer that is the best solution provided. There is
only one correct answer.
1.
A random variable that can assume only a finite number of values is referred to as a(n)
a.
infinite sequence
b.
finite sequence
c.
discrete random variable
d.
discrete probability function
ANSWER:
c
2.
The number of customers that enter a store during one day is an example of
a.
a continuous random variable
b.
a discrete random variable
c.
either a continuous or a discrete random variable, depending on the number of the
customers
d.
either a continuous or a discrete random variable, depending on the gender of the
customers
ANSWER:
b
3.
An experiment consists of measuring the speed of automobiles on a highway by the use
of radar equipment. The random variable in this experiment is speed, measured in miles
per hour. This random variable is a
a.
discrete random variable
b.
continuous random variable
c.
complex random variable
d.
None of these alternatives is correct.
ANSWER:
b
4.
A measure of the average value of a random variable is called a(n)
a.
variance
b.
standard deviation
c.
expected value
d.
None of the alternative answers is correct.
ANSWER:
c Note: So we can write E(x) = μ
5.
The expected value for a binomial probability distribution is
a.
E(x) = pn(1 - n)
b.
E(x) = p(1 - p)
c.
E(x) = np
d.
E(x) = np(1 - p)
ANSWER:
c
1
6.
The variance for the binomial probability distribution is
a.
Var(x) = p(1 - p)
b.
Var(x) = np
c.
Var(x) = n(1 - p)
d.
Var(x) = np(1 - p)
ANSWER:
d
7.
Assume that you have a binomial experiment with p = 0.4 and a sample size of 50. The
variance of this distribution is
a.
20
b.
12
c.
3.46
d.
Not enough information is given to answer this question.
ANSWER:
b
Recall that variance of the binomial is Var(x) = np(1 - p) = 50(0.4)(0.6) = 12
8.
Twenty percent of the students in a class of 100 are planning to go to graduate school.
The standard deviation of this binomial distribution is
a.
20
b.
16
c.
4
d.
2
ANSWER:
c
Recall that variance of the binomial is Var(x) = np(1 - p) = 100(0.2)(0.8) = 16
Std(x) =
=4
Exhibit 5-8
The student body of a large university consists of 60% female students. A random sample of 8
students is selected.
9.
Refer to Exhibit 5-8. What is the random variable in this experiment?
a.
the 60% of female students
b.
the random sample of 8 students
c.
the number of female students out of 8
d.
the student body size
ANSWER:
c
10.
Refer to Exhibit 5-8. What is the probability that among the students in the sample
exactly two are female?
a.
0.0896
b.
0.2936
c.
0.0413
d.
0.0007
ANSWER:
c
Recall that P( X = x) = ( nx ) px (1-p)(n-x)  P( X = 2) = ( 82 ) )0.62 (0.4)(8-2) = 0.0413
2
11.
Refer to Exhibit 5-8. What is the probability that among the students in the sample at
least 7 are female?
a.
0.1064
b.
0.0896
c.
0.0168
d.
0.8936
ANSWER:
a
Note: This is the same set up as above but now you need P (X ≥7) = P (7) + P(8)
12.
Refer to Exhibit 5-8. What is the probability that among the students in the sample at
least 6 are male?
a.
0.0413
b.
0.0079
c.
0.0007
d.
0.0499
ANSWER:
d
Note: Same set up, but now the percentage of inters is not females, but males. P (X ≥6) = P(6) +
P (7) + P(8). Make sure to use p=0.40 since the probability of interest is males.
13.
A numerical measure from a sample, such as a sample mean, is known as
a.
a statistic
b.
a parameter
c.
the mean deviation
d.
the central limit theorem
ANSWER: a
14.
A subset of a population selected to represent the population is a
a.
subset
b.
sample
c.
small population
d.
None of the alternative answers is correct.
ANSWER: b
15.
A simple random sample of 5 observations from a population containing 400 elements
was taken, and the following values were obtained.
12
18
19
20
21
A point estimate of the population mean is
a.
400
b.
18
c.
20
d.
10
ANSWER: b
Note: Since this is a sample that is supposed to be representative of the population its mean is an
_
approximate of the population mean. So we get that E ( x ) = μ.
3
16.
The sampling error is the
a.
same as the standard error
b.
absolute value of the difference between an unbiased point estimate and the
corresponding population parameter
c.
error caused by selecting a bad sample
d.
standard deviation multiplied by the sample size
ANSWER: b
17.
The probability distribution of all possible values of the sample mean is called the
a.
central probability distribution
b.
sampling distribution of the sample mean
c.
random variation
d.
standard error
ANSWER: b
18.
From a population of 200 elements, the standard deviation is known to be 14. A sample
of 49 elements is selected. It is determined that the sample mean is 56. The standard error of the
mean is
a.
3
b.
2
c.
greater than 2
d.
less than 2
ANSWER: d
Recall that σ _ = σ / n = 14 / 49 = 14 / 7 = 2
x
19.
From a population of 500 elements, a sample of 225 elements is selected. It is known
that the variance of the population is 900. The standard error of the mean is approximately
a.
1.1022
b.
2
c.
30
d.
1.4847
ANSWER: d
Note: Same process as above but now a finite population
σ _ = ( N  n) /( N  1) * (σ / n ) = (500  225) /(500  1) * (30 / 225 ) = 1.4847
x
20.
As the sample size increases, the
a.
standard deviation of the population decreases
b.
population mean increases
c.
standard error of the mean decreases
d.
standard error of the mean increases
ANSWER: c
4
21.
Random samples of size 36 are taken from an infinite population whose mean and
standard deviation are 20 and 15, respectively. The distribution of the population is unknown.
The mean and the standard error of the distribution of sample mean are
a.
36 and 15
b.
20 and 15
c.
20 and 0.417
d.
20 and 2.5
ANSWER: d
Recall that the mean of the sampling distribution and population are the same. We also know σ _
x
=σ/
n = 15 /
36 = 15 / 6 = 2.5
22.
A theorem that allows us to use the normal probability distribution to approximate the
sampling distribution of sample means and sample proportions whenever the sample size is large
is known as the
a.
approximation theorem
b.
normal probability theorem
c.
central limit theorem
d.
central normality theorem
ANSWER: c
23.
A population has a mean of 180 and a standard deviation of 24. A sample of 64
observations will be taken. The probability that the mean from that sample will be between 183
and 186 is
a.
0.1359
b.
0.8185
c.
0.3413
d.
0.4772
ANSWER: a
To make a probability statement we need to simply calculate the z-values for each.
Z183 = (X – μ ) / σ _ where σ _ = σ / n = 24 / 64 = 24/8 = 3
x
x
= (183 – 180) / 3 = 1
Z186 = (X – μ ) / σ _ = (186 – 180) / 3 = 2
x
P ( 1 < Z < 2 ) = 0.1359
24.
A population has a mean of 84 and a standard deviation of 12. A sample of 36
observations will be taken. The probability that the sample mean will be between 80.54 and 88.9
is
a.
0.0347
b.
0.7200
c.
0.9511
d.
None of the alternative ANSWERS is correct.
ANSWER: c
-same format at above.
5
25.
An estimate of a population parameter that provides an interval believed to contain the
value of the parameter is known as the
a.
confidence level
b.
interval estimate
c.
parameter value
d.
population estimate
ANSWER: b
26.
The confidence associated with an interval estimate is called the
a.
level of significance
b.
degree of association
c.
confidence level
d.
precision
ANSWER: c
27.
The z value for a 97.8% confidence interval estimation is
a.
2.02
b.
1.96
c.
2.00
d.
2.29
ANSWER: d
-simply use the z-table and find the z value that gives the middle 97.8%
28.
It is known that the variance of a population equals 1,936. A random sample of 121 has
been taken from the population. There is a .95 probability that the sample mean will provide a
margin of error of
a.
7.84 or less
b.
31.36 or less
c.
344.96 or less
d.
1,936 or less
ANSWER: a
Recall that the margin of error is the difference between the sample value and population
parameter and it is bounded by Zα/2* σ _ = (1.96*(
)/(
) < 7.84
x
Exhibit 8-1
In order to estimate the average time spent on the computer terminals per student at a local
university, data were collected from a sample of 81 business students over a one-week period.
Assume the population standard deviation is 1.2 hours.
29.
Refer to Exhibit 8-1. The standard error of the mean is
a.
7.5
b.
0.014
c.
0.160
d.
0.133
ANSWER: d
6
Recall that σ _ = σ /
x
n = 1.2/9 = 0.133
30.
Refer to Exhibit 8-1. With a 0.95 probability, the margin of error is approximately
a.
0.26
b.
1.96
c.
0.21
d.
1.64
ANSWER: a
Margin of Error = Zα/2* σ _ = 1.96*0.133≈0.26
x
31.
Refer to Exhibit 8-1. If the sample mean is 9 hours, then the 95% confidence interval is
a.
7.04 to 110.96 hours
b.
7.36 to 10.64 hours
c.
7.80 to 10.20 hours
d.
8.74 to 9.26 hours
ANSWER: d
_
Recall that x ± zα/2 σ. So just apply the formula to get 8.74 to 9.26. This means we are 95% sure
the true population mean falls within these values.
32.
If we change a 95% confidence interval estimate to a 99% confidence interval estimate,
we can expect the
a.
width of the confidence interval to increase
b.
width of the confidence interval to decrease
c.
width of the confidence interval to remain the same
d.
sample size to increase
33.
In hypothesis testing, the hypothesis tentatively assumed to be true is
a.
the alternative hypothesis
b.
the null hypothesis
c.
either the null or the alternative
d.
None of the other answers are correct.
ANSWER: b
34.
Your investment executive claims that the average yearly rate of return on the stocks she
recommends is at least 10.0%. You plan on taking a sample to test her claim.
The correct set of hypotheses is
a.
H0:  < 10.0%
Ha:   10.0%
b.
H0:   10.0%
Ha:  > 10.0%
c.
H0:  > 10.0%
Ha:   10.0%
d.
H0:   10.0%
Ha:  < 10.0%
ANSWER: d
7
35.
A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces
of soft drink. Any overfilling or underfilling results in the shutdown and
readjustment of the machine. To determine whether or not the machine is
properly adjusted, the correct set of hypotheses is
a.
H0:  < 12
Ha:   12
b.
H0:   12
Ha:  > 12
c.
H0:   12
Ha:  = 12
d.
H0:  = 12
Ha:   12
ANSWER: d
36.
If a hypothesis test leads to the rejection of the null hypothesis, a
a.
Type II error must have been committed
b.
Type II error may have been committed
c.
Type I error must have been committed
d.
Type I error may have been committed
ANSWER: d
37.
The probability of making a Type I error is denoted by
a.

b.

c.
1-
d.
1-
ANSWER: a
38.
A one-tailed test is a
a.
hypothesis test in which rejection region is in both tails of the sampling
distribution
b.
hypothesis test in which rejection region is in one tail of the sampling distribution
c.
hypothesis test in which rejection region is only in the left tail of the sampling
distribution
d.
hypothesis test in which rejection region is only in the right tail of the sampling
distribution
ANSWER: b
Exhibit 9-1
n = 36
H0:   20
Ha:  > 20
39.
Refer to Exhibit 9-1. The test statistic equals
a.
2.3
b.
0.38
c.
-2.3
d.
-0.38
ANSWER: a
Since we have that the standard deviations is unknown but we have n< 30 it is a
z-stat = ( – μ) / σ _ = (20-24.6) / (12/
) = 4.6*6 / 12 = 2.3
x = 24.6
s = 12
x
8
40.
Refer to Exhibit 9-1. The p-value is
a.
0.5107
b.
0.0214
c.
0.0107
d.
2.1
ANSWER: c
41.
Refer to Exhibit 9-1. If the test is done at a .05 level of significance, the null hypothesis
should
a.
not be rejected
b.
be rejected
c.
Not enough information is given to answer this question.
d.
None of the other answers are correct.
ANSWER: b
9
Short Answer:
Question 1: Binomial Distribution
a.
Thirty-two percent of the students in a management class are graduate students. A
random sample of 5 students is selected. Using the binomial probability function, determine the
probability that the sample contains exactly 2 graduate students?
b.
A production process produces 2% defective parts. A sample of 5 parts from the
production is selected. What is the probability that the sample contains exactly two defective
parts? Use the binomial probability function and show your computations to answer this
question.
c.
When a particular machine is functioning properly, 80% of the items produced are nondefective. If three items are examined, what is the probability that one is defective? Use the
binomial probability function to answer this question.
ANSWER:
Recall that for each we use P( X = x) = ( nx ) px (1-p)(n-x)
a. let p = 0.32 and n = 5
then P( X = 2) = ( 52 ) 0.322 (1-0.32)(5-2) = 0.322
b. Similarly n = 5, but not p = 0.02
P( X = 2) = ( 52 ) 0.022 (1-0.02)(5-2) = 0.0037648
c. Now we have n = 2 with p = 0.80
P( X = 1) = ( 13 ) 0.801 (1-0.80)(3-1) = 0.384
10
Questions 2:
A random sample of 100 credit sales in a department store showed an average sale of $120.00.
From past data, it is known that the standard deviation of the population is $40.00.
a.
Determine the standard error of the mean.
b.
With a 0.95 probability, determine the margin of error.
c.
What is the 95% confidence interval of the population mean?
ANSWERS:
a.
Recall that σ _ = σ / n = 40 / 10 = 4
x
b.
Recall that the Margin of Error = Zα/2* σ _ = 1.96*4 = 7.84
x
_
c.
x ± Zα/2* σ _  120 +/- 7.84  112.16 to 127.84
x
11
Questions 3:
In order to determine the average weight of carry-on luggage by passengers in airplanes, a
sample of 36 pieces of carry-on luggage was weighed. The average weight was 20 pounds.
Assume that we know the standard deviation of the population to be 8 pounds.
a.
Determine a 97% confidence interval estimate for the mean weight of the carry-on
luggage.
b.
Determine a 95% confidence interval estimate for the mean weight of the carry-on
luggage.
_
ANSWERS: For both of the calculations we use the following formula: x ± Zα/2* σ _
x
We know that σ _ = σ /
n = 8 / 6 ≈1.33. If we note that our Zα/2 = +/- 1.96 for the 95% CI and
+/- 2.17 for 97%. Apply the formula with the values above and you get
x
_
a.
x ± Zα/2* σ _ = 20 ± 2.17*1.33 = 17.11 to 22.89
x
_
b.
x ± Zα/2* σ _ = 20 ± 1.96*1.33 = 17.39 to 22.61
x
12
Question 4:
The price of a particular brand of jeans has a mean of $37.99 (just assume 38) and a standard
deviation of $7. A sample of 49 pairs of jeans is selected. Answer the following questions.
a.
What is the probability that the sample of jeans will have a mean price less than $40?
b.
What is the probability that the sample of jeans will have a mean price between $38 and
$39?
c.
What is the probability that the sample of jeans will have a mean price within $3 of the
population mean?
ANSWERS: Here we simply find the z-state for each one: z-stat = ( – μ) / σ _
x
Note that σ _ = σ /
x
a.
b.
c.
n = 7 / 49 = 1
P ( X > 40 ) = P ( Z < 2 ) = 0.97778
P ( 38 < X < 39 ) = P ( 0 < Z < 1) = 0.339763
P (35 < X 41 ) = P (-3 < Z < 3) 0.9973
13
Question 5:
The average gasoline price of one of the major oil companies has been $1.00 per gallon.
Because of shortages in production of crude oil, it is believed that there has been a significant
increase in the average price. In order to test this belief, we randomly selected a sample of 36 of
the company’s gas stations and determined that the average price for the stations in the sample
was $1.10. Assume that the standard deviation of the population () is $0.12.
a.
State the null and the alternative hypotheses.
b.
Test the claim at  = .05.
c.
What is the p-value associated with the above sample results?
ANSWERS:
Note that σ _ = σ /
x
n = 0.12 /
36 = 0.02
b.
H0:   1
Ha:   1
It is a one tailed test because we think that the results of the shortage is an
increase.
z = ( – μ) / σ _ = (1.10 – 1) / 0.02 = 5; therefore, reject H0, there is sufficient
c.
evidence at  = .05 to conclude that there has been an increase in the average
price. We know this for two reasons.
The critical value we compare our Z to is 1.96. If it is significantly larger than
1.96 then we think the price is much higher.
We can also use the p-value method. If our p-value of our test stat is smaller than
 = .05, then we reject Ho and conclude there is a statistically significant result
and in this case suspect the price of gas has increased. The p-value of Z = 5 is
< 0.001. So this is much less than 0.05
almost zero  < 0.001 because it is smaller than the last computed table value
a.
x
14
Question 6:
a. At a local university, a sample of 49 evening students was selected in order to determine
whether the average age of the evening students is significantly different from 21. The average
age of the students in the sample was 23 with a standard deviation of 3.5. Determine whether or
not the average age of the evening students is significantly different from 21. Use a 0.1 level of
significance.
ANSWER:
Note that σ _ = σ /
x
H0:  = 21
Ha:  =/ 21
n = 3.5 /
49 = 0.5
z =( – μ) / σ _ = (23 – 21) / 0.5 = 4; therefore, reject H0, there is
x
sufficient evidence at  = .1 to conclude that the average age of the evening
students is significantly different from 21
b. In order to determine the average price of hotel rooms in Atlanta, a sample of 64 hotels was
selected. It was determined that the average price of the rooms in the sample was $112 with
standard deviation of $16. Use a 0.05 level of significance and determine whether or not the
average room price is significantly different from $108.50.
ANSWER: Same method as above. So make sure to find σ _ .
x
H0:  = 108.50
Ha:  =/ 108.50 z = ( – μ) / σ _ = (112 – 108.5) / 2 = 1.75; therefore, do not reject H0,
x
there is not sufficient evidence at  = .05 to conclude that that the average room
price is significantly different from $108.50.
15