BUAN 6359, Spring 2022, Exam #2 Practice Questions

Question 1

The probability that the interval estimation procedure will generate an interval that
does NOT contain the actual value of the population parameter being estimated is
the

A) level of significance
B) confidence level
C) confidence coefficient
D) error factor
E) None of the above

SOLUTION: (A)

Question 2

The following data represent a random sample of 9 marks (out of 10) on a statistics
quiz:

7 9 7 5 4 8 3 10 9

Based on the above sample, the instructor finds the sample average as 6.89 and
sample standard deviation as 2.42. Estimate the population mean with 90%
confidence.
A. [4.0,8.0]
B. [5.0,6.675]
C. [4.47,9.31]
D. [5.39,8.39]
E. [5.56,8.22]
SOLUTION:
In this case, population standard deviation is not known. Therefore, we will
use 𝑥̅ ± 𝑡 / √ where n-1=9-1=8 is the degrees of freedom. 90% confidence
level means 1-=0.90, therefore, =0.1 and /2=0.05.

Further, it is given in the question that 𝑥̅ = 6.89. Also, sample standard

deviation (NOT the population standard deviation) is given. That is, s=2.42.
It follows that
BUAN 6359 Practice Questions

. .
6.89 ± 𝑡 . , = 6.89 ± 1.86 = 6.89 ± 1.5 =[5.39,8.39]
√ √

Question 3
In hypothesis testing, if the null hypothesis has been rejected when the
alternative hypothesis has been true,
A. A Type I error has been committed
B. A Type II error has been committed
C. Either a Type I or Type II error has been committed
D. The correct decision has been made
E. None of the above

SOLUTION: (D)

If the alternative hypothesis is true, then the null has to be rejected. That
is a correct decision.

Question 4

In hypothesis testing, if the null hypothesis has not been rejected when the
alternative hypothesis has been true,
A) A Type I error has been committed
B) A Type II error has been committed
C) Either a Type I or Type II error has been committed
D) The correct decision has been made
E) None of the above answers are correct.

SOLUTION: (B)

Question 5
The manager of a shoe retailer is considering a new bonus plan for his/her sales
personnel. Currently, the mean sales rate per salesperson is 100 shoes per month.
The manager thinks that the sales people should be rewarded in recognition of their
effort if the average number of shoes sold per month by a salesperson () has
increased. The correct set of hypotheses for testing if a bonus plan should be
implemented is

A. H0:   100 Ha:   100

B. H0:   100 Ha:   100

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BUAN 6359 Practice Questions

C. H0:   100 Ha:   100

D. H0:   100 Ha:   100
E. H0:  = 100 Ha:   100

SOLUTION: (B)
Note that the tests constructed in parts (A) and (C) are not even theoretically
correct. In Part (A), H0 and Ha do not contradict (THEY SHOULD). In Part
(C), the possibility of  being equal to 100 in included in Ha. Remember, that
possibility should never be part of Ha.

The manager thinks the personnel should be rewarded if the number of shoe
sales has increased above 100. That is the research question ( 100 ?),
therefore, it has to be in alternative hypothesis.

Question 6

The tuft bind strength of a synthetic material used to make carpets is known to
have a mean of 100 lb and standard deviation of 20 lb. A sample of 40 is randomly
selected and the average strength was found as 105lb. Let µ denote the average tuft
strength. Could it be that the average tuft strength is above 100 lb?

Based on the information, what should be the appropriate null and alternative
hypotheses for the purpose of your analysis?

A) H0:  = 100 Ha:  ≠ 100

B) H0:   100 Ha:  > 100
C) H0:   100 Ha:  < 100
D) H0:  = 100 Ha:  = 105
E) None of the above

SOLUTION: (B)

H0:   100 Ha:  > 100

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BUAN 6359 Practice Questions

Question 7

The time to process customer orders at an internet-based business is known to be

normally distributed with mean 3.5 days and standard deviation 0.5 day. A random
sample of twenty orders is selected, and the average processing time is found to be
4 days. The owner of the business considers hiring somebody to improve the process
if the average order processing time (i.e., µ) has increased above 3.5.

Based on the information, what should be the appropriate null and alternative
hypotheses for the purpose of the analysis?

A) H0:  ≠ 3.5 Ha:  = 4

B) H0:   3.5 Ha:  < 3.5
C) H0:   3.5 Ha:  > 3.5
D) H0:  = 3.5 Ha:   3.5
E) None of the above

SOLUTION: (C)

H0:   3.5 Ha:  > 3.5

A highway patrol officer believes that the average speed of cars traveling over a
certain stretch of highway exceeds the posted limit of 70 mph. The speeds of a
random sample of 100 cars were recorded. The sample average and the sample
standard deviation were found as 71 mph and 5 mph, respectively. The officer
develops the following test: H0:   70 Ha:   70.

Answer Questions 8 through 10 based on this information.

Question 8
The test statistic is:
A) 1.82
B) 2.0
C) 2.5
D) 0.05
E) None of the above

SOLUTION:

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BUAN 6359 Practice Questions

The test statistic is:

𝑡 = = =2
/√ /√

Question 9

When level of significance is 0.05 (=0.05), what is the critical value that
defines the rejection region?

A) 1.66
B) 2.626
C) 1.645
D) 0.6776
E) 1.96

SOLUTION: (A)
The rejection region is 𝑡 > 𝑡 , . That is, the critical value is
t0.05,99=1.66

Question 10

At .05 level of significance, it can be concluded that the average speed of cars
A) not significantly greater than 70
B) significantly greater than 70
C) not significantly less than 70
D) significantly less than 70
E) None of the above
SOLUTION: (B)
The rejection region is 𝑡 > 𝑡 , . The critical value is t0.05,99=1.66.
Since 𝑡 = 2 > 𝑡 , = 1.66, we reject the null hypothesis, and
conclude that the mean speed of cars is significantly greater than 70.

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BUAN 6359 Practice Questions

Question 11
The table below includes data regarding the service times (in minutes) of six
randomly selected customers at the checkout.

Table 1: Customer service times in minutes

5 15 10 7 5 6

Which one of the following is a good point estimate for the variance of the
customer service times?

A. 15.19 minutes
B. 12.67 minutes
C. 15.19 (minutes)2
D. 3.9 minutes
E. 3.56 (minutes)2

Solution: (C)
Sample variance is an unbiased estimator of the population variance and is
∑( ̅) ( ) ( ) ( ) ( ) ( ) ( )
given = = 15.19 (minutes)2

Consider the following hypothesis test when 𝛼 = 0.05.

𝐻 :𝜇 − 𝜇 = 0
𝐻 :𝜇 − 𝜇 ≠ 0
The following results are from independent samples taken from two populations.

Sample 1 Sample 2
𝑛 = 35 𝑛 = 40
𝑥̅ = 13.6 𝑥̅ = 10.1
𝑠 = 5.2 𝑠 = 8.5

Assuming that population variances are equal, answer Questions 12 through 17

based on the above two samples.

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BUAN 6359 Practice Questions

Question 12
The pooled estimate of the variance (𝑠 ), is:

A) 7.16
B) 6.85
C) 46.93
D) 2.928
E) 51.19

SOLUTION: (E)
(𝒏𝟏 − 𝟏)𝒔𝟐𝟏 + (𝒏𝟐 − 𝟏)𝒔𝟐𝟐 𝟑𝟒 × 𝟓. 𝟐𝟐 + 𝟑𝟗 × 𝟖. 𝟓𝟐
𝒔𝟐𝒑 = = = 𝟓𝟏. 𝟏𝟗
𝒏𝟏 + 𝒏𝟐 − 𝟐 𝟕𝟑

Question 13
The test statistic is:
A) 0.2954
B) 2.114
C) 0.489
D) 0.2147
E) 7.61

SOLUTION: (B)

( ̅ ̅ ) ( . . )
𝑡 = = = 2.114
√ .

Question 14
The rejection region is:
A) −1.993 ≤ 𝑡 ≤ 1.993
B) 𝑡 > 1.993
C) 𝑡 < 1.993
D) 𝑡 < −1.993 or 𝑡 > 1.993
E) 𝑧 < −1.993 or 𝑧 > 1.993

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BUAN 6359 Practice Questions

SOLUTION: (D)
The rejection region is
|𝑡 | > 𝑡 / ,

t0.05/2,35+40-2= t0.025,73=1.993. Therefore, our rejection region is 𝑡 < −1.993 or

𝑡 > 1.993.

> qt(0.025,73,lower.tail=FALSE)
[1] 1.992997

Question 15
What is the p-value?
A) 0.981
B) 0.038
C) 0.019
D) 0.95
E) 0.05

SOLUTION: (B)
Because t0>0, p-value=2P(T73>2.114)=0.038

> 1-pt(2.114,73,lower.tail=TRUE)
[1] 0.01896587

Question 16
What is your conclusion about the hypothesis test in consideration?
A) 𝜇 is greater than 𝜇
B) 𝜇 is no less than 𝜇
C) 𝜇 is not equal to 𝜇
D) Current evidence is not enough to reject null
E) 𝜇 is less than 𝜇

SOLUTION: (C)
Since p-value=0.038<0.05, we reject the null. This means 𝜇 is not equal to 𝜇 .

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BUAN 6359 Practice Questions

Question 17

In order to compare the responses for two treatments (those are A and B), following
sample data were collected from a randomized experiment, and summarized in file
Set3Q.xls:

Response Treatment
1202.6 A
830.1 A
372.4 A
345.5 A
321.2 A
244.3 A
163.0 A
147.8 A
95.0 A
87.0 A
2745.6 B
1697.8 B
1656.0 B
978.0 B
703.4 B
489.1 B
430.0 B
334.1 B
302.8 B
274.7 B

Assume that, a higher response is considered as a measure of a treatment’s

effectiveness. After the data is imported, the following R script is run and the
corresponding output is obtained:

> newdata=transform(Set3Q,logresponse=log(Response))
> ttestlog=t.test(logresponse~Treatment, data=newdata)
> ttestlog

Welch Two Sample t-test

data: logresponse by Treatment

t = -2.5695, df = 17.946, p-value = 0.01932
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.7535206 -0.1757552
sample estimates:
mean in group A mean in group B
5.595860 6.560498

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BUAN 6359 Practice Questions

Which one of the following is correct?

A) The median response under treatment B is 2.57 times as large as median
response under treatment A.
B) The median response under treatment B is [95% CI (-1.75,-0.17)] times as
large as the median response under treatment A.
C) Response of an experimental unit to treatment B is 2.57 times as large as
its response to treatment A.
D) Population distributions are normal.
E) The current samples are not evidence enough to conclude anything.

SOLUTION to Q17: (C)

This is a randomized experiment. Let 𝑍̅ be the average of the logged

response values under treatment A, and let 𝑍̅ be the average of the logged
response values under treatment B. In this case, 𝑍̅ =5.5959 and 𝑍̅ = 6.5606
We say “It is estimated that the response of an experimental unit to
treatment B is exp(𝑍̅ − 𝑍̅ ) = exp (6.5606 − 5.5959) = 2.57 times as large as
its response to treatment A. (See slide #20 on Unit 6 lecture notes).

Question 18

Consider the following R script and its output for a completely randomized and
balanced experimental design.

> attach(AnovaQ)
> z <- aov(Response~TreatmentType)
> summary(z)
Df Sum Sq Mean Sq F value Pr(>F)
TreatmentType 2 1488 744.0 5.498 0.0162 *
Residuals 15 2030 135.3
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

How many different treatments are there in this analysis?

A) 2
B) 9
C) 3
D) 11
E) 12

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BUAN 6359 Practice Questions

SOLUTION: (C)
k-1=2 where 2 is the degrees of freedom associated with MST.
Therefore, there k=3.

Question 19

Consider the following R script and its output for a completely randomized and
balanced experimental design.

> attach(AnovaQ)
> z <- aov(Response~TreatmentType)
> summary(z)
Df Sum Sq Mean Sq F value Pr(>F)
TreatmentType 2 1488 744.0 5.498 0.0162 *
Residuals 15 2030 135.3
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

How many observations are there in this analysis?

A) 18
B) 9
C) 3
D) 17
E) 15

SOLUTION: (A)
n-k=15 where 15 is the degrees of freedom associated with MSE. Because
k=3, it turns out that n=18.

Question 20

Consider the following R script and its output for a completely randomized and
balanced experimental design.

> attach(AnovaQ)
> z <- aov(Response~TreatmentType)
> summary(z)
Df Sum Sq Mean Sq F value Pr(>F)
TreatmentType 2 1488 744.0 5.498 0.0162 *
Residuals 15 2030 135.3
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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BUAN 6359 Practice Questions

In order to determine if there is a significant difference between the population

averages of the first and second treatments, Fisher’s Least Significant Difference
Method is used. What is the LSD value at =0.01?
A) 14.508
B) 97.433
C) 11.893
D) 19.79
E) 9.067

SOLUTION: (D)
LSD is given by

𝐿𝑆𝐷 = 𝑡 ⁄ , 𝑀𝑆𝐸 + where v=n-k.

Because it is a balanced experiment, the number of replications (sample

sizes) for each treatment is the same. Therefore, n 1=n2=18/3=6.

𝐿𝑆𝐷 = 𝑡 ⁄ , 𝑀𝑆𝐸 + =𝑡 . ⁄ , 135.3 + =2.9476.716=19.79

Question 21

Consider a completely randomized and balanced experimental design, where 𝑥̅ = 119,

𝑥̅ = 109.625, 𝑥̅ = 99.5, n=24 (n stands for number of observations), and LSD=12.49
(at =0.05). Based on Fisher’s LSD Method, which of the following can be concluded
about the population means?
A) At =0.05, there is no significant difference between µ1 and µ3.
B) At =0.05, there is no significant difference between µ1 and µ2.
C) At =0.05, there is significant difference between µ2 and µ3.
D) At =0.05, there is no significant difference among µ 1, µ2 and µ3.
E) At =0.05, all population means are significantly different than one
another.

SOLUTION: (B)

𝐿𝑆𝐷 = 𝑡 ⁄ , 𝑀𝑆𝐸 + =𝑡 . ⁄ , 144.3 + =2.07966.006=12.49

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We conclude that µi and µj differ if:

𝑥̄ − 𝑥̄ > 𝐿𝑆𝐷
In this example, |𝑥̄ − 𝑥̄ | = 9.375 < 𝐿𝑆𝐷 = 12.49, |𝑥̄ − 𝑥̄ | = 10.125 < 𝐿𝑆𝐷 =
12.49, and |𝑥̄ − 𝑥̄ | = 19.5 > 𝐿𝑆𝐷 = 12.49. Therefore, there is no significant
difference between µ1 and µ2, there is no significant difference between µ2
and µ3. But, there is a significant difference between µ1 and µ3.

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