business stat exam help

16 may 

Practice Final Exam

1

Questions 1-4 are based on the following information:

An insurance company analyst is interested in analyzing the dollar value of damage in
automobile accidents. She collects data from 115 accidents, and records the amount of
damage as well as the age of the driver. The results of her regression analysis are listed
below.

1. The regression equation is:

A) 𝑦= 10725.802 + 69.964x
B) 𝑦= 114 + 113x
C) 𝑦 = 1535.215 + 34.625x
D) 𝑦 = 113 + 114x

2. How would you best explain the y-intercept in this situation?

A) For each additional 1-year increase in the age of the driver, we would expect
damage to increase by $10,726.
B) For each additional 1-year increase in the age of the driver, we would expect
damage to increase by $70.
C) It makes no sense to explain the intercept in this situation, since we cannot
have a driver with an age of zero.
D) The average amount of damage was $10,726.

3. What would be the dollar value of an accident involving a 25-year-old driver?
A) $11,836.56
B) $10,795.47
C) $13,372.58
D) $12,474.90

Practice Final Exam

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4. Which of the following statements is the best explanation of the R2?
A) 3.5% of the accident damage can be explained by the age of the driver.
B) 3.5% of the variation in accident damage can be explained by variation in the
age of the driver.
C) 3.5% of the coefficients, t-stat, and p-value can be explained by the age of the
driver.
D) 3.5% of the total error can be explained by the SSE.

5. The residual is defined as the difference between the actual value of:

A) y and the estimated value of y.
B) x and the estimated value of x.
C) y and the estimated value of x.
D) x and the estimated value of y.

Questions 6-10 are based on the following information:
A sales manager is interested in determining the relationship between the amount spent
on advertising and total sales. The manager collects data for the past 24 months and runs
a regression of sales on advertising expenditures. The results are presented below but,
unfortunately, some values identified by asterisks are missing.

6. What are the degrees of freedom for residuals?
A) 21
B) 22
C) 23
D) 24

7. The value of mean square error (residual) is:
A) 1678.9
B) 1,554.25
C) 1,493.63
D) 1,407.35

Practice Final Exam

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8. The total degrees of freedom is:
A) 21
B) 22
C) 23
D) 24

9. A regression analysis between sales (in $1000) and advertising (in $100) resulted in
the following least squares line: 𝑦= 75 + 5x. This implies that if advertising is $800,
then the predicted amount of sales (in dollars) is:

A) $4075
B) $115,000
C) $164,000
D) $179,000

10. In a regression problem, a coefficient of determination 0.90 indicates that:

A) 90% of the y values are positive.
B) 90% of the variation in y can be explained by the regression line.
C) 90% of the x values are equal.
D) 90% of the variation in x can be explained by the regression line.

11. If the least squares regression line 𝑦= -2.88 + 1.77x and the coefficient of
determination is 0.81, the coefficient of correlation is:

A) -0.88
B) +0.88
C) +0.90
D) -0.90

12. In a simple regression analysis (where Y is a dependent and X an independent
variable), if the Y intercept is positive, then
A) there is a positive correlation between X and Y
B) if X is increased, Y must also increase
C) if Y is increased, X must also increase
D) None of these alternatives is correct.

13. In regression analysis, the independent variable is
A. used to predict other independent variables
B. used to predict the dependent variable
C. called the intervening variable
D. the variable that is being predicted

Practice Final Exam

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14. In a regression analysis if SSE = 200 and SSR = 300, then the coefficient of
determination is
A. 0.6667

B. 0.6000
C. 0.4000
D. 1.5000

15. Regression analysis was applied between sales (in €1000) and advertising (in €100)

and the following regression function was obtained.

𝑦= 500 + 4 x
Based on the above estimated regression line if advertising is €10,000, then the
point estimate for sales (in euros) is

A. €900
B. €900,000

C. €40,500
D. €505,000

16. A regression analysis between demand (Y in 1000 units) and price (X in euros)

resulted in the following equation

𝑦= 9 – 3 x

The above equation implies that if the price is increased by €1, the demand is
expected to

A. increase by 6 units
B. decrease by 3 units
C. decrease by 6,000 units
D. decrease by 3,000 units

17. Regression analysis was applied between sales (Y in €1,000) and advertising (X in
€100), and the following estimated regression equation was obtained.

𝑦= 80 + 6.2 x

Based on the above estimated regression line, if advertising is €10,000, then the
point estimate for sales (in euros) is

A. €62,080
B. €142,000
C. €700

D. €700,000

Questions 18 and 19 are based on the following information:
A large milling machine produces steel rods to certain specifications. The machine is
considered to be running normally if the standard deviation of the diameter of the rods is
at most 0.15 millimeters. The line supervisor needs to test the machine is for normal

Practice Final Exam

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functionality. The quality inspector takes a sample of 25 rods and finds that the sample
standard deviation is 0.19.

18. What are the null and alternative hypotheses for the test?

a. H0 : σ2 > 0.15 and H1 : σ2 ≤ 0.15
b. H0 : σ ≥ 0.15 and H1 : σ < 0.15 c. H0 : σ = 0.0225 and H1 : σ ≠ 0.0225 d. H0 : σ2 ≤ 0.0225 and H1 : σ2 > 0.0225

19. Which of the following statements is the most accurate?

a. Reject null hypothesis at α = 0.025
b. Fail to reject null hypothesis at α ≤ 0.10
c. Reject null hypothesis at α = 0.05
d. Reject null hypothesis at α = 0.01

20. Suppose you have the following null and alternative hypotheses: H0 : σ2 = 34.5 and

H1 : σ2 > 34.5. If you take a random sample of 15 observations from a normally
distributed population and find that s2 = 48.1, what is the most accurate statement
that can be made about the p-value for this test?
a. 0.01 < p-value < 0.025 b. 0.025 < p-value < 0.05 c. p-value < 0.01 d. 0.05 < p-value < 0.10

21. A sample of 20 cans of tomato juice showed a standard deviation of 10.0 grams. A

95% confidence interval estimate of the variance for the population is
a. 144.6 to 533.3
b. 139.0 to 495.3
c. 55.6 to 198.1
d. 57.8 to 213.3

22. The producer of bottling equipment claims that the variance of all their filled bottles

is 0.027 or less. A sample of 30 bottles showed a standard deviation of 0.2. The p-
value for the test is

a. between 0.025 to 0.05
b. between 0.05 to 0.01

c. 0.05
d. 0.025

Practice Final Exam

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23. The chi-squared value for a one-tailed (upper tail) hypothesis test of the population
standard deviation at 95% confidence and a sample size of 25 is

a. 33.196
b. 36.415

c. 39.364
d. 37.652

24. We are interested in testing a claim that the standard deviation of a population is no

more than 5.0. The null hypothesis for this test is
a. Ho: σ2 > 25.0
b. Ho: σ2 ≤ 25.0
c. Ho: σ2 ≤ 5.0

d. Ho: σ2 ≥ 5.0

25. The value of F0.025 with 15 numerator and 10 denominator degrees of freedom is
a. 3.06

b. 2.54
c. 2.85

d. 3.52

Questions 26 and 27 are based on the following information:
The standard deviation of the ages of a sample of 16 executives from large companies
was 8.2 years; while the standard deviation of the ages of a sample of 26 executives from
small companies was 12.8 years.

26. What is the test statistic to test if there any difference in the standard deviations of the

ages of all the large-company and small-company executives.
a. 2.32
b. 2.25
c. 2.21
d. 3.19

27. We would like to test to see if there is any difference in the standard deviations of the

ages of all the large-company and small-company executives. What is the critical
value of F at 10% significance level?

a. 2.12
b. 2.10
c. 2.21
d. 3.10

Practice Final Exam

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Questions 28-30 are based on the following information:
Suppose two food preservatives are extensively tested and determined safe for use in
meats. A processor wants to compare the preservatives for their effects on retarding
spoilage. Suppose 16 cuts of fresh meat are treated with preservative A and another 12
cuts of meat are treated with preservative B. The number of hours until spoilage begins is
recorded for each of the 28 cuts of meat. The results are summarized in the table below.
Preservatives A Preservatives B
Sample means 95.25 100.50
Sample standard deviation 13.45 10.55

28. The value of the test statistic for determining if there is a difference in the population

variances for preservatives A and B is equal to:
e. 0.784
f. 1.625
g. 1.275
h. 1.129

29. The numerator and denominator degrees of freedom associated with the test statistic

for determining if there is a difference in the population variances for preservatives A
and B are, respectively:
i. 16 and 12
j. 12 and 16
k. 15 and 11
l. 11 and 15

30. The most accurate statement that can be made about the p-value for testing whether

there is a difference in the population variances for preservatives A and B is:
a. p-value > 0.05
b. p-value = 0.00
c. p-value < 0.01 d. 0.01 < p-value < 0.05

31. An ANOVA procedure is used for data that was obtained from four sample groups

each comprised of five observations. The degrees of freedom for the critical value of
F are

a. 3 and 20
b. 3 and 16
c. 4 and 17
d. 3 and 19
32. The analysis of variance is a procedure that allows statisticians to compare three or

more population:
a. Means.
b. Proportions.
c. Variances.
d. Standard deviations.

Practice Final Exam

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33. In analysis of variance, between-treatments variation is measured by the:

a. SSE
b. SSTR
c. SST
d. All of the above

Questions 34-38 are based on the following information:
Consider the following ANOVA table:

Source of Variation SS df MS F

Treatments SSTR=4 2 ? ?
Error SSE= ? ? ?
Total SST=34 14

34. What are the degrees of freedom for the Errors?

a. 3
b. 4
c. 13
d. 12

35. What is the value of SSE?

a. 29
b. 20
c. 30
d. 33

36. What is the Mean-Square for Treatments?

a. 2
b. 1
c. 4
d. 0

37. What is Mean-Square Errors?

a. 3
b. 4
c. 3.5
d. 2.5

38. What is the value of the F-statistic?

a. 0.5
b. 0.8
c. 1.2
d. 1.6

Practice Final Exam

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Answer Key
1 A
2 C
3 D
4 B
5 A
6 B
7 A
8 C
9 B

10 B
11 C
12 D
13 B
14 B
15 B
16 D
17 D
18 D
19 C
20 A
21 D
22 A
23 B
24 B
25 D
26 B
27 C
28 B
29 C
30 A
31 B
32 A
33 B
34 D
35 C
36 A
37 D
38 B

1

Chapter11

• Interval Estimation of a Population Variance

𝑛 − 1 𝑠!

𝜒!
!

! ≤ 𝜎
! ≤

𝑛 − 1 𝑠!

𝜒
!!!!

!

where 𝜒! values are based on a chi-squared distribution with (n-1) degrees of freedom.

• Hypothesis Testing about the Variances of Two Populations

Test Statistic

𝑭 =
𝒔𝟏𝟐

𝒔𝟐𝟐

The critical value 𝐹! is based on an F distribution with 𝑛! − 1 (numerator) and 𝑛! − 1
(denominator) degrees of freedom

Chapter 13

Test for the Equality of k Population Means

𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 =
𝑆𝑢𝑚 𝑜𝑓 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠

𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡

𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟 =
𝑆𝑢𝑚 𝑜𝑓 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟

𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝐸𝑟𝑟𝑜𝑟

𝐹 =
𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠

𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟

The critical value 𝐹!is based on an F distribution with k-1 numerator degrees of freedom and n-k
denominator degrees of freedom.

Chapter 14 (Simple Linear Regression)
b 1 =

!!! ! (!!! !)
!!! ! ²

b1: slope coefficient in a simple regression line
b0: intercept coefficient (coefficient of the constant) in a simple regression line

𝑦: estimated value of the dependent variable y

b0 = 𝑦 − b!x
𝑦 = b0 + b 1 x (estimated equation)

2

𝑆𝑆𝐸 = (𝑌! − 𝑌)²
SST = SSR + SSE ⟹

(𝑌! − 𝑌)² = SST = ( 𝑌 −𝑌)²+ (𝑌! − 𝑌)²

𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 = 𝑆𝑆𝑅 = ( 𝑌 −𝑌)²

total variation in Y = SST = (𝑌! − 𝑌)²

R2 = !!”

!!”
=coefficient of determination,

sample correlation coefficient = 𝑟!,! = 𝑅² (carry the sign of b1)

tstat = b1 / se(b1), se(b1):standard error of the slope coefficient,

residual = error = Y-𝑌

1

Chapter11

• Interval Estimation of a Population Variance

𝑛 − 1 𝑠!

𝜒!
!

! ≤ 𝜎
! ≤

𝑛 − 1 𝑠!

𝜒
!!!!

!

where 𝜒! values are based on a chi-squared distribution with (n-1) degrees of freedom.

• Hypothesis Testing about the Variances of Two Populations

Test Statistic

𝑭 =
𝒔𝟏𝟐

𝒔𝟐𝟐

The critical value 𝐹! is based on an F distribution with 𝑛! − 1 (numerator) and 𝑛! − 1
(denominator) degrees of freedom

Chapter 13

Test for the Equality of k Population Means

𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 =
𝑆𝑢𝑚 𝑜𝑓 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠

𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡

𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟 =
𝑆𝑢𝑚 𝑜𝑓 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟

𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝐸𝑟𝑟𝑜𝑟

𝐹 =
𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠

𝑀𝑒𝑎𝑛 𝑆𝑞𝑢𝑎𝑟𝑒 𝐸𝑟𝑟𝑜𝑟

The critical value 𝐹!is based on an F distribution with k-1 numerator degrees of freedom and n-k
denominator degrees of freedom.

Chapter 14 (Simple Linear Regression)
b 1 =

!!! ! (!!! !)
!!! ! ²

b1: slope coefficient in a simple regression line
b0: intercept coefficient (coefficient of the constant) in a simple regression line

𝑦: estimated value of the dependent variable y

b0 = 𝑦 − b!x
𝑦 = b0 + b 1 x (estimated equation)

2

𝑆𝑆𝐸 = (𝑌! − 𝑌)²
SST = SSR + SSE ⟹

(𝑌! − 𝑌)² = SST = ( 𝑌 −𝑌)²+ (𝑌! − 𝑌)²

𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 = 𝑆𝑆𝑅 = ( 𝑌 −𝑌)²

total variation in Y = SST = (𝑌! − 𝑌)²

R2 = !!”

!!”
=coefficient of determination,

sample correlation coefficient = 𝑟!,! = 𝑅² (carry the sign of b1)

tstat = b1 / se(b1), se(b1):standard error of the slope coefficient,

residual = error = Y-𝑌