Review Evaluation 2 – Statistics of business and economics practice materials in English – Question – Studocu

Question 1: Given the folllowing distribution.

X 10 20 30 40
P(x) 0 0 0 0.

The expected and variance values of X (respectively) are

a/ 24 and 84
b/ 25 and 9.
c/ 30 and 85
d/ 100 and 93.

ANS: A

Question 2: Given the folllowing distribution.

X 10 20 30 40
P(x) 0 0 0 0.

Suppose that Z = -2X + 3. Then the expected and variance values of Z respectively are

a/ – 45 and 336
b/ – 45 and 339
c/ – 45 and 168
d/ 100 and 93.

E(Z) = E(-2X +3) = E(-2X) +E(3) = -2E(X) + 3 = -2*24 + 3 = -45?

Var(Z) = Var(-2X +3) = Var(-2X) + Var(3) = 4Var(X) + 0 = 484 = 336

ANS: A

Question: Given a random variable X (number of products sold

in a day) with a probability distribution table:

X 10 20 30 40
P(x) 0 0 0 0.

What is the average amount sold in a day, knowing that the selling price

of a product is 100 thousand VND per day?

a/100*24 thousand VND per day.

b/

c/

c/
d/

Question 4: X is a normally distributed random variable with a

mean of 5 and a variance of 4. The probability that X is greater

than 10 is

a/ 0.
b/ 0.
c/ 0.
d/ 0.

ANS: A

X~N(5;4)

Hint: P(X>10) = 1- P(X<10) =

#######  

5 10 5
1 1 2 1 0.
2 2

X
P P Z

   
        
 

Given that X~N(  ;
 2
)

a/ Form #1: Determine P(X< a) =?

Explain:

We have

#######    

X a
X a X a

 
 
 

   
       
 

Then

#######  

X a a
P X a P P Z

  
  

      
       
    (Based on Table 1)

where

~ 0,1 

X
Z N






.

b / Form #2 : P(X>a) = 1- P(X<a) (based on Table 1)

c/ Form#3: P(a<X<b) = P(X<b) – P(X<a) -> Form #

Chapter 6

####### Question 1: How will the confidence interval be affected

####### if the confidence level increases from 95% to 97% for the

####### same sample data?

   

2 2

1 2; ; x E x E ;

s s

x z x z

n n

   

 

    

 

   

2

z 
is determined by

P Z  z  2  12


  
(table 1)

Example: x_tb = 5; E1 = 1-(confidence interval (4; 6)

x_tb = 5; E1 = 2-(confidence interval (3; 7)
wider/larger

a/

Question 4: Consider a random sample size of 31, with sample

mean of 45 and sample standard deviation of 5. With 95%

confidence level, the margin of error is

2

( ) s
E z
n




 

a/ 1.

b/ 0.

c/ 0.

d/ 2.

Question 5:

A survey of 100 Americans reports that 65% of them own a car.

With a 95% confidence level, what is the margin of error for

confidence interval in this example?

 

1 2 2 2

(1 ) (1 )
;p p ; p

p p p p
p z z
n n

 

   
     
 

The margin of error, 2

p p (1 )
E z
n




 

Where 2

z 
is determined by

 2  1

2

P Z z 


  

a/0.

b/ 0.

c/ 0.

d/0.

Question 6 :

In a random sample of 50 intramural basketball players at a
large university, the average points per game was 8 points,
with a standard deviation of 2 points and a 95%
confidence level.

Which of the following statements about the mean points
scored by all intramural basketball payers is correct?

a/ (7; 8) points

b/ (7; 8) points.

c/ (5; 10) points.

d/ (7; 8) points.

Question 7:

####### A university planner is interested in determining the

####### percentage of spring semester students who will attend

####### summer school. She takes a pilot sample of 160 spring

####### semester students discovering that 56 will return to

####### summer school.

c/

d/

Question 8:

A random sample of 49 lunch customers was taken at a

restaurant. The average amount of time the customers in the

sample stayed in the restaurant was 45 minutes with a standard

deviation of 14 minutes.

With a 95% level of confidence, how large of a sample would

have to be taken to provide a margin of error of 2 minutes or

less?

The margin of error: 2

( ) s
E z
n




 

a/ 121
b/
c/
d/

Question: Assume that test scores from the college entrance exam are

normally distributed with a mean of 4 and a standard deviation of 1.

What is the test score (x0) so that 98% of candidates have test scores

below that score (x0)?

####### Sol: X~N(4; 1^2)

0

0

0

0
0

1. 0. 1 1
2. 0. 1

3. 2 2 4 6.
   1

( )

( 4 )

P Z

P X x

X x
P

x

x
x

 
 
 





  


     

 

  

Z(alpha) = 1, the test statistic Z = 0

•  do not reject H0 (we tend to accept H0)

a/ the proportion of the population is not greater than 42%.

b/ the proportion of the population is greater 42%.

c/ the proportion of the population equals to 42%.

d/ Not enough information given to answer this question.

Question : In a test of H 0 : 100 against Ha : 100, the sample data

yielded the test statistic Z = 2. At 5% level of significance, your

conclusion about the population mean is

Z= 2; Z(alpha) = 1.

Do not reject H0 ( we tend to accept H0).

a/ significantly greater than or equal to 100.

b/ significantly less than or equal 100.

c/ Not enough information given to answer this question.

d/ significantly equal to 100.

Question. Given the following ANOVA table.

Source of
Variation

Sum of
Squares

Degrees of
Freedom

Mean
Square F
Between

groups 36 2 —-?
Error (Within
Groups) 240 _—– —–

Total —— 42

Refer to the above table, the test statistic (F) is

a/ 3

b/ 2

c/ 10

d/ 13

Question : Given the following ANOVA table.

Source of
Variation

Sum
of
Squares

Degrees
of
Freedom

Mean
Squares
F
Between
groups 1,500 2 750 5

Within groups 4500 — 150

Total 6,000 32

Refer to the above table, the null hypothesis is to be tested at the 5% level of
significance,

The null hypothesis is to be tested at the 5% level of significance, your conclusion is

a/ one mean is different from the others.

b/

c/