Question 1

The distribution of Student's t is

Accepted Answer

A)  symmetrical. 
B)  negatively skewed. 
C)  positively skewed. 
D)  a discrete probability distribution. 
A)  symmetrical. 
B)  negatively skewed. 
C)  positively skewed. 
D)  a discrete probability distribution. A

Question 2

A bank wishes to estimate the mean balances owed by customers holding MasterCard. The
Population standard deviation is estimated to be $300. If a 98 percent conﬁdence interval is used
And an interval of $75 is desired, how many cardholders should be sampled?

Accepted Answer

A)  44 
B)  212 
C)  629 
D)  87 
A)  44 
B)  212 
C)  629 
D)  87 D

Question 3

Dottie Kleman is the "Cookie Lady." She bakes and sells cookies at 50 different locations. Ms.
Kleman is concerned about absenteeism among her workers. The information below reports the
Number of days absent for a sample of 10 workers during the last two-week pay period.  
The sample mean is calculated to be 1.8 and sample standard deviation is 1.1353. Develop a 95%
Conﬁdence interval for the population mean. Assume that the population distribution is normal. Is it
Reasonable to conclude that the typical worker misses 1 day during a pay period?

Accepted Answer

A)  [0.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 1 per employee. 
B)  [0.99, 2.61] It is reasonable to conclude that the mean number of days of work missed is 1 per employee. 
C)  [-0.99, -2.61] It is unreasonable to conclude that the mean number of days of work missed is 1 per employee. 
D)  [-0.99, -2.61] It is reasonable to conclude that the mean number of days of work missed is 1 per employee. 
A)  [0.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 1 per employee. 
B)  [0.99, 2.61] It is reasonable to conclude that the mean number of days of work missed is 1 per employee. 
C)  [-0.99, -2.61] It is unreasonable to conclude that the mean number of days of work missed is 1 per employee. 
D)  [-0.99, -2.61] It is reasonable to conclude that the mean number of days of work missed is 1 per employee. B

Question 4

i. The 95 percent conﬁdence interval states that 95 percent of the sample means of a speciﬁed
Sample size selected from a population will lie within plus and minus 1.96 standard deviations of the
Hypothesized population mean.
ii. A distribution of sample means is normally distributed with a mean equal to the population mean
And a standard deviation equal to the standard error of the mean.
iii. A sample mean is the best point estimate of a population mean.

Accepted Answer

A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements. 
A)  (i), (ii), and (iii) are all correct statements. 
B)  (i) and (ii) are correct statements but not (iii). 
C)  (i) and (iii) are correct statements but not (ii). 
D)  (ii) and (iii) are correct statements but not (i). 
E)  (i), (ii), and (iii) are all false statements.

Question 5

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard
Error of $20. What is the 99% conﬁdence interval to estimate the true cost of the DVD?

Accepted Answer

A)  $323.40 to $426.60 
B)  $328.40 to $421.60 
C)  $335.00 to $415.00 
D)  $335.80 to $414.20 
A)  $323.40 to $426.60 
B)  $328.40 to $421.60 
C)  $335.00 to $415.00 
D)  $335.80 to $414.20

Question 6

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard
Error of $20. If 95% and 98% conﬁdence intervals were developed to estimate the true cost of the
DVD, what differences would they have?

Accepted Answer

A)  Standard errors 
B)  Interval widths 
C)  Z-values 
D)  Both standard errors and interval widths 
E)  Both interval widths and Z-values 
A)  Standard errors 
B)  Interval widths 
C)  Z-values 
D)  Both standard errors and interval widths 
E)  Both interval widths and Z-values

Question 7

College X is concerned about their employees making use of their email for non-business purposes.
You have been approached to assist in this matter. College X decides on a 90% conﬁdence level
And state that the estimation proportion must be within 2 percent of the population proportion. A
Pilot survey reveals that 8 out of 50 emails sampled were not for business purposes. How many
Emails should be surveyed to meet your requirements?

Accepted Answer

A)  910 
B)  707 
C)  498 
D)  1,221 
E)  1,000 
A)  910 
B)  707 
C)  498 
D)  1,221 
E)  1,000

Question 8

Which of the following would be used as a point estimate for the population mean (µ)?

Accepted Answer

A)  σ 
B)  x/n 
C)    
D)  s 
E)  p 
A)  σ 
B)  x/n 
C)    
D)  s 
E)  p

Question 9

A consumer group would like to estimate the mean monthly electricity charge for a single family
House in July (within $5) using a 99 percent level of conﬁdence. Based on similar studies the
Standard deviation is estimated to be $20.00. How large a sample is required?

Accepted Answer

A)  105 
B)  106 
C)  107 
D)  108 
E)  109 
A)  105 
B)  106 
C)  107 
D)  108 
E)  109

Question 10

The z-value associated with an 80% level of conﬁdence is:

Accepted Answer

A)  1.96 
B)  1.645 
C)  2.33 
D)  2.575 
E)  1.28 
A)  1.96 
B)  1.645 
C)  2.33 
D)  2.575 
E)  1.28

Question 11

Dottie Kleman is the "Cookie Lady." She bakes and sells cookies at 50 different locations. Ms.
Kleman is concerned about absenteeism among her workers. The information below reports the
Number of days absent for a sample of 10 workers during the last two-week pay period.  
The sample mean is calculated to be 1.8 and sample standard deviation is 1.1353. Develop a 95%
Conﬁdence interval for the population mean. Assume that the population distribution is normal. Is it
Reasonable to conclude that the typical worker misses 2 days during a pay period?

Accepted Answer

A)  [0.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 2 per employee. 
B)  [0.99, 2.61] It is reasonable to conclude that the mean number of days of work missed is 2 per employee. 
C)  [1.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 2 per employee. 
D)  [0.99, 1.61] It is reasonable to conclude that the mean number of days of work missed is 2 per employee. 
A)  [0.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 2 per employee. 
B)  [0.99, 2.61] It is reasonable to conclude that the mean number of days of work missed is 2 per employee. 
C)  [1.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 2 per employee. 
D)  [0.99, 1.61] It is reasonable to conclude that the mean number of days of work missed is 2 per employee.

Question 12

A student wanted to quickly construct a 95% conﬁdence interval for the average age of students in
Her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a
Standard deviation of 1.5 years. What is the 95% conﬁdence interval for the population mean?

Accepted Answer

A)  [0.97, 3.27] 
B)  [15.64, 22.56] 
C)  [17.97, 20.23] 
D)  [17.95, 20.25] 
E)  [17.42, 20.78] 
A)  [0.97, 3.27] 
B)  [15.64, 22.56] 
C)  [17.97, 20.23] 
D)  [17.95, 20.25] 
E)  [17.42, 20.78]

Question 13

A ﬁnancial analyst wanted to determine the mean annual return on mutual funds. A random sample
Of 60 returns shows a mean of 12%. If the population standard deviation is assumed to be 4%,
Estimate with 95% conﬁdence the mean annual return on all mutual funds.

Accepted Answer

A)  11.325% to 13.012% 
B)  10.988% to 14.025% 
C)  11.325% to 14.025% 
D)  10.988% to 13.012% 
E)  8% to 16% 
A)  11.325% to 13.012% 
B)  10.988% to 14.025% 
C)  11.325% to 14.025% 
D)  10.988% to 13.012% 
E)  8% to 16%

Question 14

A 95% conﬁdence interval infers that the population mean is

Accepted Answer

A)  between 0 and 100%. 
B)  within ± 1.96 standard deviations of the sample mean. 
C)  within ± 1.96 standard errors of the sample mean. 
D)  within ± 1.645 standard deviations of the sample mean. 
E)  too large. 
A)  between 0 and 100%. 
B)  within ± 1.96 standard deviations of the sample mean. 
C)  within ± 1.96 standard errors of the sample mean. 
D)  within ± 1.645 standard deviations of the sample mean. 
E)  too large.

Question 15

A pharmaceutical company wanted to estimate the population mean of monthly sales for their 250
Sales people. Forty sales people were randomly selected. Their mean monthly sales were $10,000
With a standard deviation of $1,000. Construct a 95% conﬁdence interval for the population mean.

Accepted Answer

A)  [9,690.1, 10,309.9] 
B)  [9706.3, 10293.7] 
C)  [8,040, 11,960] 
D)  [8,000, 12,000] 
E)  [9,000, 11,000] 
A)  [9,690.1, 10,309.9] 
B)  [9706.3, 10293.7] 
C)  [8,040, 11,960] 
D)  [8,000, 12,000] 
E)  [9,000, 11,000]

Question 16

Dr. Patton is a professor of English. Recently she counted the number of misspelled works in a
Group of student essays. She noted the distribution of misspelled words per essay followed the
Normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 50
Students, the mean number of misspelled words per essay was 6.05. Construct a 95% conﬁdence
Interval for the mean number of misspelled words in the population of student essays.

Accepted Answer

A)  5.374 to 6.726 
B)  5.161 to 6.939 
C)  5.102 to 6.998 
D)  5.482 to 6.618 
E)  5.445 to 6.3655 
A)  5.374 to 6.726 
B)  5.161 to 6.939 
C)  5.102 to 6.998 
D)  5.482 to 6.618 
E)  5.445 to 6.3655

Question 17

A sample of union members was selected and their opinions regarding the proposed management
Union contract were recorded with 1,600 out of the 2,000 members in favour of it. Using a 95%
Conﬁdence level, the interval estimate for the population proportion was computed to be 0.78 and
0)82.

Accepted Answer

A)  This indicates that about 68 out of 100 similarly constructed intervals would include the population proportion. 
B)  This indicates that about 95 out of 100 similarly constructed intervals would include the population proportion. 
C)  This indicates that about 99 out of 100 similarly constructed intervals would include the population proportion. 
D)  This indicates that about 78 out of 100 similarly constructed intervals would include the population proportion. 
A)  This indicates that about 68 out of 100 similarly constructed intervals would include the population proportion. 
B)  This indicates that about 95 out of 100 similarly constructed intervals would include the population proportion. 
C)  This indicates that about 99 out of 100 similarly constructed intervals would include the population proportion. 
D)  This indicates that about 78 out of 100 similarly constructed intervals would include the population proportion.

Question 18

Dottie Kleman is the "Cookie Lady." She bakes and sells cookies at 50 different locations. Ms.
Kleman is concerned about absenteeism among her workers. The information below reports the
Number of days absent for a sample of 10 workers during the last two-week pay period.  
The sample mean is calculated to be 1.8 and sample standard deviation is 1.1353. Develop a 95%
Conﬁdence interval for the population mean. Assume that the population distribution is normal. Is it
Reasonable to conclude that the typical worker does not miss any days during a pay period?

Accepted Answer

A)  [0.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 0 per employee. 
B)  [0.99, 2.61] It is reasonable to conclude that the mean number of days of work missed is 0 per employee. 
C)  [-0.99, -2.61] It is unreasonable to conclude that the mean number of days of work missed is 0 per employee. 
D)  [-0.99, -2.61] It is reasonable to conclude that the mean number of days of work missed is 0 per employee. 
A)  [0.99, 2.61] It is unreasonable to conclude that the mean number of days of work missed is 0 per employee. 
B)  [0.99, 2.61] It is reasonable to conclude that the mean number of days of work missed is 0 per employee. 
C)  [-0.99, -2.61] It is unreasonable to conclude that the mean number of days of work missed is 0 per employee. 
D)  [-0.99, -2.61] It is reasonable to conclude that the mean number of days of work missed is 0 per employee.

Question 19

A manager of a local store wants to estimate the mean amount spent per shopping visit by
Customers. Summary statistics from a sample taken reveal the following:  
Determine a 95% conﬁdence interval for the mean amount spent.

Accepted Answer

A)  [45.21, 53.56] 
B)  [40.36, 58.35] 
C)  [30.54, 68.14] 
D)  [45.13, 53.56] 
E)  [29.34, 69.34] 
A)  [45.21, 53.56] 
B)  [40.36, 58.35] 
C)  [30.54, 68.14] 
D)  [45.13, 53.56] 
E)  [29.34, 69.34]

Question 20

The following summarizes the amount of snowfall in Ontario over the past number of years.   You wonder whether the population mean could have been 1,600 or 2,100.

Accepted Answer

A)  Since 1600 is within the 95% conﬁdence interval, the population mean is likely to be 1,600. 
B)  Since 1600 is not within the 95% conﬁdence interval, the population mean is not likely to be 1,600. 
C)  Since 2100 is within the 95% conﬁdence interval, the population mean is likely to be 2,100. 
D)  Since neither 1,600 nor 2,100 is within the 95% conﬁdence interval, the population mean is not likely to be 1,600 or 2000. 
E)  Since 1900 and 2000 are both within the 95% conﬁdence interval, the population mean is likely to be either value. 
A)  Since 1600 is within the 95% conﬁdence interval, the population mean is likely to be 1,600. 
B)  Since 1600 is not within the 95% conﬁdence interval, the population mean is not likely to be 1,600. 
C)  Since 2100 is within the 95% conﬁdence interval, the population mean is likely to be 2,100. 
D)  Since neither 1,600 nor 2,100 is within the 95% conﬁdence interval, the population mean is not likely to be 1,600 or 2000. 
E)  Since 1900 and 2000 are both within the 95% conﬁdence interval, the population mean is likely to be either value.

The distribution of Student's t is

A bank wishes to estimate the mean balances owed by customers holding MasterCard. The Population standard deviation is estimated to be $300. If a 98 percent conﬁdence interval is used And an interval of $75 is desired, how many cardholders should be sampled?

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard Error of $20. What is the 99% conﬁdence interval to estimate the true cost of the DVD?

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard Error of $20. If 95% and 98% conﬁdence intervals were developed to estimate the true cost of the DVD, what differences would they have?

Which of the following would be used as a point estimate for the population mean (µ)?

A consumer group would like to estimate the mean monthly electricity charge for a single family House in July (within $5) using a 99 percent level of conﬁdence. Based on similar studies the Standard deviation is estimated to be $20.00. How large a sample is required?

The z-value associated with an 80% level of conﬁdence is:

A student wanted to quickly construct a 95% conﬁdence interval for the average age of students in Her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a Standard deviation of 1.5 years. What is the 95% conﬁdence interval for the population mean?

A ﬁnancial analyst wanted to determine the mean annual return on mutual funds. A random sample Of 60 returns shows a mean of 12%. If the population standard deviation is assumed to be 4%, Estimate with 95% conﬁdence the mean annual return on all mutual funds.

A 95% conﬁdence interval infers that the population mean is

A pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 Sales people. Forty sales people were randomly selected. Their mean monthly sales were $10,000 With a standard deviation of $1,000. Construct a 95% conﬁdence interval for the population mean.

A sample of union members was selected and their opinions regarding the proposed management Union contract were recorded with 1,600 out of the 2,000 members in favour of it. Using a 95% Conﬁdence level, the interval estimate for the population proportion was computed to be 0.78 and 0)82.

A manager of a local store wants to estimate the mean amount spent per shopping visit by Customers. Summary statistics from a sample taken reveal the following: Determine a 95% conﬁdence interval for the mean amount spent.

The following summarizes the amount of snowfall in Ontario over the past number of years. You wonder whether the population mean could have been 1,600 or 2,100.

What Is Statistics

Describing Data: Frequencydistributions and Graphic Presentation

Describing Data: Numerical Measures

Describing Data: Displaying and Exploring Pata

A Sulvey of Probability Concepts

Discrete Probability Distributions

Continuous Probabilitydistributions

Estimation and Confidence Intervals

One-Sample Tests of Hypothesis

Two Sample Tests of Hypothesis

Analysis of Variance

Linear Regression and Correlation

Multiple Regression and Correlation Analysis

Chi-Square Applications

Decision Making and Payoff Tables in Investment Scenarios

Filters

Exam 8: Sampling Methods and the Central Umit Theorem

The distribution of Student's t is

A bank wishes to estimate the mean balances owed by customers holding MasterCard. The Population standard deviation is estimated to be $300. If a 98 percent conﬁdence interval is used And an interval of $75 is desired, how many cardholders should be sampled?

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard Error of $20. What is the 99% conﬁdence interval to estimate the true cost of the DVD?

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard Error of $20. If 95% and 98% conﬁdence intervals were developed to estimate the true cost of the DVD, what differences would they have?

Which of the following would be used as a point estimate for the population mean (µ)?

A consumer group would like to estimate the mean monthly electricity charge for a single family House in July (within $5) using a 99 percent level of conﬁdence. Based on similar studies the Standard deviation is estimated to be $20.00. How large a sample is required?

The z-value associated with an 80% level of conﬁdence is:

A student wanted to quickly construct a 95% conﬁdence interval for the average age of students in Her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a Standard deviation of 1.5 years. What is the 95% conﬁdence interval for the population mean?

A ﬁnancial analyst wanted to determine the mean annual return on mutual funds. A random sample Of 60 returns shows a mean of 12%. If the population standard deviation is assumed to be 4%, Estimate with 95% conﬁdence the mean annual return on all mutual funds.

A 95% conﬁdence interval infers that the population mean is

A pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 Sales people. Forty sales people were randomly selected. Their mean monthly sales were $10,000 With a standard deviation of $1,000. Construct a 95% conﬁdence interval for the population mean.

A sample of union members was selected and their opinions regarding the proposed management Union contract were recorded with 1,600 out of the 2,000 members in favour of it. Using a 95% Conﬁdence level, the interval estimate for the population proportion was computed to be 0.78 and 0)82.

A manager of a local store wants to estimate the mean amount spent per shopping visit by Customers. Summary statistics from a sample taken reveal the following: Determine a 95% conﬁdence interval for the mean amount spent.

The following summarizes the amount of snowfall in Ontario over the past number of years. You wonder whether the population mean could have been 1,600 or 2,100.

What Is Statistics

Describing Data: Frequencydistributions and Graphic Presentation

Describing Data: Numerical Measures

Describing Data: Displaying and Exploring Pata

A Sulvey of Probability Concepts

Discrete Probability Distributions

Continuous Probabilitydistributions

Estimation and Confidence Intervals

One-Sample Tests of Hypothesis

Two Sample Tests of Hypothesis

Analysis of Variance

Linear Regression and Correlation

Multiple Regression and Correlation Analysis

Chi-Square Applications

Decision Making and Payoff Tables in Investment Scenarios

Filters