Quiz about 1-21 Review - Unit 6

Question 1

The mean income of all teachers in the state of Florida for the 2013-2014 school year was $47, 800. We draw a random sample of 700 teachers and find their mean income is $43, 200.

Identify the population.

Accepted Answer

All teachers in the state of Florida

Answer

All teachers.

Answer

700 teachers in Florida

Answer

700 teachers.

Question 2

The mean income of all teachers in the state of Florida for the 2013-2014 school year was $47, 800. We draw a random sample of 700 teachers and find their mean income is $43, 200.

Identify the sample.

Accepted Answer

700 teachers in Florida

Answer

All teachers in the state of Florida

Answer

All teachers.

Answer

700 teachers.

Question 3

The mean income of all teachers in the state of Florida for the 2013-2014 school year was $47, 800. We draw a random sample of 700 teachers and find their mean income is $43, 200.

Identify the statistic.

Accepted Answer

$47,800

Answer

700

Answer

$43,200

Answer

none of these

Question 4

The mean income of all teachers in the state of Florida for the 2013-2014 school year was $47, 800. We draw a random sample of 700 teachers and find their mean income is $43, 200.

Identify the parameter.

Accepted Answer

$43,200

Answer

700

Answer

$47,800

Answer

none of these

Question 5

Using the Z-table, find the Z critical value that is associated with 98% confidence level?

Accepted Answer

2.326

Answer

2.576

Answer

1.960

Answer

1.282

Question 6

Using the Z-table, find the Z critical value that is associated with 50% confidence level?

Accepted Answer

0.674

Answer

2.576

Answer

1.282

Answer

2.326

Question 7

A college is concerned about the impact of students working while they are enrolled in classes and the effect it has on their performance. In order to do any research, they must first estimate the amount of time that students spend working. They know from previous studies that the population standard deviation of this variable is about 2.75 hours. A survey of 50 students provides a sample mean of 5.75 hours worked.

What is the 90% confidence interval for the mean working time of all college students?

Formula:
$E\ =\ Z_c\ \left(\frac{\sigma}{\sqrt{n}}\right)$
$\mu\ -\ E\ ,\ \ \mu\ +\ E$

Identify the population and sample.

Accepted Answer

Population: All students at a college.

Sample: 50 students randomly selected.

Answer

Population: All college students.

Sample: All students randomly selected.

Answer

Population: All college students in the USA.

Sample: The 50 students who are selected.

Question 8

A college is concerned about the impact of students working while they are enrolled in classes and the effect it has on their performance. In order to do any research, they must first estimate the amount of time that students spend working. They know from previous studies that the population standard deviation of this variable is about 2.75 hours. A survey of 50 students provides a sample mean of 5.75 hours worked.

What is the 90% confidence interval for the mean working time of all college students?

Formula:
$E\ =\ Z_c\ \left(\frac{\sigma}{\sqrt{n}}\right)$
$\mu\ -\ E\ ,\ \ \mu\ +\ E$

What critical value is the most appropriate to use and why?

Accepted Answer

z-value because n greater than 30

Answer

t-value because n is greater than 30

Answer

z-value since n is less than 30

Answer

t-value since n is less than 30

Question 9

A college is concerned about the impact of students working while they are enrolled in classes and the effect it has on their performance. In order to do any research, they must first estimate the amount of time that students spend working. They know from previous studies that the population standard deviation of this variable is about 2.75 hours. A survey of 50 students provides a sample mean of 5.75 hours worked.

What is the 90% confidence interval for the mean working time of all college students?

Formula:
$E\ =\ Z_c\ \left(\frac{\sigma}{\sqrt{n}}\right)$
$\mu\ -\ E\ ,\ \ \mu\ +\ E$

Identify the statistic.

Accepted Answer

5.75 hours

Answer

2.75 hours

Answer

50

Answer

none

Question 10

An observational study found that 123 out of 500 shoppers who entered a local grocery store purchased eggs during their visit. Suppose were to use this information to create a confidence interval for this local grocery store shoppers who buy eggs when they come into the store.

What is the appropriate critical value for 98% confidence interval would be?

Accepted Answer

2.326

Answer

1.282

Answer

2.576

Answer

1.960

Question 11

Using the t-table, find the t critical value that is associated with 99% confidence level with a sample size of 16.

Accepted Answer

2.947

Answer

2.326

Answer

2.576

Answer

2.921

Question 12

Using the t-table, find the t critical value that is associated with 80% confidence level with a sample size of 27.

Accepted Answer

1.315

Answer

1.318

Answer

1.376

Answer

1.314

Question 13

Suppose that you find a 98% confidence interval for the life span, in months, of a certain brand of light bulb is ( 23.5, 25.7). Which of these is a valid interpretation of this confidence interval?

Accepted Answer

We are 98% confident that the true average life span of this light bulbs is between 23.5 and 25.7 months.

Answer

There is a 98% probability that the true average life span of one of these light bulbs is between 23.5 and 25.7 months.

Answer

There is a 98% probability that a randomly selected light bulb lasted between 23.5 and 25.7 months.

Answer

We are 98% confident that the average life span of one of these light bulbs in our sample between is 23.5 and 25.7 months.

Question 14

The higher the confidence level, the ________ the distribution. If you increase the sample size, the __________ it is and if the margin of error is lower, it ____________.

Accepted Answer

wider, narrower, remains the same

Answer

narrower, wider, remains the same

Answer

wider, remains the same, narrower

Answer

narrower, remains the same, wider

1-21 Review - Unit 6

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