Question 1

Choo Choo Paper Company produces papers of various thickness. A random sample of 256 cuts had a mean thickness of 30.3 mils with a standard deviation of 4 mils. Develop a 95% confidence interval for the mean thickness of the population.

Accepted Answer

The answer of Choo Choo Paper Company produces papers of...

Question 2

Many people who bought X-Game gaming systems over the holidays have complained that the systems they purchased were defective. In a sample of 1200 units sold, 18 units were defective.
a. Determine a 95% confidence interval for the percentage of defective systems.
b. If 1.5 million X-Games were sold over the holidays, determine an interval for the number of defectives in sales.

Accepted Answer

The answer of Many people who bought X-Game gaming systems...

Question 3

Exhibit 8-6
A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of
$7.00.
-Refer to Exhibit 8-6. If we want to determine a 95% confidence interval for the average hourly income, the value of "t" statistics is
A) 1.96
B) 1.64
C) 1.28
D) 1.993

Accepted Answer

The answer of Exhibit 8-6
A sample of 75 information system...

Question 4

The average monthly electric bill of a random sample of 256 residents of a city is $90 with a standard deviation of
$24.
a. Construct a 90% confidence interval for the mean monthly electric bills of all residents.
b. Construct a 95% confidence interval for the mean monthly electric bills of all residents.

Accepted Answer

The answer of The average monthly electric bill of a...

Question 5

Exhibit 8-3
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the population of checkout times is one minute.
-Refer to Exhibit 8-3. The standard error of the mean equals
A) 0.001
B) 0.010
C) 0.100
D) 1.000

Accepted Answer

The answer of Exhibit 8-3
The manager of a grocery store...

Question 6

Exhibit 8-4
In order to estimate the average electric usage per month, a sample of 81 houses was selected, and the electric usage was determined. Assume a population standard deviation of 450-kilowatt hours.
-Refer to Exhibit 8-4. The standard error of the mean is
A) 450
B) 81
C) 500
D) 50

Accepted Answer

The formula for standard error of the mean is:
$SE=\frac{\sigma}{\sqrt{n}}$
where $\sigma$ is the population standard deviation and $n$ is the sample size.

Substituting the given values,
$SE=\frac{450}{\sqrt{81}}=50$

Therefore, the standard error of the mean is 50. The correct answer is D.

Question 7

A random sample of 81 credit sales in a department store showed an average sale of $68.00. From past data, it is known that the standard deviation of the population is $27.00.
a. Determine the standard error of the mean.
b. With a 0.95 probability, what can be said about the size of the margin of error?
c. What is the 95% confidence interval of the population mean?

Accepted Answer

The answer of A random sample of 81 credit sales...

Question 8

Exhibit 8-4
In order to estimate the average electric usage per month, a sample of 81 houses was selected, and the electric usage was determined. Assume a population standard deviation of 450-kilowatt hours.
-Refer to Exhibit 8-4. At 95% confidence, the size of the margin of error is
A) 1.96
B) 50
C) 98
D) 42

Accepted Answer

The answer of Exhibit 8-4
In order to estimate the average...

Question 9

A random sample of 87 airline pilots had an average yearly income of $99,400 with a standard deviation of $12,000.
a. If we want to determine a 95% confidence interval for the average yearly income, what is the value of t?
b. Develop a 95% confidence interval for the average yearly income of all pilots.

Accepted Answer

The answer of A random sample of 87 airline pilots...

Question 10

Exhibit 8-5
A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.
-Refer to Exhibit 8-5. The 95% confidence interval for the SAT scores is
A) 1340.05 to 1459.95
B) 1400 to 1459.95
C) 1340.05 to 1400
D) 1400 to 1600

Accepted Answer

A 95% confidence interval is calculated by using a t-distribution with 63 degrees of freedom (a sample size of 64 minus 1). The margin of error, calculated as the critical value (1.96) times the standard error (240/square root of 64), is 59.95. Therefore, the 95% confidence interval is 1400 plus or minus 59.95, which gives us a range of 1340.05 to 1459.95. Therefore, the correct choice is A.

Question 11

Exhibit 8-6
A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of
$7.00.
-Refer to Exhibit 8-6. The value of the margin of error at 95% confidence is
A) 80.83
B) 7
C) 0.8083
D) 1.611

Accepted Answer

The answer of Exhibit 8-6
A sample of 75 information system...

Question 12

You are given the following information obtained from a sample of 5 observations taken from a population that has a normal distribution.
94 72 93 54 77
Develop a 98% confidence interval estimate for the mean of the population.

Accepted Answer

The answer of You are given the following information obtained...

Question 13

Exhibit 8-6
A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of
$7.00.
-Refer to Exhibit 8-6. The 95% confidence interval for the average hourly wage of all information system managers is
A) 40.75 to 42.36
B) 39.14 to 40.75
C) 39.14 to 42.36
D) 30 to 50

Accepted Answer

To calculate the confidence interval for the population mean with a sample size of 75, we can use the formula:
CI = X̄ ± z* (σ/√n)
where X̄ is the sample mean, z* is the critical value for a 95% confidence level (which corresponds to 1.96), σ is the population standard deviation (which is unknown, so we use the sample standard deviation of $7.00), and n is the sample size.

Plugging in the values, we get:
CI = 40.75 ± 1.96 * (7/√75)
CI = 40.75 ± 1.96 * 0.806
CI = 40.75 ± 1.58

Therefore, the 95% confidence interval for the average hourly wage of all information system managers is 39.14 to 42.36. So, the best choice is C.

Question 14

Exhibit 8-3
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the population of checkout times is one minute.
-Refer to Exhibit 8-3. With a .95 probability, the sample mean will provide a margin of error of
A) 1.96
B) 0.10
C) 0.196
D) 1.64

Accepted Answer

We can use the formula: margin of error = z* (standard deviation / sqrt(sample size)). Since we want a 95% confidence interval, we use a z-score of 1.96. Plugging in the values, we get margin of error = 1.96 * (1 / sqrt(100)) = 0.196. Therefore, the best choice is C.

Question 15

Exhibit 8-5
A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.
-Refer to Exhibit 8-5. The margin of error at 95% confidence is
A) 1.998
B) 1400
C) 240
D) 59.95

Accepted Answer

The formula for margin of error is (z-value) x (standard deviation of sample mean). At 95% confidence, the z-value is 1.96. Therefore, the margin of error is 1.96 x (240/sqrt(64)) = 59.95.

Question 16

Exhibit 8-5
A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.
-Refer to Exhibit 8-5. The "t" value for this interval estimation is
A) 1.96
B) 1.998
C) 1.64
D) 1.28

Accepted Answer

Since the sample size is small (n=64) and the population standard deviation is unknown, we use a t-distribution. With a 95% confidence level, the degrees of freedom are n-1 = 63. Looking up the value for a two-tailed test in a t-table or using a calculator, we find that the t-value is 1.998.

Question 17

Exhibit 8-6
A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of
$7.00.
-Refer to Exhibit 8-6. The standard error of the mean is
A) 80.83
B) 7
C) 0.8083
D) 1.611

Accepted Answer

The formula for standard error of the mean is standard deviation / square root of sample size. Substituting the values given in the question, we get: 7 / sqrt(75) = 0.8083. Therefore, the correct answer is C.

Question 18

Exhibit 8-4
In order to estimate the average electric usage per month, a sample of 81 houses was selected, and the electric usage was determined. Assume a population standard deviation of 450-kilowatt hours.
-Refer to Exhibit 8-4. If the sample mean is 1,858 KWH, the 95% confidence interval estimate of the population mean is
A) 1,760 to 1,956 KWH
B) 1,858 to 1,956 KWH
C) 1,760 to 1,858 KWH
D) none of these alternatives is correct

Accepted Answer

To calculate the confidence interval estimate for the population mean, we use the formula:
CI = X̄ ± z(α/2) * (σ/√n)
Where:
X̄ = sample mean = 1,858 KWH
z(α/2) = z-score of the level of confidence, for a 95% confidence level, α = 0.05 and z(α/2) = 1.96
σ = population standard deviation = 450 KWH
n = sample size = 81
Plugging in the values, we get:
CI = 1,858 ± 1.96 * (450/√81)
CI = 1,858 ± 99.6
CI = (1,858 - 99.6, 1,858 + 99.6)
CI = (1,760.4, 1,956.6)
Therefore, the 95% confidence interval estimate of the population mean is 1,760 to 1,956 KWH (option A).

Question 19

Exhibit 8-5
A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.
-Refer to Exhibit 8-5. If we want to provide a 95% confidence interval for the SAT scores, the degrees of freedom for reading the critical values of "t" statistic is
A) 60
B) 61
C) 62
D) 63

Accepted Answer

The answer of Exhibit 8-5
A random sample of 64 SAT...

Question 20

Exhibit 8-3
The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the population of checkout times is one minute.
-Refer to Exhibit 8-3. The 95% confidence interval for the true average checkout time in minutes) is
A) 3:00 to 5:00
B) 1.36 to 4.64
C) 1.00 to 5.00
D) 2.804 to 3.196

Accepted Answer

We use the formula for a confidence interval of the population mean: 
CI = x̄ ± z* (σ/√n) 
where
x̄ = sample mean = 3.0 minutes
σ = population standard deviation = 1 minute
n = sample size = 100
z* = critical value for 95% confidence level = 1.96 (from the normal distribution table)
Plugging in the values, we get:
CI = 3.0 ± 1.96(1/√100)
CI = 3.0 ± 0.196
CI = (2.804, 3.196)
Therefore, the 95% confidence interval for the true average checkout time is from 2.804 to 3.196 minutes. Choice D is the best answer.

Quiz 8: Interval Estimation

Choo Choo Paper Company produces papers of various thickness. A random sample of 256 cuts had a mean thickness of 30.3 mils with a standard deviation of 4 mils. Develop a 95% confidence interval for the mean thickness of the population.

Exhibit 8-6 A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. -Refer to Exhibit 8-6. If we want to determine a 95% confidence interval for the average hourly income, the value of "t" statistics is

Exhibit 8-4 In order to estimate the average electric usage per month, a sample of 81 houses was selected, and the electric usage was determined. Assume a population standard deviation of 450-kilowatt hours. -Refer to Exhibit 8-4. The standard error of the mean is

Exhibit 8-4 In order to estimate the average electric usage per month, a sample of 81 houses was selected, and the electric usage was determined. Assume a population standard deviation of 450-kilowatt hours. -Refer to Exhibit 8-4. At 95% confidence, the size of the margin of error is

Exhibit 8-5 A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. -Refer to Exhibit 8-5. The 95% confidence interval for the SAT scores is

Exhibit 8-6 A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. -Refer to Exhibit 8-6. The value of the margin of error at 95% confidence is

You are given the following information obtained from a sample of 5 observations taken from a population that has a normal distribution. 94 72 93 54 77 Develop a 98% confidence interval estimate for the mean of the population.

Exhibit 8-6 A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. -Refer to Exhibit 8-6. The 95% confidence interval for the average hourly wage of all information system managers is

Exhibit 8-5 A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. -Refer to Exhibit 8-5. The margin of error at 95% confidence is

Exhibit 8-5 A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. -Refer to Exhibit 8-5. The "t" value for this interval estimation is

Exhibit 8-6 A sample of 75 information system managers had an average hourly income of $40.75 with a standard deviation of $7.00. -Refer to Exhibit 8-6. The standard error of the mean is