Question 1

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

The sample size is 64, so the degrees of freedom are 63. The confidence level is 95%, so the t-value is 1.998.

Question 2

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 margin of error at 95% confidence is calculated using the formula: Margin of Error = Z * (σ/√n), where Z is the Z-score for 95% confidence (approximately 1.96), σ is the standard deviation, and n is the sample size. Here, it is 1.96 * (7/√75) ≈ 1.611.

Question 3

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 4

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:

standard error of the mean = population standard deviation / square root of sample size

Plugging in the given values, we get:

standard error of the mean = 450 / square root of 81

Simplifying, we get:

standard error of the mean = 450 / 9

standard error of the mean = 50

Therefore, the correct choice is D.

Question 5

In a sample of 400 registered voters, 204 were Democrats.
a.Provide a 95% confidence for the proportion of registered Democrats voters in the population.
b.There are 169 million registered voters in the US. Determine an interval for the number (how many) of registered Democrats in the population.

Accepted Answer

The answer of In a sample of 400 registered voters,...

Question 6

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

We can use the formula for confidence interval: 
CI = X̄ ± z* (σ/√n) 
where 
X̄ = sample mean = $40.75 
σ = population standard deviation = $7.00 
n = sample size = 75 
z* = z-score for 95% confidence level = 1.96

Plugging in the values, we get: 
CI = $40.75 ± 1.96 * ($7.00 / √75) 
CI = $40.75 ± $1.81

Therefore, the 95% confidence interval for the average hourly wage of all information system managers is $39.14 to $42.36. Thus, the correct answer is C.

Question 7

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

Using a 95% confidence level and a sample size of 64, the t-value is 1.99. Therefore, the margin of error is 1.99 * (240/sqrt(64)) = 59.85. The confidence interval is (1400 - 59.85, 1400 + 59.85), which simplifies to (1340.15, 1459.85). Rounding to two decimal places gives us a 95% confidence interval of (1340.05, 1459.95). Therefore, the best choice is A.

Question 8

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 degrees of freedom for a t-distribution is calculated as the sample size minus one. Therefore, for a sample size of 64, the degrees of freedom is 64 - 1 = 63.

Question 9

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. Plugging in the values, we get standard error of the mean = 7 / sqrt(75) = 0.8083. Therefore, the answer is C.

Question 10

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 11

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 margin of error at 95% confidence is calculated as (1.96*standard deviation)/square root of sample size. 
Substituting the given values, we get (1.96*240)/square root of 64 = 59.95. Therefore, the answer is choice D.

Question 12

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 standard error of the mean is calculated as the population standard deviation divided by the square root of the sample size. Here, it is 1 minute divided by the square root of 100, which equals 0.100.

Question 13

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

A 95% confidence interval estimate can be found using the formula:

(sample mean) +/- (critical value) * (standard error)

The critical value for a 95% confidence interval with 80 degrees of freedom (81-1) is 1.990. The standard error can be found using the formula:

standard error = (population standard deviation) / sqrt(sample size)
standard error = 450 / sqrt(81)
standard error = 50

Plugging in the values, we get:

1,858 +/- 1.990 * 50
1,858 +/- 99.5

So the 95% confidence interval is 1,858 - 99.5 to 1,858 + 99.5, or 1,760 to 1,956. Therefore, the best choice is A.

Question 14

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

For a 95% confidence interval with a sample size of 75, we use the t-distribution. With 74 degrees of freedom (n-1), the t-value is approximately 1.993.

Question 15

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 margin of error is calculated using the formula: Margin of Error = Z * (σ/√n), where Z is the Z-score for the confidence level (1.96 for 95%), σ is the population standard deviation (450), and n is the sample size (81). Thus, Margin of Error = 1.96 * (450/√81) = 1.96 * 50 = 98.

Question 16

A sample of 196 bags of sugar produced by Domain sugar producers showed an average of 2 pounds and 2 ounces with a standard deviation of 7 ounces.
a.At 95% confidence, compute the margin of error and explain what it shows.
b.Determine a 95% confidence interval for the weight of the population of bags of sugar produced by the company. Give the answer in ounces and at least two significant digits.
c.The bags of sugar produced are supposed to contain 2 pounds of sugar, but the company allows 2% higher or lower than the 2 pounds. Have they reached their desired goal? Explain why or why not by giving numerical facts.

Accepted Answer

The answer of A sample of 196 bags of sugar...

Question 17

A sample of 144 commuters in Boston indicated that they commute an average of one hour and thirty minutes per day. The standard deviation of the sample was 36 minutes. Compute the 95% confidence interval for the average commute time of the population (in minutes).

Accepted Answer

The answer of A sample of 144 commuters in Boston...

Question 18

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 can use the formula for the confidence interval of a population mean:
CI = X̄ ± Z*(σ/√n)
Where X̄ is the sample mean, σ is the population standard deviation, n is the sample size, and Z is the z-score for the desired confidence level. 
For a 95% confidence interval, the z-score is 1.96.
Substituting the values given in the problem, we have:
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 2.804 to 3.196 minutes. Choice D is the correct answer.

Question 19

A local university administers a comprehensive examination to the candidates for B.S. degrees in Business Administration. Five examinations are selected at random and scored. The scores are shown below.  
a.Compute the mean and the standard deviation of the sample.
b.Compute the margin of error at 95% confidence.
c.Develop a 95% confidence interval estimate for the mean of the population. Assume the population is normally distributed.

Accepted Answer

The answer of A local university administers a comprehensive examination...

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. 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

The margin of error is given by E = z* (sigma/sqrt(n)), where z* is the critical value of the standard normal distribution corresponding to the desired level of confidence (0.95 in this case), sigma is the population standard deviation, and n is the sample size. Substituting the given values, we have E = 1.96*(1/sqrt(100)) = 0.196. Therefore, the best choice is C.

Quiz 8: Interval Estimation

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 value of the margin of error at 95% confidence 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

In a sample of 400 registered voters, 204 were Democrats. a.Provide a 95% confidence for the proportion of registered Democrats voters in the population. b.There are 169 million registered voters in the US. Determine an interval for the number (how many) of registered Democrats in 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 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 standard error of the mean 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 8-6 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. At 95% confidence, the size of the margin of error is

A sample of 144 commuters in Boston indicated that they commute an average of one hour and thirty minutes per day. The standard deviation of the sample was 36 minutes. Compute the 95% confidence interval for the average commute time of the population (in minutes).