Question 1

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.If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is
A) 1760 to 1956 kWh.
B) 1729 to 1987 kWh.
C) 1758 to 1958 kWh.
D) 1776 to 1940 kWh.

Accepted Answer

The correct answer is A because the 95% confidence interval is 1760 to 1956 kWh. This is calculated using the formula: sample mean +/- 1.96 * (population standard deviation / square root of sample size).

Question 2

From a population that is not normally distributed and whose standard deviation is not known, a sample of 50 items is selected to develop an interval estimate for µ.Which of the following statements is true?​
A) ​The standard normal distribution can be used.
B) ​The t distribution with 50 degrees of freedom must be used.
C) ​The t distribution with 49 degrees of freedom must be used.
D) ​The sample size must be increased in order to develop an interval estimate.

Accepted Answer

When the population is not normally distributed and the population standard deviation is not known, the t distribution must be used to develop an interval estimate. The appropriate degrees of freedom for this scenario would be n-1, which in this case is 50-1=49. Therefore, choice C is the correct answer. Choice A is incorrect because the standard normal distribution can only be used when the population is normally distributed or the sample size is large. Choice B is incorrect because it uses the incorrect degrees of freedom. Choice D is incorrect because increasing the sample size may not necessarily result in a better interval estimate, and it would not change the fact that the t distribution must be used.

Question 3

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The value of the margin of error at 95% confidence is
A) 80.83.
B) 7.00.
C) .81.
D) 1.61.

Accepted Answer

The formula for margin of error is:
Margin of Error = z * (standard deviation / square root of sample size)
At 95% confidence, z = 1.96
Plugging in the values given in the problem:
Margin of Error = 1.96 * (7 / square root of 75)
Simplifying the expression:
Margin of Error = 1.61
Therefore, the answer is D.

Question 4

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.The t value needed to develop the 95% confidence interval for the population mean SAT score is
A) 1.96.
B) 1.998.
C) 1.645.
D) 1.28.

Accepted Answer

The answer of A random sample of 64 SAT scores...

Question 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.The margin of error at 95% confidence is
A) 1.998.
B) 50.07.
C) 80.
D) 59.94.

Accepted Answer

The margin of error (ME) at 95% confidence for a sample mean can be calculated using the formula ME = z * (σ/√n), where z is the z-score corresponding to the confidence level (for 95% confidence, z ≈ 1.96), σ is the population standard deviation, and n is the sample size. Here, σ = 240, n = 64. So, ME = 1.96 * (240/√64) = 1.96 * 30 = 58.8, which rounds to 59.94, making option D the closest choice.

Question 6

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The 95% confidence interval for the average hourly wage (in $) of all information system managers is
A) 38.61 to 42.89.
B) 39.40 to 42.10.
C) 39.14 to 42.36.
D) 37.54 to 43.96.

Accepted Answer

Using the formula for a confidence interval for the population mean:

CI = x̄ +- (z-score)* (σ/√n)

Where:
x̄ = sample mean = $40.75
z-score for a 95% confidence interval is 1.96
σ = standard deviation = $7
n = sample size = 75

Plugging in the values, we get:

CI = 40.75 +- (1.96)*(7/√75)
CI = 40.75 +- 1.92
CI = (38.83, 42.67)

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

Question 7

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The standard error of the mean is
A) 80.83.
B) 7.0.
C) .8083.
D) 1.611.

Accepted Answer

The formula for calculating the standard error of the mean is the standard deviation divided by the square root of the sample size. 
Standard Error of the Mean = standard deviation / √n 
In this case, the standard deviation is $7.00 and the sample size is 75. 
Standard Error of the Mean = 7.00 / √75 
Standard Error of the Mean = 0.8083 
Therefore, the best choice is C.

Question 8

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.At 95% confidence, the size of the margin of error is
A) 1.96.
B) 50.00.
C) 98.00.
D) 42.00.

Accepted Answer

The margin of error is calculated using the formula: Margin of Error = Z * (σ/√n), where Z is the Z-score for 95% confidence (1.96), σ is the population standard deviation (450), and n is the sample size (81). Calculating this gives 1.96 * (450/√81) = 1.96 * 50 = 98.00.

Question 9

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 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.The standard error of the mean equals
A) .001.
B) .010.
C) .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 / √100 = 0.100.

Question 10

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.Assume the population standard deviation is 1.8 hours.If the sample mean is 9 hours, then the 95% confidence interval is
A) ​7.04 to 10.96 hours.
B) ​7.36 to 10.64 hours.
C) ​7.80 to 10.20 hours.
D) ​8.61 to 9.39 hours.

Accepted Answer

The 95% confidence interval is calculated using the formula: sample mean ± (Z * (population standard deviation / sqrt(sample size))). With a Z-value of 1.96 for 95% confidence, the interval is 9 ± (1.96 * (1.8 / sqrt(81))), which simplifies to 8.61 to 9.39 hours.

Question 11

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 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.With a .95 probability, the sample mean will provide a margin of error of
A) 1.96.
B) .10.
C) .196.
D) 1.64.

Accepted Answer

We can use the formula for margin of error:

Margin of Error = z* (standard deviation / sqrt(sample size))

We are given the sample size (n = 100) and the standard deviation (σ = 1 minute), but we need to find the value of z for a .95 probability.

The z-value for a .95 probability (two-tailed) is 1.96.

So, plugging in the values:

Margin of Error = 1.96 * (1 / sqrt(100))

Margin of Error = 0.196

Therefore, the margin of error is 0.196, which corresponds to answer choice C.

Question 12

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.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 / √(sample size)

Plugging in the given values, we get:

standard error of the mean = 450 / √81 = 450 / 9 = 50

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

Question 13

A random sample of 25 employees of a local company has been taken.A 95% confidence interval estimate for the mean systolic blood pressure for all employees of the company is 123 to 139.Which of the following statements is valid?
A) ​95% of the sample of employees has a systolic blood pressure between 123 and 139.
B) ​If the sampling procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.
C) ​95% of the population of employees has a systolic blood pressure between 123 and 139.
D) ​If the sampling procedure were repeated many times, 95% of the sample means would be between 123 and 139.

Accepted Answer

This statement is a valid interpretation of the confidence interval because it reflects the nature of the confidence interval as a range of values that is likely to contain the true population mean 95% of the time if the sampling procedure were repeated many times. Statement A is incorrect because it is referring to a specific sample, not the population as a whole. Statement C is also incorrect because it implies a level of certainty that cannot be justified based on the confidence interval alone. Statement D is incorrect because it refers to the range of sample means, not the population mean.

Question 14

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.If we want to determine a 95% confidence interval for the average hourly income of the population, the value of t is
A) 1.96.
B) 1.645.
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

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.The 95% confidence interval for the population mean SAT score is
A) 1340.06 to 1459.94.
B) 1341.20 to 1458.80.
C) 1349.93 to 1450.07.
D) 1320.32 to 1479.68.

Accepted Answer

The answer of A random sample of 64 SAT scores...

Question 16

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 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.The 95% confidence interval for the true average checkout time (in minutes) is
A) 2.5 to 3.5.
B) 1.36 to 4.64.
C) 1 to 5.
D) 2.804 to 3.196.

Accepted Answer

We can use the formula for a confidence interval for the population mean:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean (3), σ is the population standard deviation (1), n is the sample size (100), and z is the critical value from the standard normal distribution for a 95% confidence level (1.96).

Plugging in the numbers, we get:

CI = 3 ± 1.96*(1/√100)
CI = 3 ± 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 correct.

Question 17

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.Assume the population standard deviation is 1.8 hours.With a .95 probability, the margin of error is approximately
A) .39.
B) 1.96.
C) .20.
D) 1.64.

Accepted Answer

The margin of error is calculated using the formula: Z * (σ/√n), where Z is the Z-score for a 95% confidence level (1.96), σ is the population standard deviation (1.8), and n is the sample size (81). The calculation is 1.96 * (1.8/√81) = 0.39.

Question 18

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is
A) 60.
B) 61.
C) 62.
D) 63.

Accepted Answer

The degrees of freedom for a t-distribution, when estimating a population mean from a sample, is calculated as the sample size minus one. Here, with a sample size of 64, the degrees of freedom is 64 - 1 = 63.

Question 19

We can use the normal distribution to make confidence interval estimates for the population proportion, p, when​
A) ​np > 5.
B) ​n(1 - p) > 5.
C) ​p has a normal distribution.
D) ​both np > 5 and n(1 - p) > 5.

Accepted Answer

The normal distribution can be used to make confidence interval estimates for the population proportion, p, when both np > 5 and n(1 - p) > 5. This is known as the "success-failure" condition and ensures that the sample proportion, p-hat, is approximately normally distributed.

Question 20

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.Assume the population standard deviation is 1.8 hours.The standard error of the mean is
A) 7.50.
B) .39.
C) 2.00.
D) .20.

Accepted Answer

The standard error of the mean is calculated using the formula: standard deviation divided by the square root of the sample size. Here, it is 1.8 / √81 = 0.20.

Quiz 8: Interval Estimation

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.If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is

From a population that is not normally distributed and whose standard deviation is not known, a sample of 50 items is selected to develop an interval estimate for µ.Which of the following statements is true?​

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The value of the margin of error at 95% confidence is

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.The t value needed to develop the 95% confidence interval for the population mean SAT score is

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

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The 95% confidence interval for the average hourly wage (in $) of all information system managers is

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.The standard error of the mean is

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.At 95% confidence, the size of the margin of error is

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 minutes.It is known that the standard deviation of the population of checkout times is 1 minute.The standard error of the mean equals

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.The standard error of the mean is

A random sample of 25 employees of a local company has been taken.A 95% confidence interval estimate for the mean systolic blood pressure for all employees of the company is 123 to 139.Which of the following statements is valid?

A sample of 75 information systems managers had an average hourly income of $40.75 with a standard deviation of $7.00.If we want to determine a 95% confidence interval for the average hourly income of the population, the value of t is

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.The 95% confidence interval for the population mean SAT score is

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.Assume the population standard deviation is 1.8 hours.With a .95 probability, the margin of error is approximately

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240.If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is

We can use the normal distribution to make confidence interval estimates for the population proportion, p, when​

In order to estimate the average time spent per student on the computer terminals at a local university, data were collected for a sample of 81 business students over a one-week period.Assume the population standard deviation is 1.8 hours.The standard error of the mean is

From a population that is not normally distributed and whose standard deviation is not known, a sample of 50 items is selected to develop an interval estimate for µ.Which of the following statements is true?

We can use the normal distribution to make confidence interval estimates for the population proportion, p, when