Question 1

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The 95% confidence interval for the average hourly wage (in $) of all information system managers is
A) 38.684 to 41.316.
B) 32 to 48.
C) 38.432 to 41.568.
D) 39 to 41.

Accepted Answer

Using the formula for 95% confidence interval: 
CI = X̄ ± Zα/2 * (σ/√n)
where:
X̄ = sample mean = $40
Zα/2 = 1.96 (for 95% confidence interval)
σ = population standard deviation = $8
n = sample size = 100

Substituting the given values: 
CI = 40 ± 1.96 * (8/√100)
CI = 40 ± 1.568
CI = 38.432 to 41.568

Therefore, the 95% confidence interval for the average hourly wage of all information system managers is $38.432 to $41.568. Therefore, the correct choice is C.

Question 2

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 64 business students over a one-week period.Assume the population standard deviation is 1.6 hours.The standard error of the mean is
A) 1.6.
B) .025.
C) 2.00.
D) .20.

Accepted Answer

The standard error of the mean is calculated using the formula $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 1.6$ hours and $n = 64$, so $\sigma_{\bar{x}} = \frac{1.6}{\sqrt{64}} = \frac{1.6}{8} = 0.2$ hours.

Question 3

The manager of a grocery store has taken a random sample of 144 customers.The average length of time it took these 144 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) 2.863 to 3.137.
B) 1.36 to 4.64.
C) 1 to 5.
D) 2.837 to 3.163.

Accepted Answer

We can use the formula for a confidence interval of a population mean, which is:

CI = X̄ ± Zα/2 * σ/√n

Where:
- X̄ is the sample mean
- Zα/2 is the critical value for the desired confidence level (in this case, 95%). Zα/2 can be found using a standard normal distribution table or calculator. It is 1.96.
- σ is the population standard deviation
- n is the sample size

Substituting the given values, we have:

CI = 3 ± 1.96 * 1/√144
CI = 3 ± 0.16
CI = (2.84, 3.16)

Therefore, the 95% confidence interval for the true average checkout time is 2.837 to 3.163, which is closest to choice 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 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 5

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 64 business students over a one-week period.Assume the population standard deviation is 1.6 hours.With a .95 probability, the margin of error is approximately
A) .392.
B) 1.96.
C) .20.
D) 1.645.

Accepted Answer

The margin of error (ME) can be calculated using the formula ME = Z * (σ/√n), where Z is the Z-score corresponding to the confidence level (1.96 for 95%), σ is the population standard deviation (1.6 hours), and n is the sample size (64). Plugging in the values gives ME = 1.96 * (1.6/√64) = 1.96 * 0.2 = 0.392.

Question 6

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The value of the margin of error at 95% confidence is
A) 80.
B) 8.
C) .10.
D) 1.568.

Accepted Answer

To find the margin of error, we use the formula:

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

We know that our sample size is 100, the standard deviation is $8.00, and we want to find the margin of error at a 95% confidence level. To find the z-score, we can use a z-table or calculator to find the z-score that corresponds to a 95% confidence level, which is 1.96.

So, plugging in the numbers: 
Margin of Error = 1.96 * ($8.00/square root of 100) 
Margin of Error = 1.96 * $0.80 
Margin of Error = $1.568

Therefore, the margin of error at 95% confidence is $1.568, which is answer choice D.

Question 7

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.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

The value of t for a 95% confidence interval when the sample size is 100 (and thus degrees of freedom is 99) is closest to 1.984 (which rounds to 1.993 for some tables). Since the exact value can slightly vary depending on the source, D is the closest option. Standard z-values like 1.96 are used when the population standard deviation is known and the sample size is large, but for sample sizes where the population standard deviation is unknown and we use the sample's standard deviation, we refer to the t-distribution. The exact value depends on the degrees of freedom (n-1), and for large samples, it approaches the z-value. However, for a precise calculation, one would use a t-table or statistical software, and with 99 degrees of freedom, the value is very close to the z-value but slightly adjusted for the sample size and is approximately 1.984, rounded here to 1.993 in the options provided.

Question 8

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. Given σ = 240 and n = 64, ME = 1.96 * (240/√64) = 1.96 * 30 = 58.8, which rounds to approximately 59.94, making option D the closest answer.

Question 9

The manager of a grocery store has taken a random sample of 144 customers.The average length of time it took these 144 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) .008.
B) .833.
C) .083.
D) 1.000.

Accepted Answer

The formula for standard error of the mean is: standard deviation / square root of sample size
In this case, the standard deviation is given as 1 minute and the sample size is 144 customers. So, the standard error of the mean would be:
1 / sqrt(144) = 1 / 12 = 0.083

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 64 business students over a one-week period.Assume the population standard deviation is 1.6 hours.If the sample mean is 9 hours, then the 95% confidence interval is _____ hours.
A) 7.04 to 10.96
B) 7.36 to 10.64
C) 7.80 to 10.20
D) 8.608 to 9.392

Accepted Answer

The answer of In order to estimate the average time...

Question 11

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

Both conditions np > 5 and n(1 - p) > 5 must be satisfied to ensure that the normal distribution can be used for confidence interval estimates for the population proportion, p. This ensures that the sample size is large enough and the sample proportion is not too extreme.

Question 12

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The standard error of the mean is
A) 80.
B) .8.
C) 8.
D) .08.

Accepted Answer

The formula for standard error of the mean is standard deviation divided by the square root of sample size. Therefore, the standard error of the mean in this case would be $8.00 divided by the square root of 100, which is equal to $0.80 or choice C.

Question 13

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 confidence interval for the mean, is calculated as the sample size minus one. Here, the sample size is 64, so the degrees of freedom is 64 - 1 = 63.

Question 14

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

The t distribution is used instead of the standard normal distribution when the population standard deviation is unknown and the sample size is relatively small (less than 30 is a common rule, but it can be used for larger samples as well). Since the sample size is 50, we use n - 1 degrees of freedom, which is 49 in this case.

Question 15

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

A 95% confidence interval means that if we were to repeat the sampling process many times, 95% of the resulting confidence intervals would contain the true population mean. Therefore, option B is the correct answer. Option A is incorrect because we cannot make a statement about the percentage of the sample that falls within the confidence interval. Option C is incorrect because we cannot make a statement about the percentage of the population that falls within the confidence interval. Option D is incorrect because we cannot make a statement about the percentage of sample means that fall within the confidence interval.

Question 16

In order to estimate the average electric usage per month, a sample of 64 houses was selected and the electric usage was determined.Assume a population standard deviation of 320 kilowatt-hours.If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is _____ kWh.
A) 1779.6 to 1936.4
B) 1818 to 1898
C) 1792.2 to 1923.8
D) 1538 to 2178

Accepted Answer

The 95% confidence interval for the population mean is calculated using the formula: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level (for 95%, $Z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $1858 \pm 1.96 \frac{320}{\sqrt{64}} = 1858 \pm 1.96 \times 40 = 1858 \pm 78.4$, which gives the interval $1779.6$ to $1936.4$ kWh.

Question 17

In order to estimate the average electric usage per month, a sample of 64 houses was selected and the electric usage was determined.Assume a population standard deviation of 320 kilowatt-hours.The standard error of the mean is
A) 320.
B) 64.
C) 400.
D) 40.

Accepted Answer

The formula for standard error of the mean is:

SE = population standard deviation / square root of sample size

SE = 320 / square root of 64

SE = 320 / 8

SE = 40

Therefore, the standard error of the mean is 40.

Question 18

The manager of a grocery store has taken a random sample of 144 customers.The average length of time it took these 144 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) 1.63.
B) .137.
C) .163.
D) 1.37.

Accepted Answer

Using the formula for margin of error:
Margin of error = z* (standard deviation / square root of sample size)
We are given that the standard deviation of the population is 1 minute, and the sample size is 144. We need to find the value of z for a 95% confidence level. This can be found using a standard normal distribution table or calculator, and we get:
z = 1.96

Plugging in the values, we get:
Margin of error = 1.96 * (1 / sqrt(144))
Margin of error = 1.96 * (1 / 12)
Margin of error = 0.163

Therefore, the correct answer is C: 0.163.

Question 19

A random sample of 81 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 300.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 correct t value for a 95% confidence interval with 80 degrees of freedom (n-1 = 81-1 = 80) is closer to the value provided in option B, 1.998. This is because for large samples, the t-distribution approaches the z-distribution, but with a sample size of 81, the exact value from the t-distribution should be used, which is typically found in t-tables and is slightly different from the z-value of 1.96 used for the normal distribution.

Question 20

In order to estimate the average electric usage per month, a sample of 64houses was selected and the electric usage was determined.Assume a population standard deviation of 320 kilowatt-hours.At 95% confidence, the size of the margin of error is
A) 1.96.
B) 40.00.
C) 78.40.
D) 65.80.

Accepted Answer

We can use the formula for margin of error:
Margin of error = z* (standard deviation / sqrt(sample size))
Where z corresponds to the confidence level, which is 1.96 for 95% confidence level.

Plugging in the values:
Margin of error = 1.96 * (320 / sqrt(64))
Margin of error = 1.96 * 40
Margin of error = 78.40

Therefore, the size of the margin of error is 78.40 kilowatt-hours.

Quiz 8: Interval Estimation

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The 95% confidence interval for the average hourly wage (in $) of all information system managers 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 64 business students over a one-week period.Assume the population standard deviation is 1.6 hours.The standard error of the mean 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 64 business students over a one-week period.Assume the population standard deviation is 1.6 hours.With a .95 probability, the margin of error is approximately

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The value of the margin of error at 95% confidence is

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.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 margin of error at 95% confidence is

The manager of a grocery store has taken a random sample of 144 customers.The average length of time it took these 144 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

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

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00.The standard error of the mean 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.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

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 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?

In order to estimate the average electric usage per month, a sample of 64 houses was selected and the electric usage was determined.Assume a population standard deviation of 320 kilowatt-hours.If the sample mean is 1858 kWh, the 95% confidence interval estimate of the population mean is _____ kWh.

In order to estimate the average electric usage per month, a sample of 64 houses was selected and the electric usage was determined.Assume a population standard deviation of 320 kilowatt-hours.The standard error of the mean is

A random sample of 81 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 300.The t value needed to develop the 95% confidence interval for the population mean SAT score is

In order to estimate the average electric usage per month, a sample of 64houses was selected and the electric usage was determined.Assume a population standard deviation of 320 kilowatt-hours.At 95% confidence, the size of the margin of error is

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

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?