Question 1

When the level of confidence decreases, the margin of error
A)stays the same
B)becomes smaller
C)becomes larger
D)becomes smaller or larger, depending on the sample size

Accepted Answer

When the level of confidence decreases, the critical value (z or t) used to calculate the margin of error decreases, leading to a smaller margin of error.

Question 2

In general, higher confidence levels provide
A)wider confidence intervals
B)narrower confidence intervals
C)a smaller standard error
D)unbiased estimates

Accepted Answer

Higher confidence levels mean that we are more certain about our estimate, so we need wider intervals to account for the possibility of the estimate being further from the true value. Therefore, higher confidence levels result in wider confidence intervals.

Question 3

Whenever using the t distribution for interval estimation (when the sample size is very small), we must assume that
A)the sample has a mean of at least 30
B)the sampling distribution is not normal
C)the population is approximately normal
D)the finite population correction factor is necessary

Accepted Answer

When using the t distribution for interval estimation with a small sample size, we assume that the population is approximately normal.

Question 4

A 95% confidence interval for a population mean is determined to be 100 to 120. If the confidence coefficient is reduced to 0.90, the interval for $\mu$
A)becomes narrower
B)becomes wider
C)does not change
D)becomes 0.1

Accepted Answer

Reducing the confidence level from 95% to 90% decreases the margin of error, resulting in a narrower confidence interval.

Question 5

A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is
A)20.5 to 26.5
B)24.4 to 25.6
C)23.0 to 27.0
D)20.0 to 30.0

Accepted Answer

The 98% confidence interval is calculated using the formula: mean ± (z * (standard deviation/√n)). Here, the mean is 25, the standard deviation is 2, n is 64, and the z-value for 98% confidence is approximately 2.33. This gives the interval 25 ± (2.33 * (2/√64)) = 24.4 to 25.6.

Question 6

If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect
A)the size of the confidence interval to increase
B)the size of the confidence interval to decrease
C)the size of the confidence interval to remain the same
D)the sample size to increase

Accepted Answer

Increasing the confidence level from 95% to 99% means we want to be more confident in our estimate, which requires a larger margin of error. Therefore, the size of the confidence interval will increase.

Question 7

A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8. The 95.44% confidence interval for the population mean is
A)15.2 to 24.8
B)19.200 to 20.800
C)19.216 to 20.784
D)21.2 to 22.8

Accepted Answer

The answer of A random sample of 144 observations has...

Question 8

A sample of 20 items from a population with an unknown $\sigma$ is selected in order to develop an interval estimate of $\mu$. Which of the following is not necessary?
A)We must assume the population has a normal distribution.
B)We must use a t distribution.
C)Sample standard deviation must be used to estimate $\sigma$.
D)The sample must have a normal distribution.

Accepted Answer

The sample does not need to have a normal distribution in order to use a t distribution for interval estimation of the population mean. However, the other options are necessary. We do need to assume that the population has a normal distribution, use a t distribution for small sample sizes, and estimate the population standard deviation using the sample standard deviation.

Question 9

The ability of an interval estimate to contain the value of the population parameter is described by the
A)confidence level
B)degrees of freedom
C)precise value of the population mean $\mu$
D)degrees of freedom minus 1

Accepted Answer

The ability of an interval estimate to contain the value of the population parameter is described by the confidence level. The confidence level represents the probability that the interval estimate will contain the population parameter. For example, a 95% confidence level means that if the same population were sampled repeatedly and interval estimates were computed for each sample, approximately 95% of those intervals would contain the true population parameter.

Question 10

It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a .95 probability that the sample mean will provide a margin of error of
A)7.84
B)31.36
C)344.96
D)1,936

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 desired confidence level (1.96 for 95%), σ is the population standard deviation (√1,936 = 44), and n is the sample size (121). Thus, Margin of Error = 1.96 * (44/√121) = 1.96 * 4 = 7.84.

Question 11

For which of the following values of P is the value of P(1 - P) maximized?
A)P = 0.99
B)P = 0.90
C)P = 0.01
D)P = 0.50

Accepted Answer

To find the maximum value of P(1-P), we can take the derivative of the expression with respect to P and set it equal to 0.

d/dP [P(1-P)] = 1-2P

Setting this equal to 0, we get 1-2P = 0, or P = 0.5.

Therefore, the maximum is achieved at P = 0.5, which is option D.

Question 12

A sample of 225 elements from a population with a standard deviation of 75 is selected. The sample mean is 180. The 95% confidence interval for $\mu$ is
A)105.0 to 225.0
B)175.0 to 185.0
C)100.0 to 200.0
D)170.2 to 189.8

Accepted Answer

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

CI = X̄ ± Z * (σ/√n)

Where:
X̄ = sample mean = 180
σ = population standard deviation = 75
n = sample size = 225
Z = z-score for the desired confidence level = 1.96 (for 95% confidence level)

Plugging in the values, we get:

CI = 180 ± 1.96 * (75/√225) 
CI = 180 ± 1.96 * 5 
CI = 180 ± 9.8

So the 95% confidence interval for the population mean is:

CI = (180 - 9.8, 180 + 9.8)
CI = (170.2, 189.8)

Therefore, the best choice is D, with the above explanation.

Question 13

Using an $\alpha$ = 0.04 a confidence interval for a population proportion is determined to be 0.65 to 0.75. If the level of significance is decreased, the interval for the population proportion
A)becomes narrower
B)becomes wider
C)does not change
D)remains the same

Accepted Answer

Decreasing the level of significance (increasing the confidence level) results in a wider confidence interval because it requires more certainty that the interval contains the true population proportion.

Question 14

As the sample size increases, the margin of error
A)increases
B)decreases
C)stays the same
D)increases or decreases depending on the size of the mean

Accepted Answer

As the sample size increases, the margin of error decreases. This is because a larger sample size provides more data and therefore more precision in estimating the population parameter. As a result, the margin of error, which is a measure of the precision of the estimate, decreases.

Question 15

An interval estimate is a range of values used to estimate
A)the shape of the population's distribution
B)the sampling distribution
C)a sample statistic
D)a population parameter

Accepted Answer

Interval estimates are used to estimate population parameters, such as the mean or proportion. They provide a range of values within which the true parameter is believed to lie with a certain level of confidence.

Question 16

In determining the sample size necessary to estimate a population proportion, which of the following information is not needed?
A)the maximum margin of error that can be tolerated
B)the confidence level required
C)a preliminary estimate of the true population proportion P
D)the mean of the population

Accepted Answer

The mean of the population is not necessary to determine the sample size needed to estimate a population proportion. A preliminary estimate of the true population proportion, the maximum margin of error that can be tolerated, and the confidence level required are necessary inputs for determining the sample size.

Question 17

The z value for a 97.8% confidence interval estimation is
A)2.02
B)1.96
C)2.00
D)2.29

Accepted Answer

The z value for a 97.8% confidence interval estimation can be found using the standard normal distribution table or calculator.

Using the table, we can find the closest probability value to 0.978, which is 0.9793. The corresponding z-score is 2.06. However, because the table only provides values up to 2.00, we can use the fact that the standard normal distribution is symmetric to find the z-score for the remaining area.

The remaining area is (1 - 0.9793) = 0.0207, which is divided equally between the two tails of the distribution (since it is symmetric). Therefore, each tail has an area of 0.01035.

To find the z-score for an area of 0.01035, we can look up the corresponding value in the standard normal distribution table, which is approximately 2.29.

Therefore, the z value for a 97.8% confidence interval estimation is 2.29.

Question 18

The t value for a 95% confidence interval estimation with 24 degrees of freedom is
A)1.711
B)2.064
C)2.492
D)2.069

Accepted Answer

The t value for a 95% confidence interval estimation with 24 degrees of freedom is 2.064, which is the closest value to the tabled t value of 2.064 for a 95% confidence level and 24 degrees of freedom.

Question 19

A random sample of 49 statistics examinations was taken. The average score, in the sample, was 84 with a variance of 12.25. The 95% confidence interval for the average examination score of the population of the examinations is
A)76.00 to 84.00
B)77.40 to 86.60
C)83.00 to 85.00
D)68.00 to 100.00

Accepted Answer

The answer of A random sample of 49 statistics examinations...

Question 20

After computing a confidence interval, the user believes the results are meaningless because the width of the interval is too large. Which one of the following is the best recommendation?
A)Increase the level of confidence for the interval.
B)Decrease the sample size.
C)Increase the sample size.
D)Reduce the population variance.

Accepted Answer

Increasing the sample size will decrease the standard error of the mean, which in turn will narrow the confidence interval. Therefore, increasing the sample size is the best option to reduce the width of the confidence interval.

Quiz 8: Interval Estimation

When the level of confidence decreases, the margin of error

In general, higher confidence levels provide

Whenever using the t distribution for interval estimation (when the sample size is very small) , we must assume that

A 95% confidence interval for a population mean is determined to be 100 to 120. If the confidence coefficient is reduced to 0.90, the interval for $\mu$

A random sample of 64 students at a university showed an average age of 25 years and a sample standard deviation of 2 years. The 98% confidence interval for the true average age of all students in the university is

If we change a 95% confidence interval estimate to a 99% confidence interval estimate, we can expect

A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known to equal 4.8. The 95.44% confidence interval for the population mean is

A sample of 20 items from a population with an unknown $\sigma$ is selected in order to develop an interval estimate of $\mu$ . Which of the following is not necessary?

The ability of an interval estimate to contain the value of the population parameter is described by the

It is known that the variance of a population equals 1,936. A random sample of 121 has been taken from the population. There is a .95 probability that the sample mean will provide a margin of error of

For which of the following values of P is the value of P(1 - P) maximized?

A sample of 225 elements from a population with a standard deviation of 75 is selected. The sample mean is 180. The 95% confidence interval for $\mu$ is

Using an $\alpha$ = 0.04 a confidence interval for a population proportion is determined to be 0.65 to 0.75. If the level of significance is decreased, the interval for the population proportion

As the sample size increases, the margin of error

An interval estimate is a range of values used to estimate

In determining the sample size necessary to estimate a population proportion, which of the following information is not needed?

The z value for a 97.8% confidence interval estimation is

The t value for a 95% confidence interval estimation with 24 degrees of freedom is

A random sample of 49 statistics examinations was taken. The average score, in the sample, was 84 with a variance of 12.25. The 95% confidence interval for the average examination score of the population of the examinations is

After computing a confidence interval, the user believes the results are meaningless because the width of the interval is too large. Which one of the following is the best recommendation?