Question 1

It is known that the population variance equals 484.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

Accepted Answer

We can use the formula for the margin of error:

Margin of Error = z*(sigma/sqrt(n))

where z is the z-score for the desired level of confidence (0.95), sigma is the population standard deviation (sqrt(484) = 22), and n is the sample size.

We want the margin of error to be 5 or less, so we can set up the inequality:

5 >= z*(22/sqrt(n))

Solving for n:

n >= (z*22/5)^2

Using the z-score corresponding to a 95% confidence level, which is about 1.96:

n >= (1.96*22/5)^2

n >= 74.05

We need to round up to the nearest whole number, so the answer is 75. Therefore, the best choice is D.

Question 2

The following random sample from a population whose values were normally distributed was collected. 10
12
18
16
​
The 80% confidence interval for μ is

Accepted Answer

The answer of The following random sample from a population...

Question 3

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

Accepted Answer

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

Question 4

The sample size needed to provide a margin of error of 2 or less with a .95 probability when the population standard deviation equals 11 is

Accepted Answer

The formula for the sample size needed to provide a margin of error with a .95 probability when the population standard deviation is known is:

n = (Z^2 * σ^2)/(E^2)

Where:
Z = the Z-score for the chosen level of confidence (95% = 1.96)
σ = the population standard deviation
E = the desired margin of error

Plugging in the values given, we get:

n = (1.96^2 * 11^2)/(2^2) = 116.36

Rounding up, we get a sample size of 117, which is answer choice D.

Question 5

The following random sample from a population whose values were normally distributed was collected. 10
8
11
11
​
The 95% confidence interval for μ is

Accepted Answer

To find the confidence interval for the population mean, we can use the t-distribution with n-1 degrees of freedom, since the population standard deviation is unknown.

First, we need to find the sample mean and sample standard deviation:

x̄ = (10+8+11+11)/4 = 10

s = sqrt([(10-10)^2+(8-10)^2+(11-10)^2+(11-10)^2]/3) ≈ 1.24

Next, we can use the t-distribution with n-1 = 3 degrees of freedom and a 95% confidence level to find the t-value:

t* = 3.182 (from a t-distribution table or calculator)

Finally, we can calculate the confidence interval for μ:

CI = x̄ ± t* × s / sqrt(n) = 10 ± 3.182 × 1.24 / sqrt(4) ≈ 7.75 to 12.25

Therefore, the best choice is B, which gives the 95% confidence interval for μ as 7.75 to 12.25.

Question 6

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

Accepted Answer

The confidence interval is calculated using the formula: mean ± (z * (standard deviation/sqrt(n))). Here, the mean is 84, the standard deviation is sqrt(12.25) = 3.5, n = 49, and z for 95% confidence is 1.96. This gives 84 ± (1.96 * (3.5/7)), resulting in the interval 82.99 to 85.01.

Question 7

In general, higher confidence levels provide

Accepted Answer

Higher confidence levels require capturing more of the data to be more certain of the estimate, which results in wider confidence intervals.

Question 8

When the level of confidence decreases, the margin of error

Accepted Answer

The answer of When the level of confidence decreases, the...

Question 9

When constructing a confidence interval for the population mean using the standard deviation of the sample, the degrees of freedom for the t distribution equals

Accepted Answer

The degrees of freedom for the t distribution when constructing a confidence interval for the population mean using the standard deviation of the sample equals n - 1. This is because one degree of freedom is lost when estimating the sample mean.

Question 10

In a sample of 400 voters, 360 indicated they favor the incumbent governor.The 95% confidence interval of voters not favoring the incumbent is

Accepted Answer

The proportion of voters not favoring the incumbent is 1 - (360/400) = 0.1 or 10%. Using the formula for a confidence interval for a proportion, p ± Z*(sqrt[p(1-p)/n]), where p is the sample proportion (0.1), n is the sample size (400), and Z is the Z-value for a 95% confidence interval (approximately 1.96), we calculate the confidence interval for the proportion not favoring the incumbent. This calculation yields a confidence interval of approximately 0.071 to 0.129, which corresponds to choice D.

Question 11

We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor.What is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidence?

Accepted Answer

We use the formula for sample size calculation for estimating population proportion:

n = (Z^2 * p * (1-p))/E^2

where:
Z = Z-value for a 95% confidence level = 1.96
p = the estimate of the proportion of voters who would vote for the incumbent governor
(we do not know this value, so we use p = q = 0.5 for maximum variance)
E = the margin of error, which is 0.05 or 5%

Substituting these values in the formula, we get:

n = (1.96^2 * 0.5 * 0.5)/(0.05^2) = 384.16

Rounding up to the next whole number, we get a minimum sample size of 385. Therefore, the best choice is D.

Question 12

In a random sample of 144 observations, $\bar { p }$ = .6.The 95% confidence interval for p is

Accepted Answer

The 95% confidence interval for p is given by:$\bar { p } \pm 1.96 \sqrt { \frac { \bar { p } ( 1 - \bar { p } ) } { n } }$Plugging in the values we get:.6 ± 1.96 \sqrt { \frac { .6 ( 1 - .6 ) } { 144 } } = .6 ± 1.96 \sqrt { \frac { .6 ( .4 ) } { 144 } } = .6 ± 1.96 \sqrt { \frac { .24 } { 144 } } = .6 ± 1.96 \sqrt { \frac { .001667 } { } } = .6 ± 1.96 \sqrt { .001667 } = .6 ± 1.96 ( .0408 ) = .6 ± .08 = .52 to .68

Question 13

In a random sample of 100 observations, $\bar { p }$ = .2.The 95.44% confidence interval for p is

Accepted Answer

The answer of In a random sample of 100 observations,...

Question 14

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

Accepted Answer

The ability of an interval estimate to contain the value of the population parameter is determined by the confidence level. It gives a range of values within which we can say with a certain level of confidence that the true population parameter lies. So, option A is the correct choice. Options B, C, and D are not relevant to the given statement.

Question 15

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

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 16

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

Accepted Answer

When we increase the confidence level from 95% to 99%, we are increasing the level of certainty in our estimate. This means we need to be more conservative in our interval estimate and increase its width to include more possible values. Therefore, the width of the confidence interval will increase.

Question 17

Using an α = .04, a confidence interval for a population proportion is determined to be .65 to .75.For the same data, if α is decreased, the confidence interval for the population proportion

Accepted Answer

Decreasing α (alpha) increases the confidence level, which in turn increases the margin of error, making the confidence interval wider.

Question 18

A random sample of 1000 people was taken.Four hundred fifty of the people in the sample favored Candidate A.The 95% confidence interval for the true proportion of people who favor Candidate A is

Accepted Answer

The answer of A random sample of 1000 people was...

Question 19

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 μ is

Accepted Answer

The answer of A sample of 225 elements from a...

Question 20

A machine that produces a major part for an airplane engine is monitored closely.In the past, 10% of the parts produced would be defective.With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

Accepted Answer

To determine the sample size, we must first find the critical value associated with a 95% confidence level. Using a standard normal distribution table, this value is 1.96.

Next, we can use the formula:

n = (Z^2 * p * q) / E^2

where: 
- n = sample size 
- Z = the critical value 
- p = the population proportion of defective parts 
- q = (1 - p) 
- E = the desired margin of error

Plugging in the values given: 
n = (1.96^2 * 0.1 * 0.9) / 0.04^2 
n = 216.27

Rounding up to the nearest whole number, we get a sample size of 217.

Therefore, the best choice is D.

Quiz 8: Interval Estimation

It is known that the population variance equals 484.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

The following random sample from a population whose values were normally distributed was collected. 10 12 18 16 The 80% confidence interval for μ is

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

The sample size needed to provide a margin of error of 2 or less with a .95 probability when the population standard deviation equals 11 is

The following random sample from a population whose values were normally distributed was collected. 10 8 11 11 The 95% confidence interval for μ 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

In general, higher confidence levels provide

When the level of confidence decreases, the margin of error

When constructing a confidence interval for the population mean using the standard deviation of the sample, the degrees of freedom for the t distribution equals

In a sample of 400 voters, 360 indicated they favor the incumbent governor.The 95% confidence interval of voters not favoring the incumbent is

We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor.What is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidence?

In a random sample of 144 observations, $\bar { p }$ = .6.The 95% confidence interval for p is

In a random sample of 100 observations, $\bar { p }$ = .2.The 95.44% confidence interval for p is

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

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 the

Using an α = .04, a confidence interval for a population proportion is determined to be .65 to .75.For the same data, if α is decreased, the confidence interval for the population proportion

A random sample of 1000 people was taken.Four hundred fifty of the people in the sample favored Candidate A.The 95% confidence interval for the true proportion of people who favor Candidate A is

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 μ is

A machine that produces a major part for an airplane engine is monitored closely.In the past, 10% of the parts produced would be defective.With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

Quiz 8: Interval Estimation

It is known that the population variance equals 484.With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

The following random sample from a population whose values were normally distributed was collected. 10 12 18 16 ​ The 80% confidence interval for μ is

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

The sample size needed to provide a margin of error of 2 or less with a .95 probability when the population standard deviation equals 11 is

The following random sample from a population whose values were normally distributed was collected. 10 8 11 11 ​ The 95% confidence interval for μ 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

In general, higher confidence levels provide

When the level of confidence decreases, the margin of error

When constructing a confidence interval for the population mean using the standard deviation of the sample, the degrees of freedom for the t distribution equals

In a sample of 400 voters, 360 indicated they favor the incumbent governor.The 95% confidence interval of voters not favoring the incumbent is

We are interested in conducting a study to determine the percentage of voters of a state would vote for the incumbent governor.What is the minimum sample size needed to estimate the population proportion with a margin of error of .05 or less at 95% confidence?

In a random sample of 144 observations, pˉ\bar { p }pˉ​ = .6.The 95% confidence interval for p is

In a random sample of 100 observations, pˉ\bar { p }pˉ​ = .2.The 95.44% confidence interval for p is

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

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 the

Using an α = .04, a confidence interval for a population proportion is determined to be .65 to .75.For the same data, if α is decreased, the confidence interval for the population proportion

A random sample of 1000 people was taken.Four hundred fifty of the people in the sample favored Candidate A.The 95% confidence interval for the true proportion of people who favor Candidate A is

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 μ is

A machine that produces a major part for an airplane engine is monitored closely.In the past, 10% of the parts produced would be defective.With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

The following random sample from a population whose values were normally distributed was collected. 10 12 18 16 The 80% confidence interval for μ is

The following random sample from a population whose values were normally distributed was collected. 10 8 11 11 The 95% confidence interval for μ is

In a random sample of 144 observations, $\bar { p }$ = .6.The 95% confidence interval for p is

In a random sample of 100 observations, $\bar { p }$ = .2.The 95.44% confidence interval for p is