Question 1

Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is invested in a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of eight private colleges in the United States revealed the following endowments (in millions of dollars): 64.5, 55.1, 232.8, 496.1, 127.6, 186.4, 104.7, and 212.2. What value will be used as the point estimate for the mean endowment of all private colleges in the United States?
A) 184.925
B) 1479.4
C) 211.343
D) 8

Accepted Answer

The point estimate for the mean endowment is the average of the given endowments. Adding the values and dividing by the number of colleges (8) gives 184.925 million dollars.

Question 2

$$\text { Let } t _ { 0 } \text { be a particular value of } t \text {. Find a value of } t _ { 0 } \text { such that } P \left( t \leq t _ { 0 } \right) = .005 \text { where } d f = 9 \text {. }$$

Accepted Answer

The answer of \[\text { Let } t _ {...

Question 3

How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled results show that the sample mean and sample standard deviation were $70.00 and $17.50, respectively. Use this information to create a 95 percent confidence interval for the population mean. A) $70 \pm 2.228 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

B) $70 \pm 1.960 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

C) $70 \pm 1.833 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

D) $70 \pm 2.262 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

Accepted Answer

The answer of How much money does the average professional...

Question 4

Suppose you selected a random sample of n = 7 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 90% confidence interval.

Accepted Answer

The answer of Suppose you selected a random sample of...

Question 5

Find the value of $$t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .90 \text { where } \mathrm { df } = 14 \text {. }$$
A) 1.761
B) 1.345
C) 2.145
D) 2.624

Accepted Answer

The value of $ t_0 $ for which $ P(-t_0 \leq t \leq t_0) = 0.90 $ with 14 degrees of freedom is approximately 1.761, which corresponds to the 90% confidence interval for a t-distribution with 14 degrees of freedom.

Question 6

Let t0 be a particular value of t. Find a value of $$t _ { 0 } \text { such that } P \left( t \leq t _ { 0 } \text { or } t \geq t _ { 0 } \right) = .1 \text { where } d f = 14 \text {. }$$

Accepted Answer

The answer of Let t0 be a particular value of...

Question 7

Fifteen SmartCars were randomly selected and the highway mileage of each was noted. The analysis yielded a mean of 47 miles per gallon and a standard deviation of 5 miles per gallon. Which of the following would represent a 90% confidence interval for the average highway mileage of all SmartCars? A) $47 \pm 1.761 \frac { 5 } { \sqrt { 15 } }$

B) $47 \pm 1.345 \frac { 5 } { \sqrt { 15 } }$

C) $47 \pm 1.753 \frac { 5 } { \sqrt { 15 } }$

D) $47 \pm 1.645 \frac { 5 } { \sqrt { 15 } }$

Accepted Answer

The answer of Fifteen SmartCars were randomly selected and the...

Question 8

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 90% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180). Explain what the phrase "90% confident" means.
A) In repeated sampling, 90% of the intervals constructed would contain μ.
B) 90% of the population values will fall within the interval.
C) 90% of the sample means from similar samples fall within the interval.
D) 90% of the similarly constructed intervals would contain the value of the sample mean.

Accepted Answer

In repeated sampling, 90% of the intervals constructed would contain μ. This means that if we were to take many samples and construct a confidence interval from each, 90% of those intervals would contain the true population mean.

Question 9

Find the value of $$t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .95 \text { where } \mathrm { df } = 15 \text {. }$$
A) 2.131
B) 1.753
C) 2.602
D) 2.947

Accepted Answer

For a t-distribution with 15 degrees of freedom, the critical value for a two-tailed test with 95% confidence is approximately 2.131.

Question 10

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 97% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180). What would happen to the confidence interval if the confidence level were changed to 95%?
A) The interval would get narrower.
B) The interval would get wider.
C) There would be no change in the width of the interval.
D) It is impossible to tell until the 95% interval is constructed.

Accepted Answer

A lower confidence level results in a narrower confidence interval because there is less certainty required, which reduces the margin of error.

Question 11

You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 years. Since you are particularly interested in a certain foreign sedan, you decide to estimate the resale value of this car with a 99% confidence interval. You manage to obtain data on 17 recently resold 5-year-old foreign sedans of the same model. These 17 cars were resold at an average price of $12,380 with a standard deviation of $800. What is the 99% confidence interval for the true mean resale value of a 5- year-old car of this model? A) $12,380 \pm 2.921 ( 800 / \sqrt { 17 } )$

B) $12,380 \pm 2.898 ( 800 / \sqrt { 17 } )$

C) $12,380 \pm 2.575 ( 800 / \sqrt { 17 } )$

D) $12,380 \pm 2.921 ( 800 / \sqrt { 16 } )$

Accepted Answer

The answer of You are interested in purchasing a new...

Question 12

How much money does the average professional football fan spend on food at a single football game? That question was posed to 10 randomly selected football fans. The sample results provided a sample mean and standard deviation of $19.00 and $2.95, respectively. Use this information to construct a 98% confidence interval for the mean. A) $19 \pm 2.821 ( 2.95 / \sqrt { 10 } )$

B) $19 \pm 2.764 ( 2.95 / \sqrt { 10 } )$

C) $19 \pm 2.718 ( 2.95 / \sqrt { 10 } )$

D) $19 \pm 2.262 ( 2.95 / \sqrt { 10 } )$

Accepted Answer

The answer of How much money does the average professional...

Question 13

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 98% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180). What additional assumption is necessary for this confidence interval to be valid?
A) The population of total compensations of CEOs in the service industry is approximately normally distributed.
B) None. The Central Limit Theorem applies.
C) The sample standard deviation is less than the degrees of freedom.
D) The distribution of the sample means is approximately normal.

Accepted Answer

The population of total compensations of CEOs in the service industry is approximately normally distributed.

Question 14

Let t0 be a specific value of t. Find t0 such that the following statement is true: $$P \left( t \geq t _ { 0 } \right) = .025 \text { where df } = 20 \text {. }$$
A) 2.086
B) -2.086
C) 2.093
D) -2.093

Accepted Answer

The correct answer is A because the t-value corresponding to a probability of 0.025 with 20 degrees of freedom is 2.086.

Question 15

Suppose you selected a random sample of n = 29 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 95% confidence interval.

Accepted Answer

The answer of Suppose you selected a random sample of...

Question 16

A random sample of n = 144 measurements was selected from a population with unknown mean μ and standard deviation σ. Calculate a 90% confidence interval if $\bar { x } = 3.55 \text { and } s = .49$

Accepted Answer

The answer of A random sample of n = 144...

Question 17

An educator wanted to look at the study habits of university students. As part of the research, data was collected for three variables - the amount of time (in hours per week) spent studying, the amount of time (in hours per week) spent playing video games and the GPA - for a sample of 20 male university students. As part of the research, a 95% confidence interval for the average GPA of all male university students was calculated to be: (2.95, 3.10). Which of the following statements is true?
A) In construction of the confidence interval, a t-value with 19 degrees of freedom was used.
B) In construction of the confidence interval, a t-value with 20 degrees of freedom was used.
C) In construction of the confidence interval, a z-value was used.
D) In construction of the confidence interval, a z-value with 20 degrees of freedom was used. 2 Use t-Distribution

Accepted Answer

The t-distribution is used when the sample size is small (n < 30) and the population standard deviation is unknown. With a sample size of 20, the degrees of freedom would be 19 (n-1).

Question 18

A random sample of 80 observations produced a mean $\bar { x } = 35.4 \text { and a standard deviation } s = 3.1 \text {. }$ a. Find a 90% confidence interval for the population mean μ. b. Find a 95% confidence interval for μ. c. Find a 99% confidence interval for μ. d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed? 7.3 Confidence Interval for a Population Mean: Student's t-Statistic 1 Compare t-Distribution to Normal Distribution

Accepted Answer

The answer of A random sample of 80 observations produced...

Question 19

Find the value of $$t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .99 \text { where } \mathrm { df } = 9 \text {. }$$
A) 3.250
B) 2.2821
C) 2.262
D) 1.833

Accepted Answer

The correct answer is A because the t-value with 9 degrees of freedom and a cumulative probability of 0.99 is 3.250.

Question 20

You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 years. Since you are particularly interested in a certain foreign sedan, you decide to estimate the resale value of this car with a 90% confidence interval. You manage to obtain data on 17 recently resold 5-year-old foreign sedans of the same model. These 17 cars were resold at an average price of $12,320 with a standard deviation of $600. Suppose that the interval is calculated to be ($12,065.92, $12,574.08). How could we alter the sample size and the confidence coefficient in order to guarantee a decrease in the width of the interval?
A) Increase the sample size but decrease the confidence coefficient.
B) Increase the sample size and increase the confidence coefficient.
C) Keep the sample size the same but increase the confidence coefficient.
D) Decrease the sample size but increase the confidence coefficient.

Accepted Answer

Increasing the sample size reduces the standard error, which narrows the confidence interval. Decreasing the confidence coefficient (confidence level) also narrows the interval, as it reduces the range needed to capture the true mean with the specified confidence.

Quiz 7: Inferences Based on a Single Sample: Estimation With Confidence Intervals

$\text { Let } t _ { 0 } \text { be a particular value of } t \text {. Find a value of } t _ { 0 } \text { such that } P \left( t \leq t _ { 0 } \right) = .005 \text { where } d f = 9 \text {. }$

Suppose you selected a random sample of n = 7 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 90% confidence interval.

Find the value of $t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .90 \text { where } \mathrm { df } = 14 \text {. }$

Let t0 be a particular value of t. Find a value of $t _ { 0 } \text { such that } P \left( t \leq t _ { 0 } \text { or } t \geq t _ { 0 } \right) = .1 \text { where } d f = 14 \text {. }$

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 90% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180) . Explain what the phrase "90% confident" means.

Find the value of $t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .95 \text { where } \mathrm { df } = 15 \text {. }$

Let t0 be a specific value of t. Find t0 such that the following statement is true: $P \left( t \geq t _ { 0 } \right) = .025 \text { where df } = 20 \text {. }$

Suppose you selected a random sample of n = 29 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 95% confidence interval.

A random sample of n = 144 measurements was selected from a population with unknown mean μ and standard deviation σ. Calculate a 90% confidence interval if $\bar { x } = 3.55 \text { and } s = .49$

Find the value of $t _ { 0 } \text { such that the following statement is true: } P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .99 \text { where } \mathrm { df } = 9 \text {. }$

Quiz 7: Inferences Based on a Single Sample: Estimation With Confidence Intervals

Suppose you selected a random sample of n = 7 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 90% confidence interval.

Let t0 be a particular value of t. Find a value of t0 such that P(t≤t0 or t≥t0)=.1 where df=14. t _ { 0 } \text { such that } P \left( t \leq t _ { 0 } \text { or } t \geq t _ { 0 } \right) = .1 \text { where } d f = 14 \text {. }t0​ such that P(t≤t0​ or t≥t0​)=.1 where df=14.

A marketing research company is estimating the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 90% confidence interval for the mean was calculated to be ($2,181,260, $5,836,180) . Explain what the phrase "90% confident" means.

Let t0 be a specific value of t. Find t0 such that the following statement is true: P(t≥t0) =.025 where df =20. P \left( t \geq t _ { 0 } \right) = .025 \text { where df } = 20 \text {. }P(t≥t0​) =.025 where df =20.

Suppose you selected a random sample of n = 29 measurements from a normal distribution. Compare the standard normal z value with the corresponding t value for a 95% confidence interval.

A random sample of n = 144 measurements was selected from a population with unknown mean μ and standard deviation σ. Calculate a 90% confidence interval if xˉ=3.55 and s=.49\bar { x } = 3.55 \text { and } s = .49xˉ=3.55 and s=.49

Let t0 be a particular value of t. Find a value of $t _ { 0 } \text { such that } P \left( t \leq t _ { 0 } \text { or } t \geq t _ { 0 } \right) = .1 \text { where } d f = 14 \text {. }$

Let t0 be a specific value of t. Find t0 such that the following statement is true: $P \left( t \geq t _ { 0 } \right) = .025 \text { where df } = 20 \text {. }$

A random sample of n = 144 measurements was selected from a population with unknown mean μ and standard deviation σ. Calculate a 90% confidence interval if $\bar { x } = 3.55 \text { and } s = .49$