Question 1

A random sample of 80 observations produced a mean $\bar { x } = 35.4$ and a standard deviation $s$ $= 3.1$.
a. Find a $90 \%$ confidence interval for the population mean $\mu$.
b. Find a $95 \%$ confidence interval for $\mu$.
c. Find a $99 \%$ confidence interval for $\mu$.
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?

Accepted Answer

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

Question 2

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

Accepted Answer

The answer of Find the value of \(t _ {...

Question 3

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

Accepted Answer

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

Question 4

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 $12,760 with a standard deviation of $700. Suppose that the interval is
Calculated to be ($12,264.09, $13,255.91). 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 and increase the confidence coefficient.
B)Keep the sample size the same but increase the confidence coefficient.
C)Increase the sample size but decrease the confidence coefficient.
D)Decrease the sample size but increase the confidence coefficient.

Accepted Answer

Increasing the sample size decreases the standard error, which narrows the interval. Decreasing the confidence coefficient reduces the z-score or t-score multiplier, also narrowing the interval.

Question 5

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): 88.9, 56, 249.3, 497,
124.7, 157.6, 108.9, and 222.8. What value will be used as the point estimate for the mean
Endowment of all private colleges in the United States?
A)215.029
B)8
C)188.15
D)1505.2

Accepted Answer

The point estimate for the mean endowment of all private colleges in the United States is calculated by finding the average of the given endowments. Adding all the endowments together and dividing by the number of colleges (8) gives the mean endowment, which is 1505.2 divided by 8, resulting in 188.15 million dollars.

Question 6

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

Accepted Answer

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

Question 7

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 99% confidence interval for
The mean was calculated to be ($2,181,260, $5,836,180). Explain what the phrase ʺ99% confidentʺ
Means. 
A) $99 \%$ of the similarly constructed intervals would contain the value of the sample mean.
B) $99 \%$ of the sample means from similar samples fall within the interval.
C) In repeated sampling, $99 \%$ of the intervals constructed would contain $\mu$.
D) $99 \%$ of the population values will fall within the interval.

Accepted Answer

The phrase "99% confident" in the context of a confidence interval means that if we were to take many samples and construct a confidence interval in the same way for each sample, 99% of these intervals would contain the true population mean ($\mu$). This is a fundamental concept in inferential statistics, reflecting the reliability of the confidence interval method rather than the variability of the sample itself or the population values.

Question 8

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 $12,580 with a standard deviation of $700. What is the 90% confidence
Interval for the true mean resale value of a 5- year-old car of this model? 
A) $12,580 \pm 1.746 ( 700 / \sqrt { 17 } )$
B) $12,580 \pm 1.746 ( 700 / \sqrt { 16 } )$
C) $12,580 \pm 1.645 ( 700 / \sqrt { 17 } )$
D) $12,580 \pm 1.740 ( 700 / \sqrt { 17 } )$

Accepted Answer

The correct formula for a confidence interval is $\bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right)$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score for the desired confidence level with $n-1$ degrees of freedom. For a 90% confidence interval with 17 samples (16 degrees of freedom), the closest t-score is approximately 1.746. Therefore, choice A correctly applies this formula with the given data.

Question 9

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) $70 \pm 1.833 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
B) $70 \pm 2.262 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
C) $70 \pm 2.228 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$
D) $70 \pm 1.960 \left( \frac { 17.50 } { \sqrt { 60 } } \right)$

Accepted Answer

The correct choice is B because it uses the t-distribution critical value for a 95% confidence interval with 9 degrees of freedom (n-1, where n is the sample size of 10). The value 2.262 is the appropriate t-score for a 95% confidence level with 9 degrees of freedom. Choices A, C, and D use incorrect critical values for this scenario.

Question 10

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 $3.25, respectively. Use this information to
Construct a 99% confidence interval for the mean. 
A) $19 \pm 3.169 ( 3.25 / \sqrt { 10 } )$
B) $19 \pm 3.25 ( 3.25 / \sqrt { 10 } )$
C) $19 \pm 3.106 ( 3.25 / \sqrt { 10 } )$
D) $19 \pm 2.821 ( 3.25 / \sqrt { 10 } )$

Accepted Answer

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

Question 11

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

Accepted Answer

The value of $t_0$ is the critical value from the t-distribution with 15 degrees of freedom that captures the central 95% of the distribution. For df = 15, this value is approximately 2.131.

Question 12

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 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 90% 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 distribution of the sample means is approximately normal.
B)The sample standard deviation is less than the degrees of freedom.
C)None. The Central Limit Theorem applies.
D)The population of total compensations of CEOs in the service industry is approximately normally distributed.

Accepted Answer

The confidence interval assumes that the population distribution is approximately normal, especially with a small sample size like 18.

Question 14

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) $47 \pm 1.645 \frac { 5 } { \sqrt { 15 } }$
C) $47 \pm 1.345 \frac { 5 } { \sqrt { 15 } }$
D) $47 \pm 1.753 \frac { 5 } { \sqrt { 15 } }$

Accepted Answer

The correct confidence interval uses the t-distribution since the population standard deviation is unknown and the sample size is small (n < 30). For a 90% confidence interval and 14 degrees of freedom (n-1 = 15-1), the closest t-value from tables is approximately 1.761. Thus, option A correctly applies this t-value in the formula for the confidence interval.

Question 15

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 95% 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 90%?
A)There would be no change in the width of the interval.
B)The interval would get narrower.
C)It is impossible to tell until the 90% interval is constructed.
D)The interval would get wider.

Accepted Answer

Lowering the confidence level from 95% to 90% decreases the critical value, which in turn reduces the margin of error, resulting in a narrower confidence interval.

Question 16

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 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 z-value with 20 degrees of freedom was used.
B)In construction of the confidence interval, a t-value with 19 degrees of freedom was used.
C)In construction of the confidence interval, a t-value with 20 degrees of freedom was used.
D)In construction of the confidence interval, a z-value was used.

Accepted Answer

When constructing a confidence interval for the mean with a sample size smaller than 30 and the population standard deviation unknown, a t-distribution is used. The degrees of freedom for a sample of size n is n-1, hence for 20 students, it's 19 degrees of freedom.

Question 18

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

Accepted Answer

The answer of Let $t _ { 0 }$ be...

Question 19

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

Accepted Answer

The value of $t_0$ is the critical value for a two-tailed t-distribution with 14 degrees of freedom that captures 90% of the probability. This corresponds to a 5% tail on each side, and the critical value is approximately 1.761.

Question 20

Let $t _ { 0 }$ be a specific value of $t$. Find $t _ { 0 }$ such that the following statement is true: $P \left( t \leq t _ { 0 } \right) = .005$ where $\mathrm { df } = 20$.
A) $2.845$
B) $2.861$
C) $- 2.861$
D) $- 2.845$

Accepted Answer

The answer of Let $t _ { 0 }$ be...

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

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

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

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

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 99% confidence interval for The mean was calculated to be ($2,181,260, $5,836,180) . Explain what the phrase ʺ99% confidentʺ Means.

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

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.

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?

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

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

Let $t _ { 0 }$ be a specific value of $t$ . Find $t _ { 0 }$ such that the following statement is true: $P \left( t \leq t _ { 0 } \right) = .005$ where $\mathrm { df } = 20$ .

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

Find the value of t0t _ { 0 }t0​ such that the following statement is true: P(−t0≤t≤t0) =.99P \left( - t _ { 0 } \leq t \leq t _ { 0 } \right) = .99P(−t0​≤t≤t0​) =.99 where df=9\mathrm { df } = 9df=9 .

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