Exam 3

Question 1

Suppose a 95% confidence interval for turns out to be (1,000, 2,100). Give a definition of what it means to be “95% confident” in an inference.

95% of the observations in the sample fall in the given interval.

In repeated sampling, 95% of the intervals constructed would contain the population mean.

95% of the observations in the entire population fall in the given interval.

In repeated sampling, the population parameter would fall in the given interval 95% of the time.

Answer 1

In repeated sampling, 95% of the intervals constructed would contain the population mean.

Question 2

It is desired to estimate the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and the 97% confidence interval was calculated to be ($2,181,260, $5,836,180). Which of the following interpretations is correct?

In the population of service-industry CEOs, 97% of them will have total compensations that fall in the interval $2,181,260 to $5,836,180.

97% of the sampled total compensation values fell between $2,181,260 and $5,836,180.

We are 97% confident that the mean of the sampled CEOs falls in the interval $2,181,260 to $5,836,180.

We are 97% confident that the average total compensation of all CEOs in the service industry falls in the interval $2,181,260 to $5,836,180.

Answer 2

We are 97% confident that the average total compensation of all CEOs in the service industry falls in the interval $2,181,260 to $5,836,180.

Question 3

Which of the following is NOT true about the Student’s t distribution?

It is used to construct confidence intervals for the population mean when the population standard deviation is known.

As the number of degrees of freedom increases, the t distribution approaches the normal distribution.

It is bell-shaped and symmetrical.

It has more area in the tails and less in the center than does the normal distribution.

Answer 3

It is used to construct confidence intervals for the population mean when the
population standard deviation is known.

Question 4

A 99% confidence interval estimate can be interpreted to mean that

both of these

we have 99% confidence that we have selected a sample whose interval does include the population mean.

if all possible samples are taken and confidence interval estimates are developed, 99% of them would include the true population mean somewhere within their interval.

none of these

Answer 4

both of these

Question 5

A confidence interval was used to estimate the proportion of statistics students that are females. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Based on the interval above, is the population proportion of females equal to 0.60?

Yes, and we are 90% sure of it.

No. The proportion is 54.17%.

No, and we are 90% sure of it.

Maybe. 0.60 is a believable value of the population proportion based on the information above.

Answer 5

Maybe. 0.60 is a believable value of the population proportion based on the information above.

Question 6

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

597

150

420

105

Answer 6

150

Question 7

When determining the sample size necessary for estimating the true population mean, which factor is not considered when sampling with replacement?

The population size

The level of confidence desired in the estimate

The population standard deviation

The allowable or tolerable sampling error

Answer 7

The population size

Question 8

In the construction of confidence intervals, if all other quantities are unchanged, an increase in the sample size will lead to a ________ interval.

narrower

biased

less significant

wider

Answer 8

narrower

Question 9

Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars): 60.2, 47.0, 235.1, 490.0, 122.6, 177.5, 95.4, and 220.0. Summary statistics yield and . Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments.

$180.975 ± $116.621

$180.975 ± $94.066

$180.975 ± $99.123

$180.975 ± $119.586

Answer 9

$180.975 ± $119.586

Question 10

The managers of a company are worried about the morale of their employees. In order to determine if a problem in this area exists, they decide to evaluate the average attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points. Thus, they should sample ________ employees if they want to estimate the mean score of the employees within 5 points with 90% confidence.

52

63

85

74

Answer 10

63

Question 11

The managers of a company are worried about the morale of their employees. In order to determine if a problem in this area exists, they decide to evaluate the average attitudes of their employees with a standardized test. They select the Fortunato test of job satisfaction, which has a known standard deviation of 24 points. Due to financial limitations, the managers decide to take a sample of 45 employees. This yields a mean score of 88.0 points. A 90% confidence interval would then go from ________ to ________.

78.24 to 100.04

95.65 to 120.45

82.12 to 93.88

62.12 to 85.23

Answer 11

82.12 to 93.88

Question 12

To become an actuary, it is necessary to pass a series of 10 exams, including the most important one, an exam in probability and statistics. An insurance company wants to estimate the mean score on this exam for actuarial students who have enrolled in a special study program. They take a sample of 8 actuarial students in this program and determine that their scores are: 2, 5, 8, 8, 7, 6, 5, and 7 (S = 2). This sample will be used to calculate a 90% confidence interval for the mean score for actuarial students in the special study program. The critical value used in constructing a 90% confidence interval is ________.

1.3968

1.8595

1.8946

1.4149

Answer 12

1.8946

Question 13

To become an actuary, it is necessary to pass a series of 10 exams, including the most important one, an exam in probability and statistics. An insurance company wants to estimate the mean score on this exam for actuarial students who have enrolled in a special study program. They take a sample of 8 actuarial students in this program and determine that their scores are: 2, 5, 8, 8, 7, 6, 5, and 7 (S = 2). This sample will be used to calculate a 90% confidence interval for the mean score for actuarial students in the special study program. A 90% confidence interval for the mean score of actuarial students in the special program is from ________ to ________.

4.66 to 7.34

2.66 to 8.34

5.66 to 8.34

3.66 to 7.34

Answer 13

4.66 to 7.34

Question 14

After an extensive advertising campaign, the manager of a company wants to estimate the proportion of potential customers that recognize a new product. She samples 120 potential consumers and finds that 54 recognize this product. She uses this sample information to obtain a 95% confidence interval that goes from 0.36 to 0.54.

True or False: Referring to the information above, it is possible that the true proportion of people that recognize the product is between 0.36 and 0.54.

True

False

Answer 14

True

Question 15

The president of a university would like to estimate the proportion of the student population who owns a personal computer. In a sample of 500 students, 417 own a personal computer.

True or False: Referring to the information above, we are 99% confidence that between 79.11% and 87.69% of the student population own a personal computer.

True

False

Answer 15

True

Question 16

A university wanted to find out the percentage of students who felt comfortable reporting cheating by their fellow students. A survey of 2,800 students was conducted and the students were asked if they felt comfortable reporting cheating by their fellow students. The results were 1,344 answered “Yes” and 1,456 answered “no”.

Referring to above, a 99% confidence interval for the proportion of student population who feel comfortable reporting cheating by their fellow students is from ________ to ________.

0.3557 to 0.8043

0.4557 to 0.5043

0.2557 to 0.7043

0.3557 to 0.6043

Answer 16

0.4557 to 0.5043

Question 17

The confidence interval obtained will always correctly estimate the population parameter.

True

False

Answer 17

False

Question 18

The t distribution is used to develop a confidence interval estimate of the population proportion when the population standard deviation is unknown.

True

False

Answer 18

False

Question 19

The standardized normal distribution is used to develop a confidence interval estimate of the population proportion when the sample size is sufficiently large.

True

False

Answer 19

True

Question 20

A sample of 100 fuses from a very large shipment is found to have 10 that are defective. The 95% confidence interval would indicate that, for this shipment, the proportion of defective fuses is between 0 and 0.28.

True

False

Answer 20

False

Question 21

A point estimate consists of a single sample statistic that is used to estimate the true population parameter.

True

False

Answer 21

True

Question 22

The sample mean is a point estimate of the population mean.

True

False

Answer 22

True

Question 23

You have been asked to find the critical value for a confidence interval of the mean where you do not know the standard deviation. The sample selected contains 45 people from a population that contains 1000. Which of the following Excel commands would be correct if you wanted to be 94% confident with your results.

=NORMSINV(.03)

=T.INV(.03, 44)

=NORMSINV(.06)

=T.INV(.06, 44)

=NORMSDIST(.03)

Answer 23

=T.INV(.03, 44)

Question 24

We have created a 95% confidence interval for with the result (10, 15). What conclusion will we make if we test H0: = 16 versus H1: 16 at = 0.05?

Fail to reject H0 in favor of H1.

We cannot tell what our decision will be
from the information given.

Accept H0 in favor of H1.

Reject H0 in favor of H1.

Answer 24

Reject H0 in favor of H1.

Question 25

A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors results in 83 who indicate that they recommend aspirin. The value of the test statistic in this problem is approximately equal to

-0.07

-4.12

-2.33

-1.86

Answer 25

-2.33

Question 26

A major videocassette rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with videocassette recorders (VCRs). It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have VCRs. State the test of interest to the rental chain.

H0: p 0.25 versus H1: p > 0.25

H0: p 0.32 versus H1: p > 0.32

H0: p 5,000 versus H1: p > 5,000

H0: p 5,000 versus H1: p > 5,000

Answer 26

H0: p 0.25 versus H1: p > 0.25

Question 27

The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. The mean for unstressed furniture is 650 psi. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p value of 0.080 for the test.

Suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. What would be the p value of this one-tailed test?

0.040

0.840

0.160

0.960

Answer 27

0.960

Question 28

If you were conducting an upper tailed test where the level of significance was 0.05 and the population standard deviation was not known, then which of the following Excel commands would give you the critical value? The sample size is 26.

=T.INV(0.95, 25)

=NORMSINV(.95)

=T.INV(0.025, 26)

=T.INV(0.5, 25)

=1-NORMSINV(.05)

Answer 28

=T.INV(0.95, 25)

Question 29

To reduce the chance of rejecting a true null hypothesis, which of the following is true?

increase alpha

decrease alpha

Answer 29

decrease alpha

Question 30

The AAA would like to find out if the average horsepower of cars in the population is more than 100 (H1). The following information indicates the descriptive statistics for the horsepower of cars in a particular sample.

n = 100
sample mean = 104.83
standard deviation of the sample = 38.52
level of significance = 0.01

What is the critical value for the above scenario?

2.3642

2.6213

2.3646

2.6264

Answer 30

2.3646

Question 31

A major home improvement store conducted its biggest brand recognition campaign in the company’s history. A series of new television advertisements featuring well-known entertainers and sports figures were launched. A key metric for the success of television advertisements is the proportion of viewers who “like the ads a lot”. A study of 1,189 adults who viewed the ads reported that 230 indicated that they “like the ads a lot.” The percentage of a typical television advertisement receiving the “like the ads a lot” score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e.. if there is evidence that the population proportion of “like the ads a lot” for the company’s ads is less than 0.22) at a 0.01 level of significance.

Referring to the above, the null hypothesis will be rejected if the test statistic is:

(HINT: Draw this out and you will hopefully find there is only one answer that makes sense.)

less than 2.3263

less than -2.3263

greater than -2.3263

greater than 2.3263

Answer 31

less than -2.3263

Question 32

If a researcher felt it was a very bad idea to incorrectly reject the null hypothesis, he would probably address this concern by

choosing alpha as close to beta as possible

choosing a high value for alpha

choosing a low value for alpha

setting alpha equal to 1 – c, where c is the
desired confidence level

Answer 32

choosing a low value for alpha

Question 33

In choosing a 5% level of significance for a hypothesis test, a researcher is saying that

she is willing to reject a true null hypothesis 5% of the time

she is 95% confident that her null hypothesis is correct

she is willing to fail to reject a true null hypothesis 5% of the time

she is willing to draw the wrong conclusion in her test 5% of the time

Answer 33

she is willing to reject a true null hypothesis 5% of the time

Question 34

Which of the following would be an appropriate alternative hypothesis?

The mean of a population is greater than 55.

The mean of a sample is equal to 55.

The mean of a sample is greater than 55.

The mean of a population is equal to 55.

Answer 34

The mean of a population is greater than 55.

Question 35

A Type II error is committed when

you reject a null hypothesis that is true.

you reject a null hypothesis that is false.

you don’t reject a null hypothesis that is true.

you don’t reject a null hypothesis that is false.

Answer 35

you don’t reject a null hypothesis that is false.

Question 36

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is greater than 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Suppose she found that the sample mean was 30.45 years and the sample standard deviation was 5 years. What is the p-value associated with the test statistic?

0.1423

0.0780

0.3577

0.02

Answer 36

0.0780

Question 37

How many tissues should the Kimberly Clark Corporation package of Kleenex contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X = 52, S = 22. Suppose the test statistic does fall in the rejection region at = 0.05. Which of the following decision is correct?
At = 0.10, you do not reject H0.

At = 0.05, you do not reject H0.

At = 0.10, you reject H0.

At = 0.05, you reject H0.

Answer 37

At = 0.05, you reject H0.

Question 38

A major DVD rental chain is considering opening a new store in an area that currently does not have any such stores. The chain will open if there is evidence that more than 5,000 of the 20,000 households in the area are equipped with DVD players. It conducts a telephone poll of 300 randomly selected households in the area and finds that 96 have DVD players. State the test of hypothesis that is of interest to the rental chain.
H0: p 5,000 versus H1: p > 5,000

H0: p 5,000 versus H1: p > 5,000

H0: p 0.32 versus H1: p > 0.32

H0: p 0.25 versus H1: p > 0.25

Answer 38

H0: p 0.25 versus H1: p > 0.25

Question 39

Microsoft Excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the Microsoft Excel output for the sample of 46 cases:

n: 46
arthmetic mean: 28.00
SD: 25.92
Standard Error: 3.82
Null Hypothesis: H0: mu <= 20
alpha = 0.10
df: 45
T stat: 2.09
One-tail test upper critical value: 1.3006
p-value: 0.021
Desicion = REJECT

Referring to the above, the null hypothesis would be rejected if a 1% probability of committing a Type I error is allowed.

True
OR
False

Answer 39

False

Question 40

The larger the p-value, the more likely you are to reject the null hypothesis.
True
OR
False

Answer 40

False

Question 41

A student claims that he can correctly identify whether a person is a business major or an agriculture major by the way the person dresses. Suppose in actuality that if someone is a business major, he can correctly identify that person as a business major 87% of the time. When a person is an agriculture major, the student will incorrectly identify that person as a business major 16% of the time. Presented with one person and asked to identify the major of this person (who is either a business or an agriculture major), he considers this to be a hypothesis test with the null hypothesis being that the person is a business major and the alternative that the person is an agriculture major.

Referring to the above, what would be a Type I error?

Saying that the person is a business major when in fact the person is an agriculture major

Saying that the person is a business major when in fact the person is a business major

Saying that the person is an agriculture major when in fact the person is a business major

Saying that the person is an agriculture major when in fact the person is an agriculture major

Answer 41

Saying that the person is an agriculture major when in fact the person is a business major

Question 42

An appliance manufacturer claims to have developed a compact microwave oven that consumes a mean of no more than 250 W. From previous studies, it is believed that power consumption for microwave ovens is normally distributed with a population standard deviation of 15 W. A consumer group has decided to try to discover if the claim appears true. They take a sample of 20 microwave ovens and find that they consume a mean of 257.3 W.

Referring to the above, the null hypothesis will be rejected at 5% level of significance.
True
OR
False

Answer 42

true

Question 43

A major home improvement store conducted its biggest brand recognition campaign in the company’s history. A series of new television advertisements featuring well-known entertainers and sports figures was launched. A key metric for the success of television advertisements is the proportion of viewers who “like the ads a lot.” A study of 1,189 adults who viewed the ads reported that 230 indicated that they “like the ads a lot.” The percentage of a typical television advertisement receiving the “like the ads a lot” score is believed to be 22%. Company officials wanted to know if there is evidence that the series of television advertisements are less successful than the typical ad (i.e. if there is evidence that the population proportion of “like the ads a lot” for the company’s ads is less than 0.22) at a 0.01 level of significance.

Referring to the above, the null hypothesis would be rejected.
True
OR
False

Answer 43

False

Question 44

The president of a university claimed that the entering class this year appeared to be larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year’s entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The university’s record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance.

Referring to the above, the null hypothesis would be rejected.
True
OR
False

Answer 44

False

Question 45

Assume a researcher is conducting an upper tail test with a sample size of 42 and a level of significance of 0.05. You know that the population standard deviation is not known. The researcher states that the Z score for this test is 2.3. What is the p-value?

=1 - T.DIST(2.3, 41, TRUE)

=T.INV(0.95, 41)

= -T.INV(0.05, 41)

=T.DIST(2.3, 41, TRUE)

=1 - T.INV(0.05, 41)

Answer 45

=1 - T.DIST(2.3, 41, TRUE)