Question 1 of 40

2.5

Points

In the past, the mean running time

for a certain type of flashlight battery has been 8.0 hours. The manufacturer

has introduced a change in the production method and wants to perform a

hypothesis test to determine whether the mean running time has increased as a

result. The hypotheses are:

H0 : Âµ = 8.0 hours

Ha : Âµ > 8.0 hours

Explain the meaning of a Type II

error.

A. Concluding that Âµ > 8.0

hours when in fact Âµ > 8.0 hours

B. Failing to reject the

hypothesis that Âµ = 8.0 hours when in fact Âµ >

8.0 hours

C. Concluding that Âµ > 8.0

hours

D. Failing to reject the

hypothesis that Âµ = 8.0 hours when in fact Âµ = 8.0 hours

Question 2 of 40

2.5

Points

.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%2022.JPG”>

without computing a P-value,

determine whether the alternate hypothesis is supported and give a reason for

your conclusion.

A.

.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%2022%20_A_.JPG”>is less than 1 standard deviation above the claimed mean.

B.

.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%2022%20_A_.JPG”>is more than 4 standard deviations above the claimed mean.

C.

.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%2022%20_C_.JPG”>is less than 1 standard deviation above the claimed mean.

D.

is more than 4 standard deviations above the claimed mean.

Question 3 of 40

2.5

Points

A poll of 1,068 adult Americans

reveals that 52% of the voters surveyed prefer the Democratic candidate for the

presidency. At the 0.05 significance level, test the claim that more than half

of all voters prefer the Democrat.

A. Reject the null hypothesis.

Conclude that there is insufficient evidence that more than half of all

voters prefer Democrats.

B. Do not reject the null

hypothesis. Conclude that there is sufficient evidence that more than half of

all voters prefer Democrats.

C. Reject the null hypothesis.

Conclude that there is sufficient evidence that more than half of all voters

prefer Democrats.

D. Do not reject the null

hypothesis. Conclude that there is insufficient evidence that more than half

of all voters prefer Democrats.

Question 4 of 40

2.5

Points

A manufacturer claims that the mean

amount of juice in its 16 ounce bottles is 16.1 ounces. A consumer advocacy

group wants to perform a hypothesis test to determine whether the mean amount

is actually less than this. The mean volume of juice for a random sample of 70

bottles was 15.94 ounces. Do the data provide sufficient evidence to conclude

that the mean amount of juice for all 16-ounce bottles, Âµ, is less than 16.1

ounces? Perform the appropriate hypothesis test using a significance level of

0.10. Assume thats = 0.9 ounces. ?

A.

The z of-1.49

provides sufficient evidence to conclude that the mean amount of juice is

less than 16.1 oz.

B.

The z of-1.49 does not provide sufficient evidence to conclude that the

mean amount of juice is less than 16.1 oz.

C.

The z of-0.1778 does not provide sufficient evidence to conclude that the

mean amount of juice is less than 16.1 oz.

D.

The z of-0.1778 provides sufficient evidence to conclude that the mean

amount of juice is less than 16.1 oz.

Question 5 of 40

2.5

Points

A researcher wants to check

the claim that convicted burglars spend an average of 18.7 months in jail. She

takes a random sample of 35 such cases from court files and finds that.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%2029.JPG”> months.

Assume that the population standard deviation is 7 months. Test the null

hypothesis that Âµ = 18.7 at the 0.05 significance level.

A.

Do not reject the null hypothesis

and conclude that the claim that the mean is different from 18.7 months is

supported.

B.

Do not reject the null hypothesis

and conclude that the claim that the mean is different from 18.7 months

cannot be supported.

C.

Reject the null hypothesis and

conclude that the claim that the mean is different from 18.7 months is

supported.

D.

Reject the null hypothesis and

conclude that the claim that the mean is different from 18.7 months cannot be

supported.

Question 6 of 40

2.5

Points

In the past, the mean running time

for a certain type of flashlight battery has been 9.8 hours. The manufacturer

has introduced a change in the production method and wants to perform a

hypothesis test to determine whether the mean running time has increased as a

result. The hypotheses are:

H0 : Âµ = 9.8 hours

Ha : Âµ > 9.8 hours

Suppose that the results of the

sampling lead to rejection of the null hypothesis. Classify that conclusion as

a Type I error, a Type II error, or a correct decision, if in fact the mean

running time has not increased.

A. Type I error

B. Type II error

C. Correct decision

D. Can not be determined from this

information

Question 7 of 40

2.5

Points

A two-tailed test is conducted at

the 0.10 significance level. What is the P-value required to reject the null

hypothesis?

A. Greater than or equal to .010

B. Greater than or equal to 0.05

C. Less than or equal to 0.10

D. Less than or equal to 0.05

Question 8 of 40

2.5

Points

A study of a brand of âin the shell

peanutsâ gives the following results:

.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%2017.JPG”>

A significant event at the 0.01

level is a fan getting a bag with how many peanuts?

A. 30 peanuts

B. 25 or 30 peanuts

C. 25 or 55 peanuts

D. 25 peanuts

Question 9 of 40

2.5

Points

.jpg” alt=”https://study.ashworthcollege.edu/access/content/group/8ef8b2f7-197d-41de-a4c4-db81a717c013/v9/Images/Lesson%207%20exam/MA260%20Lesson%207%20exam%20question%202.JPG”>

If a fan purchased a bag with 30

peanuts, what is the lowest level at which this would be a significant event?

A. 0.05

B. 0.025

C. 0.01

D. It is not significant at any of

the levels given

Question 10 of 40

2.5

Points

A researcher claims that the amounts

of acetaminophen in a certain brand of cold tablets have a mean different from

the 600 mg claimed by the manufacturer. Test this claim at the 0.02 level of

significance. The mean acetaminophen content for a random sample of n = 41

tablets is 603.3 mg. Assume that the population standard deviation is 4.9 mg.

A. Since the test statistic is

greater than the critical z, there is sufficient evidence to accept the null

hypothesis and to support the claim that the mean content of acetaminophen is

600 mg.

B. Since the test statistic is

greater than the critical z, there is sufficient evidence to reject the null

hypothesis and to support the claim that the mean content of acetaminophen is

not 600 mg.

C. Since the test statistic is

less than the critical z, there is sufficient evidence to reject the null

hypothesis and to support the claim that the mean content of acetaminophen is

not 600 mg.

D. Since the test statistic is

greater than the critical z, there is insufficient evidence to reject the

null hypothesis and to support the claim that the mean content of

acetaminophen is not 600 mg.

Question 11 of 40

2.5

Points

A right-tailed test is conducted at

the 5% significance level. Which of the following z-scores is the smallest one

in absolute value that leads to rejection of the null hypothesis? ?

A. 1.61

B. 1.85

C. -1.98

D. -2.06

Question 12 of 40

2.5

Points

A supplier of DVDs claims that no

more than 1% of the DVDs are defective. In a random sample of 600 DVDs, it is

found that 3% are defective, but the supplier claims that this is only a sample

fluctuation. At the 0.01 level of significance, test the supplierâs claim that

no more than 1% are defective.

A. Do not reject the null

hypothesis and conclude that there is evidence to support the claim that more

than 1% of the DVDs are defective.

B. Reject the null hypothesis and

conclude that there is insufficient evidence to support the claim that more

than 1% of the DVDs are defective.

C. Do not reject the null

hypothesis and conclude that there is insufficient evidence to support the

claim that more than 1% of the DVDs are defective.

D. Reject the null hypothesis and

conclude that there is sufficient evidence to support the claim that more

than 1% of the DVDs are defective.

Question 13 of 40

2.5

Points

A consumer advocacy group claims

that the mean amount of juice in a 16

ounce bottled drink is not 16 ounces, as stated by the bottler.

Determine the null and alternative hypotheses for the test described.

A.

H0: Âµ = 16

ounces Ha: Âµ < 16=”” ounces=”” b.=”” h0:=”” âµâ¹16=”” ounces=”” ha:=”” âµ=”16″ ounces=”” c.=”” h0:=”” âµ=”16″ ounces=”” ha:=”” âµ=””> 16 ounces

D.

H0: Âµ = 16

ounces

Ha:

ÂµÂ¹

16

ounces

Question 14 of 40

2.5

Points

A consumer group claims that the

mean running time for a certain type of flashlight battery is not the same as

the manufacturerâs claims. Determine the null and alternative hypotheses for

the test described.

A.

H0: Âµ = Manufacturerâs

claims Ha: Âµ < manufacturerâs=”” claims=”” b.=”” h0:=”” âµ=”Manufacturerâs” claims=”” ha:=”” âµâ¹manufacturerâs=”” claims=”” c.=”” h0:=”” âµ=”Manufacturerâs” claims=”” ha:=”” âµ=””> Manufacturerâs claims

D.

H0: ÂµÂ¹Manufacturerâs

claims Ha: Âµ = Manufacturerâs claims

Question 15 of 40

2.5

Points

A two-tailed test is conducted at

the 5% significance level. Which of the z-scores below is the smallest one that

leads to rejection of the null hypothesis?

A. 1.12

B. 1.48

C. 1.84

D. 2.15

Question 16 of 40

2.5

Points

A nationwide study of American

homeowners revealed that 65% have one or more lawn mowers. A lawn equipment

manufacturer, located in Omaha, feels the estimate is too low for households in

Omaha. Find the P-value for a test of the claim that the proportion with lawn

mowers in Omaha is higher than 65%. Among 497 randomly selected homes in Omaha,

340 had one or more lawn mowers. Use Table 5.1 to find the best answer.

A. 0.0559

B. 0.1118

C. 0.0252

D. 0.0505

Question 17 of 40

2.5

Points

In 1990, the average duration of

long-distance telephone calls originating in one town was 9.3 minutes. A

long-distance telephone company wants to perform a hypothesis test to determine

whether the average duration of long-distance phone calls has changed from the

1990 mean of 9.3 minutes. Formulate the null and alternative hypotheses for the

study described.

A.

Ho: Âµ = 9.3

minutes Ha: Âµ < 9.3=”” minutes=”” b.=”” ho:=”” âµ=”9.3″ minutes=”” ha:=”” âµ=””> 9.3 minutes

C.

Ho: Âµ = 9.3

minutes Ha

:

ÂµÂ¹

9.3

minutes

D.

Ho: ÂµÂ¹9.3

minutes Ha: Âµ = 9.3 minutes

Question 18 of 40

2.5

Points

A long-distance telephone company

claims that the mean duration of long-distance telephone calls originating in

one town was greater than 9.4 minutes, which is the average for the state.

Determine the conclusion of the hypothesis test assuming that the results of

the sampling do not lead to rejection of the null hypothesis.

A. Conclusion: Support the claim

that the mean is less than 9.4 minutes.

B. Conclusion: Support the claim

that the mean is greater than 9.4 minutes.

C. Conclusion: Support the claim

that the mean is equal to 9.4 minutes.

D. Conclusion: Do not support the

claim that the mean is greater than 9.4 minutes.

Question 19 of 40

2.5

Points

A consumer advocacy group claims

that the mean amount of juice in a 16 ounce bottled drink is not 16 ounces, as

stated by the bottler. Determine the conclusion of the hypothesis test assuming

that the results of the sampling lead to rejection of the null hypothesis.

A. Conclusion: Support the claim

that the mean is equal to 16 ounces.

B. Conclusion: Support the claim

that the mean is greater than 16 ounces.

C. Conclusion: Support the claim

that the mean is not equal to 16 ounces.

D. Conclusion: Support the claim

that the mean is less than 16 ounces.

Question 20 of 40

2.5

Points

At one school, the mean amount of

time that tenth-graders spend watching television each week is 18.4 hours. The

principal introduces a campaign to encourage the students to watch less

television. One year later, the principal wants to perform a hypothesis test to

determine whether the average amount of time spent watching television per week

has decreased.

Formulate the null and alternative hypotheses for the study described.

A.

Ho: Âµ = 18.4

hours H

a: ÂµÂ¹18.4 hours

B.

Ho: Âµ = 18.4 hours Ha: Âµ < 18.4=”” hours=”” c.=”” ho:=”” âµâ³18.4=”” hours=”” h=”” a:=”” âµ=””>< 18.4=”” hours=”” d.=”” ho:=”” âµ=”18.4″ hours=”” h=”” a:=”” âµ=””> 18.4 hours

Question 21 of 40

2.5

Points

A large test statistic F tells us

that the sample means __________ the data within the individual samples, which

would be unlikely if the populations means really were equal (as the null

hypothesis claims).

A. differ more than

B. differ less than

C. are equal to

D. do not vary with

Question 22 of 40

2.5

Points

One hundred people are selected at

random and tested for colorblindness to determine whether gender and

colorblindness are independent. The following counts were observed.

Colorblind

Not

Colorblind

Total

Male

8

52

60

Female

2

38

40

Total

10

90

100

Find the value of the X2statistic

for the data above.

A. 1.463

B. 1.852

C. 1.947

D. 1.949

Question 23 of 40

2.5

Points

A simple random sample from a normal

distribution is taken in order to obtain a 95% confidence interval for the

population mean. If the sample size is 8, the sample mean x? is 22, and

the sample standard deviation sis 6.3, what is the margin of error? Show

your answer to 2 decimal places.

A. df = 7; E = 3.3445.38 = 5.6566

B. df = 8; E = 3.3445.38 = 5.6566

C. df = 6; E = 2.3656.38 = 5.769

D. df = 7; E = 2.3656.38 = 5.869

Question 24 of 40

2.5

Points

One hundred people are selected at random

and tested for colorblindness to determine whether gender and colorblindness

are independent.

The critical value of X2

for a 2 x 2 table using a 0.05 significance level is 3.841. If the value of the

X2 statistic is 3.179, state your conclusion about the

relationship between gender and colorblindness.

A.

Do not reject H0.

B.

Reject H0.

C.

There is sufficient evidence to

support the claim that gender and colorblindness are not related.

D.

There is not sufficient evidence

to accept or reject H0.

Question 25 of 40

2.5

Points

One hundred people are selected at

random and tested for colorblindness to determine whether gender and

colorblindness are independent. The following counts were observed.

Colorblind

Not

Colorblind

Total

Male

8

52

60

Female

2

38

40

Total

10

90

100

State the null and alternative

hypothesis for the test associated with this data.

A.

H0: Colorblindness and

gender are dependent characteristics.

Ha: Colorblindness and gender are not related in any way.

B.

H0:

Colorblindness and gender are dependent characteristics.

Ha: Colorblindness and gender are related in some way.

C.

H0: Colorblindness and

gender are independent characteristics.

Ha: Colorblindness and gender are not related in any way.

D.

H0: Colorblindness and

gender are independent characteristics.

Ha: Colorblindness and gender are related in some way.

Question 26 of 40

2.5

Points

One hundred people are selected at

random and tested for colorblindness to determine whether gender and

colorblindness are independent. The following counts were observed.

Colorblind

Not

Colorblind

Total

Male

7

53

60

Female

1

39

40

Total

8

92

100

Find the value of the X2

statistic for the data above.

A. 1.325

B. 1.318

C. 1.286

D. 1.264

Question 27 of 40

2.5

Points

A 95% confidence interval for the

mean of a normal population is found to be 13.2 < âµ=””>< 22.4.=”” what=”” is=”” the=”” margin=”” of=”” error?=”” a.=”” 4.6=”” b.=”” 4.4=”” c.=”” 4.2=”” d.=”” 5.6=”” question=”” 28=”” of=”” 40=”” 2.5=”” points=”” the=”” following=”” data=”” were=”” analyzed=”” using=”” one-way=”” analysis=”” of=”” variance.=”” a=”” b=”” c=”” 34=”” 27=”” 19=”” 26=”” 23=”” 21=”” 31=”” 29=”” 22=”” 28=”” 21=”” 12=”” which=”” one=”” of=”” the=”” following=”” statements=”” is=”” correct?=”” a.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” determine=”” whether=”” the=”” groups=”” a,=”” b,=”” and=”” c=”” are=”” independent.=”” b.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” test=”” the=”” hypothesis=”” that=”” the=”” population=”” means=”” of=”” the=”” three=”” groups=”” are=”” equal.=”” c.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” test=”” the=”” hypothesis=”” that=”” the=”” population=”” variances=”” of=”” the=”” three=”” groups=”” are=”” equal.=”” d.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” test=”” the=”” hypothesis=”” that=”” the=”” sample=”” means=”” of=”” the=”” three=”” groups=”” are=”” equal.=”” question=”” 29=”” of=”” 40=”” 2.5=”” points=”” a=”” 95%=”” confidence=”” interval=”” for=”” the=”” mean=”” of=”” a=”” normal=”” population=”” is=”” found=”” to=”” be=”” 15.6=””>< âµ=””>< 25.2.=”” what=”” is=”” the=”” margin=”” of=”” error?=”” a.=”” 3.9=”” b.=”” 4.8=”” c.=”” 4.9=”” d.=”” 3.7=”” question=”” 30=”” of=”” 40=”” 2.5=”” points=”” one=”” hundred=”” people=”” are=”” selected=”” at=”” random=”” and=”” tested=”” for=”” colorblindness=”” to=”” determine=”” whether=”” gender=”” and=”” colorblindness=”” are=”” independent.=”” the=”” critical=”” value=”” of=”” x2=”” for=”” a=”” 2=”” x=”” 2=”” table=”” using=”” a=”” 0.05=”” significance=”” level=”” is=”” 3.841.=”” if=”” the=”” value=”” of=”” the=”” x2=”” statistic=”” is=”” 3.427,=”” state=”” your=”” conclusion=”” about=”” the=”” relationship=”” between=”” gender=”” and=”” colorblindness.=”” a.=”” do=”” not=”” reject=”” h0.=”” there=”” is=”” not=”” sufficient=”” evidence=”” to=”” support=”” the=”” claim=”” that=”” gender=”” and=”” colorblindness=”” are=”” related.=”” b.=”” do=”” not=”” reject=”” h0.=”” there=”” is=”” sufficient=”” evidence=”” to=”” support=”” the=”” claim=”” that=”” gender=”” and=”” colorblindness=”” are=”” related.=”” c.=”” reject=”” h0.=”” there=”” is=”” not=”” sufficient=”” evidence=”” to=”” support=”” the=”” claim=”” that=”” gender=”” and=”” colorblindness=”” are=”” related.=”” d.=”” reject=”” h0.=”” there=”” is=”” sufficient=”” evidence=”” to=”” support=”” the=”” claim=”” that=”” gender=”” and=”” colorblindness=”” are=”” related.=”” question=”” 31=”” of=”” 40=”” 2.5=”” points=”” the=”” following=”” data=”” were=”” analyzed=”” using=”” one-way=”” analysis=”” of=”” variance.=”” a=”” b=”” c=”” 34=”” 27=”” 19=”” 26=”” 23=”” 31=”” 31=”” 29=”” 22=”” 28=”” 21=”” 22=”” which=”” one=”” of=”” the=”” following=”” statements=”” is=”” correct?=”” a.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” determine=”” whether=”” the=”” groups=”” a,=”” b,=”” and=”” c=”” are=”” independent.=”” b.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” test=”” the=”” hypothesis=”” that=”” the=”” population=”” means=”” of=”” the=”” three=”” groups=”” are=”” equal.=”” c.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” test=”” the=”” hypothesis=”” that=”” the=”” population=”” variances=”” of=”” the=”” three=”” groups=”” are=”” equal.=”” d.=”” the=”” purpose=”” of=”” the=”” analysis=”” is=”” to=”” test=”” the=”” hypothesis=”” that=”” the=”” sample=”” means=”” of=”” the=”” three=”” groups=”” are=”” equal.=”” question=”” 32=”” of=”” 40=”” 2.5=”” points=”” which=”” of=”” the=”” following=”” statements=”” is=”” true?=”” a.=”” the=”” p=”” distribution=”” cannot=”” be=”” used=”” when=”” finding=”” a=”” confidence=”” interval=”” for=”” the=”” population=”” mean=”” with=”” a=”” small=”” sample=”” anytime=”” the=”” population=”” standard=”” deviation=”” is=”” unknown.=”” b.=”” the=”” t=”” distribution=”” can=”” be=”” used=”” when=”” finding=”” a=”” confidence=”” interval=”” for=”” the=”” population=”” mean=”” with=”” a=”” small=”” sample=”” anytime=”” the=”” population=”” standard=”” deviation=”” is=”” unknown.=”” c.=”” the=”” t=”” distribution=”” cannot=”” be=”” used=”” when=”” finding=”” a=”” confidence=”” interval=”” for=”” the=”” population=”” mean=”” with=”” a=”” small=”” sample=”” anytime=”” the=”” population=”” standard=”” deviation=”” is=”” unknown.=”” d.=”” the=”” p=”” distribution=”” can=”” be=”” used=”” when=”” finding=”” a=”” confidence=”” interval=”” for=”” the=”” population=”” mean=”” with=”” a=”” small=”” sample=”” anytime=”” the=”” population=”” standard=”” deviation=”” is=”” unknown.=”” question=”” 33=”” of=”” 40=”” 2.5=”” points=”” one=”” hundred=”” people=”” are=”” selected=”” at=”” random=”” and=”” tested=”” for=”” colorblindness=”” to=”” determine=”” whether=”” gender=”” and=”” colorblindness=”” are=”” independent.=”” the=”” following=”” counts=”” were=”” observed.=”” colorblind=”” not=”” colorblind=”” total=”” male=”” 8=”” 52=”” 60=”” female=”” 2=”” 38=”” 40=”” total=”” 10=”” 90=”” 100=”” if=”” gender=”” and=”” colorblindness=”” are=”” independent,=”” find=”” the=”” expected=”” values=”” corresponding=”” to=”” the=”” four=”” combinations=”” of=”” gender=”” and=”” colorblindness,=”” and=”” enter=”” them=”” in=”” the=”” following=”” table=”” along=”” with=”” row=”” and=”” column=”” totals.=”” colorblind=”” not=”” colorblind=”” total=”” male=”” female=”” total=”” a.=”” male=”” colorblind=”” 6.0;=”” male=”” not=”” colorblind=”” 54.0=”” b.=”” male=”” colorblind=”” 7.0;=”” male=”” not=”” colorblind=”” 53.0=”” c.=”” male=”” colorblind=”” 8.0;=”” male=”” not=”” colorblind=”” 52.0=”” d.=”” male=”” colorblind=”” 6.0;=”” male=”” not=”” colorblind=”” 53.0=”” question=”” 34=”” of=”” 40=”” 2.5=”” points=”” a=”” golfer=”” wished=”” to=”” find=”” a=”” ball=”” that=”” would=”” travel=”” more=”” than=”” 180=”” yards=”” when=”” hit=”” with=”” his=”” 5-iron=”” with=”” a=”” club=”” speed=”” of=”” 90=”” miles=”” per=”” hour.=”” he=”” had=”” a=”” golf=”” equipment=”” lab=”” test=”” a=”” low=”” compression=”” ball=”” by=”” having=”” a=”” robot=”” swing=”” his=”” club=”” 7=”” times=”” at=”” the=”” required=”” speed.=”” data=”” from=”” this=”” test=”” resulted=”” in=”” a=”” sample=”” mean=”” of=”” 184.2=”” yards=”” and=”” a=”” sample=”” standard=”” deviation=”” of=”” 5.8=”” yards.=”” assuming=”” normality,=”” carry=”” out=”” a=”” hypothesis=”” test=”” at=”” the=”” 0.05=”” significance=”” level=”” to=”” determine=”” whether=”” the=”” ball=”” meets=”” the=”” golferâs=”” requirements.=”” use=”” the=”” partial=”” t-table=”” below.=”” area=”” in=”” one=”” tail=”” 0.025=”” 0.05=”” area=”” in=”” two=”” tails=”” degrees=”” of=”” freedom=”” n=”” -=”” 1=”” 0.05=”” 0.10=”” 6=”” 2.447=”” 1.943=”” 7=”” 2.365=”” 1.895=”” 8=”” 2.306=”” 1.860=”” 9=”” 2.262=”” 1.833=”” a.=”” reject=”” the=”” null=”” hypothesis.=”” the=”” data=”” do=”” not=”” provide=”” sufficient=”” evidence=”” that=”” the=”” average=”” distance=”” is=”” greater=”” than=”” 180=”” yards.=”” b.=”” reject=”” the=”” null=”” hypothesis.=”” the=”” data=”” do=”” provide=”” sufficient=”” evidence=”” that=”” the=”” average=”” distance=”” is=”” greater=”” than=”” 180=”” yards.=”” c.=”” do=”” not=”” reject=”” the=”” null=”” hypothesis.=”” the=”” data=”” do=”” provide=”” sufficient=”” evidence=”” that=”” the=”” average=”” distance=”” is=”” greater=”” than=”” 180=”” yards.=”” d.=”” do=”” not=”” reject=”” the=”” null=”” hypothesis.=”” the=”” data=”” do=”” not=”” provide=”” sufficient=”” evidence=”” that=”” the=”” average=”” distance=”” is=”” greater=”” than=”” 180=”” yards.=”” question=”” 35=”” of=”” 40=”” 2.5=”” points=”” a=”” golfer=”” wished=”” to=”” find=”” a=”” ball=”” that=”” would=”” travel=”” more=”” than=”” 170=”” yards=”” when=”” hit=”” with=”” his=”” 6-iron=”” with=”” a=”” club=”” head=”” speed=”” of=”” 90=”” miles=”” per=”” hour.=”” he=”” had=”” a=”” golf=”” equipment=”” lab=”” test=”” a=”” low=”” compression=”” ball=”” by=”” having=”” a=”” robot=”” swing=”” his=”” club=”” 12=”” times=”” at=”” the=”” required=”” speed.=”” state=”” the=”” null=”” and=”” alternative=”” hypotheses=”” for=”” this=”” test.=”” a.=”” h0:=”” âµ=””> 170; Ha:

Âµ = 170

B.

H0: Âµ < 170;=”” ha:=”” âµ=”170″ c.=”” h0:=”” âµ=”170;” ha:=”” âµ=””> 170

D.

H0: Âµ = 160; Ha:

Âµ > 160

Question 36 of 40

2.5

Points

random and tested for colorblindness to determine whether gender and

colorblindness are independent. The following counts were observed.

Colorblind

Not

Colorblind

Total

Male

7

53

60

Female

1

39

40

Total

8

92

100

If gender and colorblindness are

independent, find the expected values corresponding to the male combinations of

gender and colorblindness.

A. Colorblind Male 4.8; Not

Colorblind Male 55.2

B. Colorblind Male 6.8; Not

Colorblind Male 53.2

C. Colorblind Male 4.8; Not

Colorblind Male 55.4

D. Colorblind Male 4.8; Not

Colorblind Male 56.2

Question 37 of 40

2.5

Points

The margin of error in estimating

the population mean of a normal population is E = 9.3 when the sample size is

15. If the sample size had been 25 and the sample standard deviation did not

change, would the margin of error be larger or smaller than 9.3?

A. Smaller. E increases as the

square root of the sample size gets larger.

B. Smaller. E decreases as the

square root of the sample size gets larger.

C. Larger. E decreases as the

square root of the sample size gets larger.

D. Larger. E increases as the

square root of the sample size gets larger.

Question 38 of 40

2.5

Points

Which of the following statements is

true?

A.

The t distribution can be used

when finding a confidence interval for the population mean whenever the

sample size is small.

B. The p distribution can be used

when finding a confidence interval for the population mean whenever the

sample size is small.

C. The t distribution cannot be

used when finding a confidence interval for the population mean whenever the

sample size is small.

D. The p distribution cannot be

used when finding a confidence interval for the sample mean whenever the

sample size is small.

Question 39 of 40

2.5

Points

random and tested for colorblindness to determine whether gender and

colorblindness are independent. The following counts were observed.

Colorblind

Not

Colorblind

Total

Male

7

53

60

Female

1

39

40

Total

8

92

100

If gender and colorblindness are

independent, find the expected values corresponding to the female combinations

of gender and colorblindness.

A. Colorblind Female 4.8; Not

Colorblind Female 55.2

B. Colorblind Female 3.2; Not

Colorblind Female 36.8

C. Colorblind Female 4.8; Not

Colorblind Female 35.2

D. Colorblind Female 3.8; Not

Colorblind Female 36.2

Question 40 of 40

2.5

Points

A 95% confidence interval for the

mean of a normal population is found to be 15.6 < âµ=””>< 24.8.=”” what=”” is=”” the=”” margin=”” of=”” error?=”” a.=”” 4.4=”” b.=”” 4.6=”” c.=”” 4.8=”” d.=”” 5.0=””>