Question 1 of 40 2.5 Points In the past, the mean running time for a certain type of flashlight batt

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 : µ = 8.0 hours
Ha : µ > 8.0 hours
Explain the meaning of a Type II
error.

A. Concluding that µ > 8.0
hours when in fact µ > 8.0 hours

B. Failing to reject the
hypothesis that µ = 8.0 hours when in fact µ >
8.0 hours

C. Concluding that µ > 8.0
hours

D. Failing to reject the
hypothesis that µ = 8.0 hours when in fact µ = 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, µ, 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 µ = 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 : µ = 9.8 hours
Ha : µ > 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 “in the shell
peanuts” 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 supplier’s 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: µ = 16
ounces Ha: µ < 16=”” ounces=”” b.=”” h0:=”” âµâ¹16=”” ounces=”” ha:=”” âµ=”16″ ounces=”” c.=”” h0:=”” âµ=”16″ ounces=”” ha:=”” âµ=””> 16 ounces

D.
H0: µ = 16
ounces
Ha:
µ¹
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 manufacturer’s claims. Determine the null and alternative hypotheses for
the test described.

A.
H0: µ = Manufacturer’s
claims Ha: µ < manufacturer’s=”” claims=”” b.=”” h0:=”” âµ=”Manufacturer’s” claims=”” ha:=”” âµâ¹manufacturer’s=”” claims=”” c.=”” h0:=”” âµ=”Manufacturer’s” claims=”” ha:=”” âµ=””> Manufacturer’s claims

D.
H0: µ¹Manufacturer’s
claims Ha: µ = Manufacturer’s 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: µ = 9.3
minutes Ha: µ < 9.3=”” minutes=”” b.=”” ho:=”” âµ=”9.3″ minutes=”” ha:=”” âµ=””> 9.3 minutes

C.
Ho: µ = 9.3
minutes Ha
:
µ¹
9.3
minutes

D.
Ho: µ¹9.3
minutes Ha: µ = 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: µ = 18.4
hours H
a: µ¹18.4 hours

B.
Ho: µ = 18.4 hours Ha: µ < 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=”” golfer’s=”” 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:
µ = 170

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

D.
H0: µ = 160; Ha:
µ > 160

Question 36 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

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

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

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=””>