1.A study of 420 comma 012420,012 cell phone users found that 138138 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found t

1.A study of 420 comma 012420,012 cell phone users found that 138138 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be

0.03160.0316% for those not using cell phones. Complete parts (a) and (b).

a. Use the sample data to construct a 9090% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system.

nothing %less than(Round to three decimal places as needed.)

b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? Why or why not?

A NoNo, because 0.03160.0316% isis included in the confidence interval.

B.YesYes, because 0.03160.0316% is notis not included in the confidence interval.

C.YesYes, because 0.03160.0316% isis included in the confidence interval.

D.NoNo, because 0.03160.0316% is notis not included in the confidence interval.

2. A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 350350 babies were born, and 315315 of them were girls. Use the sample data to construct a 9999%

confidence interval estimate of the percentage of girls born. Based on the result, does the method appear to be effective?

nothing less than< pless than

(Round to three decimal places as needed.)

Does the method appear to be effective?

YesYes, the proportion of girls isis significantly different from 0.5.

NoNo, the proportion of girls is notis not significantly different from 0.5.

3. A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before treatment, 2525 subjects had a mean wake time of 105.0105.0 min. After treatment, the 2525 subjects had a mean wake time of 83.583.5 min and a standard deviation of 44.844.8 min. Assume that the 2525 sample values appear to be from a normally distributed population and construct a 9090% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 105.0105.0 min before the treatment? Does the drug appear to be effective?Construct the 9090% confidence interval estimate of the mean wake time for a population with the treatment.

nothing minless thanmin

(Round to one decimal place as needed.)

What does the result suggest about the mean wake time of 105.0105.0

min before the treatment? Does the drug appear to be effective?

The confidence interval

â¼

does not include

includes the mean wake time of 105.0105.0 min before the treatment, so the means before and after the treatment

â¼

are different.

could be the same.

This result suggests that the drug treatment

â¼

does not have

has

a significant effect.

4. In a test of the effectiveness of garlic for lowering cholesterol, 4343 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes

(beforeminusâafter) in their levels of LDL cholesterol (in mg/dL) have a mean of 4.14.1 and a standard deviation of 17.517.5. Construct a 9090% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?

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What is the confidence interval estimate of the population mean

muμ?

nothing mg/dLless thanmg/dL

(Round to two decimal places as needed.)

What does the confidence interval suggest about the effectiveness of the treatment?

A.The confidence interval limits containcontain 0, suggesting that the garlic treatment

did notdid not affect the LDL cholesterol levels.

B.The confidence interval limits do not containdo not contain 0, suggesting that the garlic treatment

diddid affect the LDL cholesterol levels.

C.The confidence interval limits do not containdo not contain 0, suggesting that the garlic treatment did notdid not affect the LDL cholesterol levels.

D.The confidence interval limits containcontain 0, suggesting that the garlic treatment

diddid affect the LDL cholesterol levels.

5.The test statistic of zequals=2.722.72 is obtained when testing the claim that p not equalsâ 0.1350.135.

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value.

c. Using a significance level of alphaαequals=0.100.10, should we reject Upper H 0H0 or should we fail to reject Upper H 0H0?

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a. This is a

â¼

right-tailed

two-tailed

left-tailed

test.

b.

P-valueequals=nothing

(Round to three decimal places as needed.)

c. Choose the correct conclusion below.

A.Fail to rejectFail to reject Upper H 0H0. There isis sufficient evidence to support the claim that pnot equalsâ 0.1350.135.

B.RejectReject Upper H 0H0. There is notis not sufficient evidence to support the claim that

pnot equalsâ 0.1350.135.

C.Fail to rejectFail to reject Upper H 0H0. There is notis not sufficient evidence to support the claim that pnot equalsâ 0.1350.135.

D.RejectReject Upper H 0H0. There isis sufficient evidence to support the claim that pnot equalsâ 0.1350.135.

6. The test statistic of zequals=2.722.72 is obtained when testing the claim that pnot equalsâ 0.1350.135.

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value.

c. Using a significance level of alphaαequals=0.100.10, should we reject Upper H 0H0 or should we fail to reject Upper H 0H0?

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a. This is a

â¼

right-tailed

two-tailed

left-tailed

test.

b.

P-valueequals=nothing

(Round to three decimal places as needed.)

c. Choose the correct conclusion below.

A.

Fail to rejectFail to reject Upper H 0H0. There isis sufficient evidence to support the claim that pnot equalsâ 0.1350.135.

B.RejectReject Upper H 0H0. There is notis not sufficient evidence to support the claim that

pnot equalsâ 0.1350.135.

C.Fail to rejectFail to reject Upper H 0H0. There is notis not sufficient evidence to support the claim that pnot equalsâ 0.1350.135.

D RejectReject Upper H 0H0. There isis sufficient evidence to support the claim that

pnot equalsâ 0.1350.135.

7. A random sample of 874874 births included 429429 boys. Use a 0.010.01 significance level to test the claim that 50.750.7% of newborn babies are boys. Do the results support the belief that 50.750.7% of newborn babies are boys?Identify the null and alternative hypotheses for this test. Choose the correct answer below.

A.Upper H 0H0: pequals=0.5070.507Upper H 1H1:

pless than

B.Upper H 0H0: pequals=0.5070.507Upper H 1H1: pgreater than>0.5070.507

C.Upper H 0H0: pnot equalsâ 0.5070.507Upper H 1H1: pequals=0.5070.507

D.Upper H 0H0: pequals=0.5070.507Upper H 1H1: pnot equalsâ 0.5070.507

Identify the test statistic for this hypothesis test.

The test statistic for this hypothesis test is

nothing .

(Round to two decimal places as needed.)

Identify the P-value for this hypothesis test.

The P-value for this hypothesis test is

nothing .

(Round to three decimal places as needed.)

Identify the conclusion for this hypothesis test.

A.Fail to rejectFail to reject Upper H 0H0. There is notis not sufficient evidence to warrant rejection of the claim that 50.750.7% of newborn babies are boys.

B. RejectReject Upper H 0H0. There isis sufficient evidence to warrant rejection of the claim that 50.750.7% of newborn babies are boys.

C.RejectReject Upper H 0H0.

There is notis not sufficient evidence to warrant rejection of the claim that 50.750.7%

of newborn babies are boys.

D.Fail to rejectFail to reject Upper H 0H0. There isis sufficient evidence to warrant rejection of the claim that 50.750.7% of newborn babies are boys.Do the results support the belief that

50.750.7% of newborn babies are boys?

A.The results support the belief that 50.750.7% of newborn babies are boys because there was no evidence to show that the belief is untrue.

B.The results support the belief that 50.750.7% of newborn babies are boys because there was sufficient evidence to show that the belief is true.

C.The results do not support the belief that 50.750.7% of newborn babies are boys; the results merely show that there is not strong evidence against the rate of 50.750.7%.

D.The results do not support the belief that 50.750.7% of newborn babies are boys because there was sufficient evidence to show that the belief is untrue.

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8. A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are listed below. Use a 0.010.01 significance level to test the claim that these times are from a population with a mean equal to 60 seconds. Does it appear that students are reasonably good at estimating one minute?

76

88

50

72

48

33 66 73 72 54 72 76 101 76 What are the null and alternative hypotheses?

A.Upper H 0H0: muμ equals=6060 secondsUpper H 1H1: muμnot equalsâ 6060

seconds

B.Upper H 0H0: muμ equals=6060 secondsUpper H 1H1: muμless than

C.Upper H 0H0: muμnot equalsâ 6060 secondsUpper H 1H1: muμequals=6060 seconds

D.Upper H 0H0: muμequals=6060 seconds Upper H 1H1: muμgreater than>6060

seconds

Determine the test statistic.

nothing

(Round to two decimal places as needed.)

Determine the P-valuenothing

(Round to three decimal places as needed.)

State the final conclusion that addresses the original claim.

â¼

Reject

Fail to reject

Upper H 0H0.

There is

â¼

sufficient

not sufficient

evidence to conclude that the mean of the population of estimates

â¼

is not

is greater than

is less than

6060

seconds. It

â¼

appears

does not appear

that, as a group, the students are reasonably good at estimating one minute.

9. Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

A safety administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a

0.050.05

significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement?

717717âââââ

588588âââââ

10411041âââââ

555555âââââ

530530âââââ

529529

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+

What are the hypotheses?

A.Upper H 0H0: muμequals=10001000 hicUpper H 1H1: muμgreater than or equalsâ¥10001000 hic

B.Upper H 0H0: muμless than

C.Upper H 0H0: muμgreater than>10001000 hicUpper H 1H1: muμless thanIdentify the test statistic.

tequals=nothing

(Round to three decimal places as needed.)

Identify the P-value.

The P-value is

nothing .

(Round to four decimal places as needed.)

State the final conclusion that addresses the original claim.

â¼

Reject

Fail to reject

Upper H 0H0.

There is

â¼

Sufficient, insufficient evidence to support the claim that the sample is from a population with a mean less than 1000 hic.

What do the results suggest about the child booster seats meeting the specified requirement?

A.There is not strong evidence that the mean is less than 1000 hic, and one of the booster seats has a measurement that is greater than 1000 hic.

B.There is strong evidence that the mean is less than 1000 hic, but one of the booster seats has a measurement that is greater than 1000 hic.

C.The requirement is met since most sample measurements are less than 1000 hic.

D.

The results are inconclusive regarding whether one of the booster seats could have a measurement that is greater than 1000 hic.

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10. Rhino viruses typically cause common colds. In a test of the effectiveness of echinacea, 4040

of the 4747 subjects treated with echinacea developed rhinovirus infections. In a placebo group, 7575 of the 9090 subjects developed rhinovirus infections. Use a 0.010.01

significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts (a) through (c) below.

a. Test the claim using a hypothesis test.

Consider the first sample to be the sample of subjects treated with echinacea and the second sample to be the sample of subjects treated with a placebo. What are the null and alternative hypotheses for the hypothesis test?

A.Upper H 0H0: p 1p1less than or equalsâ¤p 2p2 Upper H 1H1:

p 1p1not equalsâ p 2p2

B.Upper H 0H0: p 1p1equals=p 2p2Upper H 1H1: p 1p1greater than>p 2p2

C.Upper H 0H0: p 1p1equals=p 2p2Upper H 1H1: p 1p1not equalsâ p 2p2

D.Upper H 0H0: p 1p1greater than or equalsâ¥p 2p2Upper H 1H1: p 1p1not equalsâ p 2p2

E.Upper H 0H0: p 1p1not equalsâ p 2p2Upper H 1H1: p 1p1equals=p 2p2

F.Upper H 0H0: p 1p1equals=p 2p2Upper H 1H1: p 1p1less than

Identify the test statistic.

zequals=nothing

(Round to two decimal places as needed.)

Identify the P-value.

P-valueequals=nothing

(Round to three decimal places as needed.)

What is the conclusion based on the hypothesis test?

The P-value is

â¼

less than

greater than

the significance level of

alphaαequals=0.010.01,

so

â¼

reject

fail to reject

the null hypothesis. There

â¼

is not

is

sufficient evidence to support the claim that echinacea treatment has an effect.

b. Test the claim by constructing an appropriate confidence interval.

The

9999%

confidence interval is

nothing less than(Round to three decimal places as needed.)

What is the conclusion based on the confidence interval?

Because the confidence interval limits

â¼

include

do not include

0, there

â¼

does

does not

appear to be a significant difference between the two proportions. There

â¼

is

is not

evidence to support the claim that echinacea treatment has an effect.

c. Based on the results, does echinacea appear to have any effect on the infection rate?

A.

Echinacea does appear to have a significant effect on the infection rate. There is evidence that it lowers the infection rate.

B.

Echinacea does appear to have a significant effect on the infection rate. There is evidence that it increases the infection rate.

C.

Echinacea does not appear to have a significant effect on the infection rate.

D.

The results are inconclusive.

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11A person's body mass index (BMI) is computed by dividing the weight (kg) by the square of height (m). The accompanying table contains the BMI statistics for random samples of males and females. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Let population 1 be females. Complete parts (a) through (c) below.

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Click the icon to view the table of statistics.

a. Use a

0.100.10

significance level to test the claim that females and males have the same mean BMI.

Identify the null and alternative hypotheses.

A.

Upper H 0H0:

mu 1μ1equals=mu 2μ2

Upper H 1H1:

mu 1μ1less than

B.

Upper H 0H0:

mu 1μ1equals=mu 2μ2

Upper H 1H1:

mu 1μ1not equalsâ mu 2μ2

C.

Upper H 0H0:

mu 1μ1equals=mu 2μ2

Upper H 1H1:

mu 1μ1greater than>mu 2μ2

D.

Upper H 0H0:

mu 1μ1greater than>mu 2μ2

Upper H 1H1:

mu 1μ1equals=mu 2μ2

E.

Upper H 0H0:

mu 1μ1not equalsâ mu 2μ2

Upper H 1H1:

mu 1μ1equals=mu 2μ2

F.

Upper H 0H0:

mu 1μ1less than

Upper H 1H1:

mu 1μ1equals=mu 2μ2

The test statistic is

nothing .

(Round to two decimal places as needed.)

The P-value is

nothing .

(Round to three decimal places as needed.)

State the conclusion for the test.

â¼

Fail to reject

Reject

the null hypothesis. There

â¼

is not

is

sufficient evidence to support the claim that females and males have the same mean BMI.

b. Construct a confidence interval appropriate for testing the claim in part (a).

The

nothing %

confidence interval estimate is

nothing less than(Round to two decimal places as needed.)

c. Do females and males appear to have the same mean BMI?

Since

nothing

â¼

is not

is

contained within the confidence interval, there

â¼

does not appear

appears

to be evidence that females and males do not have the same mean BMI.

(Type an integer or a decimal. Do not round.)

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12When subjects were treated with a drug, their systolic blood pressure readings (in mm Hg) were measured before and after the drug was taken. Results are given in the table below. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Using a

0.050.05

significance level, is there sufficient evidence to support the claim that the drug is effective in lowering systolic blood pressure?

Before

192192

157157

183183

205205

179179

210210

167167

188188

166166

158158

169169

164164

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After

149149

152152

144144

143143

147147

179179

177177

179179

182182

159159

186186

148148

In this example,

mu Subscript dμd

is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the systolic blood pressure reading before the drug was taken minus the reading after the drug was taken. What are the null and alternative hypotheses for the hypothesis test?

A.

Upper H 0H0:

mu Subscript dμd equals=0

Upper H 1H1:

mu Subscript dμd greater than>0

B.

Upper H 0H0:

mu Subscript dμd not equalsâ 0

Upper H 1H1:

mu Subscript dμd greater than>0

C.

Upper H 0H0:

mu Subscript dμd equals=0

Upper H 1H1:

mu Subscript dμd less than

D.

Upper H 0H0:

mu Subscript dμd not equalsâ 0

Upper H 1H1:

mu Subscript dμd equals=0

Identify the test statistic. T equals=nothing

(Round to two decimal places as needed.)

Identify the P-value.

P-value equals=nothing

(Round to three decimal places as needed.)

Since the P-value is

â¼

greater

less

than the significance level,

â¼

reject

fail to reject

Upper H 0H0.

There is

â¼

insufficient

sufficient

evidence to support the claim that the drug is effective in lowering systolic blood pressure.