Written Hypothesis Practice End Solutions

Exam

Name___________________________________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

A hypothesis test is to be performed. Determine the null and alternative hypotheses.1) In the past, the mean running time for a certain type of flashlight battery has been 9.0 hours. Themanufacturer has introduced a change in the production method and wants to perform ahypothesis test to determine whether the mean running time has changed as a result.A) H0 : µ = 9.0 hours

Ha : µ ≠ 9.0 hoursB) H0 : µ ≥ 9.0 hoursHa : µ = 9.0 hours

C) H0 : µ ≠ 9.0 hoursHa : µ = 9.0 hours

D) H0 : µ = 9.0 hoursHa : µ > 9.0 hours

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2) The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.6 partsper million (ppm). A consumer health group measured the level of the chemical in a randomsample of tomatoes obtained from one producer to determine whether the mean level of thechemical in these tomatoes exceeds the recommended limit.A) H0 : µ = 0.6 ppm

Ha : µ > 0.6 ppmB) H0 : µ > 0.6 ppmHa : µ = 0.6 ppm

C) H0 : µ < 0.6 ppmHa : µ > 0.6 ppm

D) H0 : µ = 0.6 ppmHa : µ ≥ 0.6 ppm

2)

3) A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.1 ounces. Aconsumer advocacy group wants to perform a hypothesis test to determine whether the meanamount is actually less than this.A) H0 : µ = 16.1 ounces

Ha : µ < 16.1 ouncesB) H0 : µ < 16.1 ouncesHa : µ = 16.1 ounces

C) H0 : µ > 16.1 ouncesHa : µ < 16.1 ounces

D) H0 : µ = 16.1 ouncesHa : µ ≤ 16.1 ounces

3)

Classify the hypothesis test as two-tailed, left-tailed, or right-tailed.

4) The recommended dietary allowance (RDA) of vitamin C for women is 75 milligrams per day. Ahypothesis test is to be performed to decide whether adult women are, on average, getting lessthan the RDA of 75 milligrams per day.A) Two-tailed B) Left-tailed C) Right-tailed

4)

5) The mean credit card debt among households in one state is $8400. A hypothesis test is to beperformed to decide whether the mean credit card debt for households in the formerly affluenttown of Rich-No-More differs from the mean credit card debt for the state.A) Two-tailed B) Left-tailed C) Right-tailed

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6) A psychologist has designed a test to measure stress levels in adults. She has determined thatnationwide the mean score on her test is 27. A hypothesis test is to be conducted to determinewhether the mean score for trial lawyers exceeds the national mean score.A) Two-tailed B) Left-tailed C) Right-tailed

6)

For the given hypothesis test, explain the meaning of a Type I error, a Type II error, or a correct decision as specified.7) At one school, in 2005, the average amount of time that tenth-graders spent watching televisioneach week was 21.6 hours. The principal introduced a campaign to encourage the students towatch less television. One year later, in 2006, the principal performed a hypothesis test todetermine whether the average amount of time spent watching television per week haddecreased. The hypotheses are:

H 0 : µ = 21.6 hoursHa : µ < 21.6 hours

where µ is the mean amount of time, in 2006, that tenth-graders spend watching television eachweek

Explain the meaning of a correct decision.A) A correct decision would occur if, in fact, µ = 21.6 hours, and the results of the samplinglead to rejection of the null hypothesis; or if, in fact, µ < 21.6 hours and the results of thesampling lead to that conclusion.

B) A correct decision would occur if, in fact, µ = 21.6 hours, and the results of the sampling donot lead to rejection of that fact; or if, in fact, µ < 21.6 hours and the results of the samplingdo not lead to rejection of the null hypothesis that µ = 21.6 hours.

C) A correct decision would occur if, in fact, µ = 21.6 hours, and the results of the sampling donot lead to rejection of that fact; or if, in fact, µ < 21.6 hours and the results of the samplinglead to that conclusion.

D) A correct decision would occur if, in fact, µ < 21.6 hours and the results of the sampling donot lead to rejection of the null hypothesis that µ = 21.6 hours.

7)

8) A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.1 ounces. Aconsumer advocacy group wants to perform a hypothesis test to determine whether the meanamount is actually less than this. The hypotheses are:

H 0 : µ = 16.1 ouncesHa : µ < 16.1 ounces

where µ is the mean amount of juice in the manufacturer's 16 ounce bottles. Explain the meaningof a Type I error.A) A Type I error would occur if, in fact, µ < 16.1 ounces, but the results of the sampling fail tolead to that conclusion.

B) A Type I error would occur if, in fact, µ = 16.1 ounces, but the results of the sampling leadto the conclusion that µ < 16.1 ounces.

C) A Type I error would occur if, in fact, µ < 16.1 ounces, but the results of the sampling leadto the conclusion that µ > 16.1 ounces.

D) A Type I error would occur if, in fact, µ = 16.1 ounces, but the results of the sampling donot lead to rejection of that fact.

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9) A health insurer has determined that the "reasonable and customary" fee for a certain medicalprocedure is $1200. They suspect that the mean fee charged by one particular clinic for thisprocedure is higher than $1200. The insurer wants to perform a hypothesis test to determinewhether their suspicion is correct. The hypotheses are:

H 0 : µ = $1200Ha : µ > $1200

where µ is the mean amount charged by the clinic for this procedure. Explain the meaning of acorrect decision.A) A correct decision would occur if, in fact, µ = $1200, and the results of the sampling lead torejection of the null hypothesis; or if, in fact, µ > $1200 and the results of the sampling leadto that conclusion.

B) A correct decision would occur if, in fact, µ > $1200 and the results of the sampling do notlead to rejection of the null hypothesis that µ = $1200.

C) A correct decision would occur if, in fact, µ = $1200, and the results of the sampling do notlead to rejection of that fact; or if, in fact, µ > $1200 and the results of the sampling lead tothat conclusion.

D) A correct decision would occur if, in fact, µ = $1200, and the results of the sampling do notlead to rejection of that fact; or if, in fact, µ > $1200 and the results of the sampling do notlead to rejection of the null hypothesis that µ = $1200.

9)

10) In 2000, the mean math SAT score for students at one school was 485. Five years later, in 2005, ateacher performed a hypothesis test to determine whether the average math SAT score ofstudents at the school had changed from the 2000 mean of 485. The hypotheses were:

H 0 : µ = 485Ha : µ ≠ 485

where µ is the mean math SAT score, in 2005, for students at the schoolExplain the meaning of a Type I error.A) A Type I error would occur if, in fact, µ ≠ 485, and the results of the sampling lead to thatconclusion.

B) A Type I error would occur if, in fact, µ = 485, but the results of the sampling lead to theconclusion that µ ≠ 485

C) A Type I error would occur if, in fact, µ ≠ 485, but the results of the sampling fail to lead tothat conclusion.

D) A Type I error would occur if, in fact, µ = 485, but the results of the sampling do not lead torejection of that fact

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11) A man is on trial accused of murder in the first degree. The prosecutor presents evidence that hehopes will convince the jury to reject the hypothesis that the man is innocent. This situation canbe modeled as a hypothesis test with the following hypotheses:

H 0 : The defendant is innocent.H a : The defendant is guilty.

Explain the meaning of a Type II error.A) A Type II error would occur if, in fact, the defendant is guilty and the jury rejects the nullhypothesis that he is innocent.

B) A Type II error would occur if, in fact, the defendant is innocent but the jury concludesthat he is guilty.

C) A Type II error would occur if, in fact, the defendant is guilty but the jury fails to concludethat he is guilty.

D) A Type II error would occur if, in fact, the defendant is innocent and the jury fails to rejectthe null hypothesis that he is innocent.

11)

12) In 2000, the mean math SAT score for students at one school was 487. Five years later, in 2005, ateacher performed a hypothesis test to determine whether the average math SAT score ofstudents at the school had changed from the 2000 mean of 487. The hypotheses were:

H 0 : µ = 487Ha : µ ≠ 487

where µ is the mean math SAT score, in 2005, for students at the schoolExplain the meaning of a Type II error.A) A Type II error would occur if, in fact, µ = 487, but the results of the sampling do not leadto rejection of that fact

B) A Type II error would occur if, in fact, µ ≠ 487, and the results of the sampling lead to thatconclusion.

C) A Type II error would occur if, in fact, µ = 487, but the results of the sampling lead to theconclusion that µ ≠ 487

D) A Type II error would occur if, in fact, µ ≠ 487, but the results of the sampling fail to leadto that conclusion.

12)

13) The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4 partsper million (ppm). A consumer health group measured the level of the chemical in a randomsample of tomatoes obtained from one producer to determine whether the mean level of thechemical in these tomatoes exceeds the recommended limit. The hypotheses are H0 : µ = 0.4 ppmHa : µ > 0.4 ppmwhere µ is the mean level of the chemical in tomatoes from this producer. Explain the meaning ofa Type I error.A) A Type I error would occur if, in fact, µ = 0.4 ppm, but the results of the sampling fail tolead to rejection of that fact.

B) A Type I error would occur if, in fact, µ > 0.4 ppm, and the results of the sampling lead torejection of the null hypothesis that µ = 0.4 ppm.

C) A Type I error would occur if, in fact, µ = 0.4 ppm, but the results of the sampling lead tothe conclusion that µ > 0.4 ppm

D) A Type I error would occur if, in fact, µ > 0.4 ppm, but the results of the sampling fail tolead to that conclusion.

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14) The recommended dietary allowance (RDA) of vitamin C for women is 75 milligrams per day. Ahypothesis test is to be performed to decide whether adult women are, on average, getting lessthan the RDA of 75 milligrams per day. The hypotheses are

H0 : µ = 75 mg

Ha : µ < 75 mg

where µ is the mean vitamin C intake (per day) of all adult females. Explain the meaning of aType II error.

A) A Type II error would occur if, in fact, µ = 75 mg, and the results of the sampling do notlead to rejection of that fact.

B) A Type II error would occur if, in fact, µ < 75 mg, and the results of the sampling lead torejection of the null hypothesis that µ = 75 mg.

C) A Type II error would occur if, in fact, µ = 75 mg, but the results of the sampling lead tothe conclusion that µ < 75 mg

D) A Type II error would occur if, in fact, µ < 75 mg, but the results of the sampling fail tolead to that conclusion.

14)

15) A psychologist has designed a test to measure stress levels in adults. She has determined thatnationwide the mean score on her test is 20. A hypothesis test is to be conducted to determinewhether the mean score for trial lawyers exceeds the national mean score. The hypotheses are H0 : µ = 20Ha : µ > 20where µ is the mean score for all trial lawyers. Explain the meaning of a Type I error.A) A Type I error would occur if, in fact, µ > 20, but the results of the sampling fail to lead tothat conclusion.

B) A Type I error would occur if, in fact, µ = 20, but the results of the sampling fail to lead torejection of that fact.

C) A Type I error would occur if, in fact, µ = 20, but the results of the sampling lead to theconclusion that µ > 20.

D) A Type I error would occur if, in fact, µ > 20, but the results of the sampling lead torejection of the null hypothesis that µ = 20.

15)

Classify the conclusion of the hypothesis test as a Type I error, a Type II error, or a correct decision.16) In the past, the mean running time for a certain type of flashlight battery has been 9.9 hours. The

manufacturer has introduced a change in the production method and wants to perform ahypothesis test to determine whether the mean running time has increased as a result. Thehypotheses are:

H 0 : µ = 9.9 hoursHa : µ > 9.9 hours

where µ is the mean running time of the new batteries Suppose that the results of the sampling lead to nonrejection of the null hypothesis. Classify thatconclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean runningtime has increased.A) Type I error B) Type II error C) Correct decision

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17) In the past, the mean running time for a certain type of flashlight battery has been 8.1 hours. Themanufacturer has introduced a change in the production method and wants to perform ahypothesis test to determine whether the mean running time has increased as a result. Thehypotheses are:

H 0 : µ = 8.1 hoursHa : µ > 8.1 hours

where µ is the mean running time of the new batteriesSuppose that the results of the sampling lead to rejection of the null hypothesis. Classify thatconclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean runningtime has not increased.A) Correct decision B) Type I error C) Type II error

17)

18) A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.1 ounces. Aconsumer advocacy group wants to perform a hypothesis test to determine whether the meanamount is actually less than this. The hypotheses are:

H 0 : µ = 16.1 ouncesHa : µ < 16.1 ounces

where µ is the mean amount of juice in the manufacturer's 16 ounce bottles.Suppose that the results of the sampling lead to rejection of the null hypothesis. Classify thatconclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean amount ofjuice, µ, is less than 16.1 ounces.A) Type I error B) Correct decision C) Type II error

18)

19) A health insurer has determined that the "reasonable and customary" fee for a certain medicalprocedure is $1200. They suspect that the average fee charged by one particular clinic for thisprocedure is higher than $1200. The insurer wants to perform a hypothesis test to determinewhether their suspicion is correct. The hypotheses are:

H 0 : µ = $1200Ha : µ > $1200

where µ is the mean amount charged by the clinic for this procedure.Suppose that the results of the sampling lead to rejection of the null hypothesis. Classify thatconclusion as a Type I error, a Type II error, or a correct decision, if in fact the average feecharged by the clinic is $1200 .A) Correct decision B) Type I error C) Type II error

19)

20) The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.7 partsper million (ppm). A consumer health group measured the level of the chemical in a randomsample of tomatoes obtained from one producer to determine whether the mean level of thechemical in these tomatoes exceeds the recommended limit. The hypotheses are H0 : µ = 0.7 ppmHa : µ > 0.7 ppmwhere µ is the mean level of the chemical in tomatoes from this producer. Suppose that theresults of the sampling lead to nonrejection of the null hypothesis. Classify that conclusion as aType I error, a Type II error, or a correct decision, if in fact the mean level of the chemical in thesetomatoes is greater than 0.7 ppm.A) Correct decision B) Type I error C) Type II error

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SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Provide an appropriate response.21) You wish to test the hypotheses shown below.

H0 : µ = 40Ha : µ > 40

Would you be inclined to reject the null hypothesis if the sample mean turned out to bemuch smaller than 40? Explain your thinking.

21)

22) Jenny is conducting a hypothesis test concerning a population mean. The hypotheses areas follows.

H 0 : µ = 50Ha : µ > 50

She selects a sample and finds that the sample mean is 54.2. She then does somecalculations and is able to make the following statement:If H0 were true, the chance that the sample mean would have come out as big ( orbigger) than 54.2 is 0.3. Do you think that she should reject the null hypothesis? Why orwhy not?

22)

23) Robert is conducting a hypothesis test concerning a population mean. The hypothesesare as follows.

H 0 : µ = 50Ha : µ > 50

He selects a sample of size 35 and finds that the sample mean is 60. He then does somecalculations and finds that for samples of size 35, the standard deviation of the samplemeans is 3.2. Do you think that he should reject the null hypothesis? Why or why not?

23)


24) Traditionally in hypothesis testing the null hypothesis represents the "status quo" which will beoverturned only if there is evidence against it. Which of the statements below might represent anull hypothesis?A) The defendant is guilty. B) The teaching method raises SAT scores.C) The treatment has no effect. D) None of the above

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25) True or false: If it is important to reject a false null hypothesis, the β probability should be small.A) True B) False

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26) For a fixed sample size, how will increasing the significance level of a hypothesis testaffect the probability of a Type I error? How will it affect the probability of a Type IIerror?

26)

27) Give an example of a hypothesis test for which it is important to have a small αprobability. Explain why it is important to have a small value for α.

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28) Give an example of a hypothesis test for which it is important to have a small βprobability. Explain why it is important to have a small value for β.

28)

29) A pharmaceutical company has a new drug which relieves headaches. However, there issome indication that the drug may have the side effect of increasing blood pressure.Suppose the drug company conducts a hypothesis test to determine whether themedication raises blood pressure. The hypotheses are:

H 0 : The drug does not increase blood pressure.H a : The drug increases blood pressure.

Do you think that for doctors and patients it is more important to have a small αprobability or a small β probability? Why? Do you think that the pharmaceuticalcompany would prefer to have a small α probability or a small β probability? Why?

29)

30) A man is on trial accused of murder in the first degree. The prosecutor presents evidencethat he hopes will convince the jury to reject the hypothesis that the man is innocent.This situation can be modeled as a hypothesis test with the following hypotheses:

H 0 : The defendant is innocent.H a : The defendant is guilty.

If convicted, the defendant will receive the death penalty. Do you think that a Type Ierror or a Type II error would be more serious? Why?

30)


The graph portrays the decision criterion for a one-mean z-test. The curve in the graph is the normal curve for the teststatistic under the assumption that the null hypothesis is true. Use the graph to solve the problem.

31) A graphical display of the decision criterion follows.

Determine the rejection region.A) z = 2.575 B) z ≤ 2.575 C) z ≥ 0.005 D) z ≥ 2.575

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Determine the rejection region.A) z ≤ -2.33 or z ≥ 2.33 B) -2.33 ≤ z ≤ 2.33C) z ≥ 0.01 D) z = -2.33 or z = 2.33

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Determine the nonrejection region.A) α ≤ 0.10 B) z ≥ 0.10 C) z ≤ -1.28 D) z ≥ -1.28

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Determine the significance level.A) α = 0.01 B) α = 0.98C) α = -2.33 or α = 2.33 D) α = 0.02

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Identify the hypothesis test as two-tailed, left-tailed, or right-tailed.A) Right-tailedB) Left-tailedC) Two-tailedD) There is not enough information to answer the question.

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Construct a graph portraying the decision criterion for the specified hypothesis test.36) A hypothesis test for a population mean is conducted. The hypotheses are:

H 0 : µ = µ 0Ha : µ ≠ µ 0

The significance level is 0.01 and the critical values are -2.575 and 2.575. Sketch a normalcurve displaying the decision criterion. This curve will represent the normal curve forthe test statistic under the assumption that the null hypothesis is true. On your graphindicate the area in each tail, the critical values, the rejection region, and the nonrejectionregion.

36)

37) A hypothesis test for a population mean is conducted. The hypotheses are:H 0 : µ = µ 0Ha : µ > µ 0

The significance level is 0.01 and the critical value is 2.33. Sketch a normal curvedisplaying the decision criterion. This curve will represent the normal curve for the teststatistic under the assumption that the null hypothesis is true. On your graph indicatethe area in the tail, the critical value, the rejection region, and the nonrejection region.

37)


Determine the critical value(s) for a one-mean z-test.38) A two-tailed test with α = 0.1.

A) ±1.4805 B) ±1.645 C) ±2.052 D) ±2.3338)

39) A left-tailed test with α = 0.02 .A) ±2.054 B) -2.054 C) -1.96 D) ±1.96

39)

The significance level and P-value of a hypothesis test are given. Decide whether the null hypothesis should berejected.

40) α = 0.10, P-value = 0.06A) Reject the null hypothesis. B) Do not reject the null hypothesis.

40)

41) α = 0.01, P-value = 0.02A) Reject the null hypothesis. B) Do not reject the null hypothesis.

41)

The P-value for a hypothesis test is given. Determine whether the strength of the evidence against the null hypothesisis weak/none, moderate, strong, or very strong.

42) P = 0.15A) Weak or none B) Moderate C) Very strong D) Strong

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43) P = 0.057A) Strong B) Weak or none C) Very strong D) Moderate

43)

44) P = 0.002A) Moderate B) Weak or none C) Strong D) Very strong

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45) P = 0.05A) Strong B) Weak or none C) Very strong D) Moderate

45)

The value obtained for the test statistic, z, in a one-mean z-test is given. Also given is whether the test is two tailed,left tailed, or right tailed. Determine the P-value.

46) A right-tailed test: z = 2.38A) 0.0174 B) 0.0087 C) 0.9826 D) 0.9913

46)

47) A left-tailed test: z = -2.65A) 0.9920 B) 0.0080 C) 0.0040 D) 0.9960

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48) A two-tailed test: z = 1.31A) 0.8098 B) 0.0951 C) 0.1902 D) 0.9049

48)

Provide an appropriate response.49) A one-sample z-test for a population mean is to be performed. Let z 0 denote the observed

value of the test statistic, z. True or false: For a left-tailed test, the P-value is the area under thestandard normal curve to the left of z 0 .A) True B) False

49)

50) A one-sample z-test for a population mean is to be performed. Let z 0 denote the observedvalue of the test statistic, z. Assume that a two-tailed test is being performed. True or false: If z0is negative, the P-value is twice the area under the standard normal curve to the right of z0.A) True B) False

50)

51) A one-sample z-test for a population mean is performed. Suppose that the P-value for the test is0.04. For what significance levels (values of α) can the null hypothesis be rejected?A) For all values of α smaller than 0.04B) For α = 0.05, 0.10C) For α = 0.04D) For all values of α greater than or equal to 0.04

51)


52) A one-sample z-test for a population mean is to be performed. The hypotheses areH0 : µ = 100Ha : µ ≠ 100.

Let z 0 denote the observed value of the test statistic, z. Suppose that z 0 is positiveand the P-value for the test is 0.07. Is the following a correct interpretation of theP-value? If not, give a correct interpretation of the P-value. If the null hypothesis weretrue, the probability of observing a value of the test statistic as large or larger than thatobserved would be 0.07.

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53) A one-sample z-test for a population mean is to be performed. True or false: The larger theP-value, the stronger the evidence against the null hypothesis?A) True B) False

53)


54) A manufacturer claims that the mean weight of flour in its 32-ounce bags is 32.1 ounces.A z-test is performed to determine whether the mean weight is actually less than this.The hypotheses are

H0 : µ = 32.1 ouncesHa : µ < 32.1 ounces.

The mean weight for a sample of 45 bags of flour was 30.7 ounces. Suppose that theP-value corresponding to this sample data is 0.001. Give an interpretation of theP-value. Would you feel confident in concluding that the mean weight is less than32.1 ounces?

54)


A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform therequired hypothesis test at the given significance level. Use the critical -value approach.

55) x = 20, n = 60, σ = 1.5, H0: µ = 22; Ha: µ ≠ 22, α = 0.05A) z = -10.33; critical values = ±1.96; reject H0B) z = -10.33; critical values = ±1.645; reject H0C) z = -10.33; critical values = ±1.645; do not reject H0D) z = -10.33; critical values = ±1.96; do not reject H0

55)

56) x = 54, n = 37, σ = 5.6, H0: µ = 56; Ha: µ < 56, α = 0.05A) z = -2.17; critical value = -1.645; reject H0B) z = -0.36; critical value = -1.645; do not reject H0C) z = -2.17; critical value = -1.645; do not reject H0D) z = -2.17; critical value = -1.96; reject H0

56)

57) x = 51, n = 52 , σ = 3.6, H0: µ = 50; Ha: µ > 50, α = 0.01A) z = 2.00; critical value = 2.33; do not reject H0B) z = 2.00; critical value = 1.33; reject H0C) z = 0.28; critical value = 2.33; do not reject H0D) z = 2.00; critical value = 2.33; reject H0

57)

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A sample mean, sample size, and population standard deviation are given. Use the one-mean z-test to perform therequired hypothesis test at the given significance level. Use the P-value approach.

58) x = 137, n = 20, σ = 14.2, H0: µ = 132 , Ha: µ ≠ 132 , α = 0.05A) z = 1.57; P-value = 0.1164; do not reject H0B) z = 2.57; P-value = 0.0101; reject H0C) z = 0.35; P-value = 0.7263; do not reject H0.D) z = 1.57; P-value = 0.0582; do not reject H0

58)

59) x = 94, n = 27, σ = 14, H0: µ = 89 , Ha: µ ≠ 89, α = 0.10A) z = 0.36; P-value = 0.7188; do not reject H0B) z = 1.86; P-value = 0.0628; reject H0C) z = 1.86; P-value = 0.0314; reject H0D) z = 0.36; P-value = 0.3594; do not reject H0

59)

60) x = 3.7, n = 32, σ = 1.8, H0: µ = 4.2 , Ha: µ < 4.2 , α = 0.05A) z = -1.57; P-value = 0.0582; reject H0B) z = -1.57; P-value = 0.1164; reject H0C) z = -1.57; P-value = 0.0582; do not reject H0.D) z = -1.57; P-value = 0.1164; do not reject H0

60)

61) x = 78, n = 28, σ = 11, H0: µ = 72 , Ha: µ > 72 , α = 0.01A) z = 2.89; P-value = 0.0039; reject H0B) z = 0.55; P-value = 0.2912; do not reject H0C) z = 2.89; P-value = 0.0019; reject H0D) z = 0.55; P-value = 0.2912; reject H0

61)


Perform a hypothesis test for the population mean. Assume that preliminary data analyses indicate that it isreasonable to apply the z-test. Use the critical-value approach.

62) The National Weather Service says that the mean daily high temperature for October ina large midwestern city is 56°F. A local weather service suspects that this value is notaccurate and wants to perform a hypothesis test to determine whether the mean isactually lower than 56°F. A sample of mean daily high temperatures for October overthe past 34 years yields x = 54°F. Assume that the population standard deviation is 5.6°F. Perform the hypothesis test at the 1% significance level.

62)

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63) The maximum acceptable level of a certain toxic chemical in vegetables has been set at0.4 parts per million (ppm). A consumer health group measured the level of thechemical in a random sample of tomatoes obtained from one producer. The levels, inppm, are shown below. 0.31 0.47 0.19 0.72 0.56 0.91 0.29 0.83 0.49 0.28 0.31 0.46 0.25 0.34 0.17 0.58 0.19 0.26 0.47 0.81

At the 5% significance level, do the data provide sufficient evidence to conclude that themean level of the chemical in tomatoes from this producer is greater than therecommended level of 0.4 ppm? Assume that the population standard deviation oflevels of the chemical in these tomatoes is 0.21 ppm.

63)

64) A researcher claims that the amounts of acetaminophen in a certain brand of cold tabletshave a mean different from the 600 mg claimed by the manufacturer. Test this claim atthe 2% level of significance. The mean acetaminophen content for a random sample ofn = 42 tablets is 604.1 mg. Assume that the population standard deviation is 4.7 mg.

64)

Perform a one-sample z-test for a population mean using the P-value approach. Be sure to state the hypotheses andthe significance level, to compute the value of the test statistic, to obtain the P-value, and to state your conclusion.Also, assess the strength of the evidence against the null hypothesis.

65) A manufacturer claims that the mean amount of juice in its 16-ounce bottles is16.1 ounces. A consumer advocacy group wants to perform a hypothesis test todetermine whether the mean amount is actually less than this. The mean volume of juicefor a random sample of 70 bottles was 15.94 ounces. Do the data provide sufficientevidence to conclude that the mean amount of juice for the 16-ounce bottles, µ, is lessthan 16.1 ounces? Perform the appropriate hypothesis test using a significance level of10%. Assume that σ = 0.9 ounces.

65)

66) A health insurer has determined that the "reasonable and customary" fee for a certainmedical procedure is $1200. They suspect that the average fee charged by one particularclinic for this procedure is higher than $1200. The insurer wants to perform a hypothesistest to determine whether their suspicion is correct. The mean fee charged by the clinicfor a random sample of 65 patients receiving this procedure was $1280. Do the dataprovide sufficient evidence to conclude that the mean fee charged by this clinic for thisprocedure is higher than $1200? Perform the appropriate hypothesis test using asignificance level of 1%. Assume that σ = $220.

66)

67) Five years ago, the average math SAT score for students at one school was 475. Ateacher wants to perform a hypothesis test to determine whether the mean math SATscore of students at the school has changed. The mean math SAT score for a randomsample of 40 students from this school is 469. Do the data provide sufficient evidence toconclude that the mean math SAT score for students at the school has changed from theprevious mean of 475? Perform the appropriate hypothesis test using a significance levelof 10%. Assume that σ = 73.

67)

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68) The forced vital capacity (FVC) is often used by physicians to assess a person's ability tomove air in and out of their lungs. It is the maximum amount of air that can be exhaledafter a deep breath. For adult males, the mean FVC is 5.0 liters. A researcher wants toperform a hypothesis test to determine whether the mean forced vital capacity forwomen differs from this value. The mean forced vital capacity for a random sample of85 women was 4.8 liters. Do the data provide sufficient evidence to conclude that themean forced vital capacity for women differs from 5.0 liters, the mean value for men?Perform the appropriate hypothesis test using a significance level of 5%. Assume thatσ = 0.9 liters.

68)

69) The maximum acceptable level of a certain toxic chemical in vegetables has been set at0.4 parts per million (ppm). A consumer health group measured the level of thechemical in a random sample of tomatoes obtained from one producer. The levels, inppm, are shown below. 0.31 0.47 0.19 0.72 0.56 0.91 0.29 0.83 0.49 0.28 0.31 0.46 0.25 0.34 0.17 0.58 0.19 0.26 0.47 0.81

At the 5% significance level, do the data provide sufficient evidence to conclude that themean level of the chemical in tomatoes from this producer is greater than therecommended level of 0.4 ppm? Assume that the population standard deviation oflevels of the chemical in these tomatoes is 0.21 ppm.

69)


Provide an appropriate response.70) A hypothesis test for a population mean is to be performed at the 1% level of significance. True

or false: If the null hypothesis is true, the probability that the test statistic will fall in the rejectionregion is 0.01.A) True B) False

70)

71) A two-tailed hypothesis test for a population mean is to be performed at the 1% level ofsignificance. The population standard deviation is known. True or false: The critical values arethe two z-scores which divide the area under the standard normal curve into a middle 0.98 areaand two outside areas of 0.01.A) True B) False

71)


72) In 1995, the mean math SAT score for students at one school was 488. A teacherintroduces a new teaching method to prepare students for the SAT. One year later, heperforms a hypothesis test to determine whether the mean math SAT score hasincreased. The hypotheses are

H0: µ = 488Ha: µ > 488.

If the null hypothesis is rejected at the 10% level of significance, do you think the teacherwould feel confident that his teaching method works? What about if the null hypothesisis rejected at the 1% level of significance? Which of these two results would constitutestronger evidence that his teaching method works? Explain your thinking.

72)

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Answer KeyTestname: WRITTEN HYPOTHESIS PRACTICE

1) A2) A3) A4) B5) A6) C7) C8) B9) C10) B11) C12) D13) C14) D15) C16) B17) B18) B19) B20) C21) No. The alternative hypothesis is that the true mean is greater than 40. A sample mean much smaller than 40 does

not provide evidence in favor of this alternative hypothesis. The null hypothesis should be rejected only if thesample mean turns out much larger than 40.

22) Answers will vary. Possible answer. No, she should not reject the null hypothesis. If H0 were true, the samplemean could easily be as big as 54.2 by chance. So the observed sample mean is consistent with the null hypothesis

23) Answers will vary. Possible answer. Yes, he should reject the null hypothesis. If H0 were true, it is not very likelythat the sample mean would be as big as 60, since this is more than three standard deviations from 50. So theobserved sample mean is inconsistent with the null hypothesis.

24) C25) A26) The probability of a Type I error will increase. The probability of a Type I error will decrease.27) Examples will vary. Students should give an example of a test in which the consequences of rejecting a true null

hypothesis would be serious.28) Examples will vary. Students should give an example of a test in which the consequences of failing to reject a false

null hypothesis would be serious.29) For doctors and patients it is more important to have a small β probability. If the drug really does increase blood

pressure, it is important to detect that. So the probability of a Type II error should be small. The pharmaceuticalcompany may prefer to have a small α probability since they would lose money if the test concluded that the drugincreased blood pressure when in fact it didn't.

30) A Type II error would mean failing to convict a guilty man. A Type I error would mean convicting an innocent man,which would be more serious, especially since this is a death penalty case.

31) D32) A33) D34) D35) B

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36)

37)38) B39) B40) A41) B42) A43) D44) D45) A46) B47) C48) C49) A50) B51) D

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52) The interpretation is not correct. The correct interpretation is:If the null hypothesis were true, the probability of observing a value of the test statistic as extreme (i.e., as far fromthe null-hypothesis mean) or more extreme than that observed would be 0.07.

53) B54) If the null hypothesis were true (i.e., if the mean weight really were 32.1 ounces), the probability of observing a

sample mean as small or smaller than 30.7 ounces would be 0.001. Since the P-value is so small, the evidenceagainst the null hypothesis is overwhelming.

55) A56) A57) A58) A59) B60) C61) C62) H0 : µ = 56° F

Ha : µ < 56° Fα = 0.01Test statistic: z = -2.08.Critical value z = -2.33. Since -2.08. > -2.33, do not reject H0: µ = 56°F. At the 1% significance level, the data do not provide sufficient evidence to conclude that the mean daily hightemperature for October is less than 56°F.

63) H 0 : µ = 0.4 ppmHa : µ > 0.4 ppmα = 0.05Test statistic: z = 0.95Critical value z = 1.645 Since 0.95 < 1.645, do not reject H0: µ = 0.4 ppmAt the 5% significance level, the data do not provide sufficient evidence to conclude that the mean level of thechemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm.

64) H0 : µ = 600 mgHa : µ ≠ 600 mgα = 0.02Test statistic: z = 5.65.Critical values: z = ± 2.33. Since 5.65 > 2.33, reject H0. At the 2% significance level, the data do provide sufficient evidence to conclude that the mean acetaminophencontent differs from 600 mg.

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65) H0: µ = 16.1 ouncesHa: µ < 16.1 ouncesα = 0.10z = -1.49P-value = 0.0681Since 0.0681 < 0.10, reject H0. At the 10% significance level, the data provide sufficient evidence to conclude that the mean amount of juice for the16-ounce bottles is less than 16.1 ounces. The evidence against the null hypothesis is moderate.

66) H0: µ = $1200Ha: µ > $1200α = 0.01z = 2.93P-value = 0.0017Since 0.0017 < 0.01, reject H0. At the 1% significance level, the data provide sufficient evidence to conclude that themean fee charged by this clinic for this procedure is higher than $1200.The evidence against the null hypothesis is very strong.

67) H0: µ = 475Ha: µ ≠ 475α = 0.10z = -0.52P-value = 0.6030Since 0.6030 > 0.10, do not reject H0. At the 10% significance level, the data do not provide sufficient evidence toconclude that the mean math SAT score for students at the school has changed from the previous mean of 475. Theevidence against the null hypothesis is weak or none.

68) H0: µ = 5.0 litersHa: µ ≠ 5.0 litersα = 0.05z = -2.05P-value = 0.0404Since 0.0404 < 0.05, reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that themean forced vital capacity for women differs from 5.0 liters, the mean value for men. The evidence against the nullhypothesis is strong.

69) H 0 : µ = 0.4 ppmHa : µ > 0.4 ppmα = 0.05Test statistic: z = 0.95P-value = 0.1711Since 0.1711 > 0.05, do not reject H0.At the 5% significance level, the data do not provide sufficient evidence to conclude that the mean level of thechemical in tomatoes from this producer is greater than the recommended level of 0.4 ppm. The evidence againstthe null hypothesis is weak or none.

70) A71) B

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72) If the null hypothesis is rejected at the 1% level of significance this provides much stronger evidence that µ > 488than if the null hypothesis is rejected at the 10% level of significance. Suppose that his teaching method actuallydoes not work and that µ = 488. Then the chance that the null hypothesis would be rejected at the 10% level is 10%.This is not so unlikely. However, the chance that the null hypothesis would be rejected at the 1% level is only 1%.

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