Self Assessment for Lesson 52 1

1. Click on “Begin Assessment” button.

2. Scroll down to begin quiz.

Self Assessment for Lesson 52 2

1. TRUE or FALSE: If the alternate hypothesis is Ha:µ > 0, then we reject the null hypothesis when theP-value is large.

True

False

2. If a researcher is interested in testing whether themean is different from some claimed value, then

a “6=” will be found in the null hypothesis.

a “6=” will be found in the alternative hypothesis.

the P-value will be found by dividing the teststatistic by 2.

the P-value will be found by multiplying the teststatistic by 2.

Self Assessment for Lesson 52 3

3. If we test H0: µ = 40 vs. Ha: µ < 40, this test is

one-sided (left tail).

one-sided (right tail).

two-sided.

4. If we test H0: µ = 10 vs. Ha: µ 6= 10, this test is

one-sided (left tail).

one-sided (right tail).

two-sided.

Self Assessment for Lesson 52 4

Use the following information for questions 5-7.

The mean weight of all new-born infants deliveredat a local hospital is 7.35 pounds with a standarddeviation of 1.1 pounds. A researcher is interested indetermining if the mean weight of new-born infantsdelivered by cesarian section in the hospital is greaterthan the hospital’s overall mean infant weight of 7.35.A sample of 40 cesarean delivery records was obtainedand the sample mean weight for these infants was 7.56pounds.

Self Assessment for Lesson 52 5

5. What are the null and alternative hypotheses for thisstudy?

H0: µ = 7.35 vs. Ha: µ < 7.35

H0: µ = 7.35 vs. Ha: µ 6= 7.35

H0: µ = 7.35 vs. Ha: µ > 7.35

H0: x̄ = 7.56 vs. Ha: x̄ < 7.56

H0: x̄ = 7.56 vs. Ha: x̄ 6= 7.56

H0: x̄ = 7.56 vs. Ha: x̄ > 7.56

Self Assessment for Lesson 52 6

6. The value of the test statistic for this problem is z =1.21. Which of the following best describes of theP-value for this test?

The area under the standard normal curve lessthan z = 1.21.

The area under the standard normal curve greaterthan z = 1.21.

The area under the standard normal curve betweenz = −1.21 and z = 1.21.

The area under the standard normal curve lessthan z = −1.21 plus the area under the standardnormal curve greater than z = 1.21.

Self Assessment for Lesson 52 7

7. Using z = 1.21, what is the P-value for this test?

0.1131

0.2262

0.7738

0.8869

Self Assessment for Lesson 52 8

Use the following information for questions 8-13.

An endocrinologist is interested in the effects of de-pression on the thyroid. It is believed that healthysubjects have a mean thyroxin (a hormone related tothyroid function) level of 7.0 micrograms/100 ml anda standard deviation of 1.6 micrograms/100 ml. Theendocrinologist wants to assess whether the mean thy-roxin level is different for those with depression. Shegathers a sample of 35 subjects with depression andobtains a sample mean of 7.82 micrograms/100 ml.Use a significance level of α = 0.05.

Self Assessment for Lesson 52 9

8. What are the null and alternative hypotheses for thisstudy?

H0: µ = 7.0 vs. Ha: µ < 7.0

H0: µ = 7.0 vs. Ha: µ 6= 7.0

H0: µ = 7.0 vs. Ha: µ > 7.0

H0: x̄ = 7.82 vs. Ha: x̄ < 7.82

H0: x̄ = 7.82 vs. Ha: x̄ 6= 7.82

H0: x̄ = 7.82 vs. Ha: x̄ > 7.82

9. What is the value of the test statistic?

0.23

0.51

1.52

3.03

Self Assessment for Lesson 52 10

10. Ignore your answer to the previous question and as-sume the value of the test statistic is z = 2.50. Whichof the following best describes the P-value for thistest.

The area under the standard normal curve lessthan z = 2.50.

The area under the standard normal curve greaterthan z = 2.50.

The area under the standard normal curve betweenz = −2.50 and z = 2.50.

The area under the standard normal curve lessthan z = −2.50 plus the area under the standardnormal curve greater than z = 2.50.

Self Assessment for Lesson 52 11

11. Using z = 2.50, what is the P-value for this test?

0.0062

0.0124

0.9876

0.9938

Self Assessment for Lesson 52 12

12. Ignore your answers in the previous problems and as-sume that the P-value is 0.003. What is the appro-priate decision?

Reject H0 and declare statistical significance.

Fail to reject H0 and declare statistical signifi-cance.

Reject H0 and do not declare statistical signifi-cance.

Fail to reject H0 and do not declare statistical sig-nificance.

Self Assessment for Lesson 52 13

13. Assuming the P-value is 0.003, what is the most ap-propriate conclusion?

The mean thyroxin level for persons with depres-sion is 7.0 micrograms/100 ml.

The mean thyroxin level for persons with depres-sion is 7.82 micrograms/100 ml.

We have insufficient evidence to conclude the meanthyroxin level for persons with depression is differ-ent from 7.0 micrograms/100 ml.

We have insufficient evidence to conclude the meanthyroxin level for persons with depression is differ-ent from 7.82 micrograms/100 ml.

The mean thyroxin level for persons with depres-sion is different from 7.0 micrograms/100 ml.

The mean thyroxin level for persons with depres-sion is different from 7.82 micrograms/100 ml.

Self Assessment for Lesson 52 14

Click on the “Grade Quiz” button to see how you did.

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Click on the “Correct” button to highlight your correctanswers in green and wrong answers in red.For question solutions either:

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Solutions to Quizzes 15

Solutions to Quizzes

Solution to Quiz: 1. We always reject H0 when theP-value is less than α.

End Quiz

Solutions to Quizzes 16

Solution to Quiz: 2. We never double the test statistic.We get the P-value by doubling the area in one tail for atwo-sided test. End Quiz

Solutions to Quizzes 17

Solution to Quiz: 3. Because we are interested in testingwhether µ is less than 40, the test is one-sided (left tail).

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Solutions to Quizzes 18

Solution to Quiz: 4. Because we are interested in testingwhether µ is different from 10, this is a two-sided test.

End Quiz

Solutions to Quizzes 19

Solution to Quiz: 5. Because the researcher is interestedin testing whether the mean weight of new-born infantsdelivered by cesarian section in the hospital is greater than7.35 pounds, the alternative hypothesis is Ha: µ > 7.35.

End Quiz

Solutions to Quizzes 20

Solution to Quiz: 6. Because the alternative hypothesisis Ha: µ > 7.35, the P-value is an upper tail region.

End Quiz

Solutions to Quizzes 21

Solution to Quiz: 7. Because the alternative hypothesisis Ha: µ > 7.35, the P-value is the area under the stan-dard normal curve greater than z = 1.21. The area of thestandard normal distribution to the right of z = 1.21 is 1- 0.8869 = 0.1131. Because the standard normal distribu-tion is symmetric about 0, we could also find the area tothe right of z = 1.21 by looking up the area to the left ofz = −1.21.

End Quiz

Solutions to Quizzes 22

Solution to Quiz: 8. Because the endocrinologist isinterested in testing whether the mean thyroxin level forthose with depression is different than 7.0 micrograms/100ml, the alternative hypothesis is Ha: µ 6= 7.0. End Quiz

Solutions to Quizzes 23

Solution to Quiz: 9. z = 7.82−71.6/

√35

= 3.03 End Quiz

Solutions to Quizzes 24

Solution to Quiz: 10. Because the alternative hypothesisis Ha: µ 6= 7.0, the P-value equals the lower tail plus theupper tail.

End Quiz

Solutions to Quizzes 25

Solution to Quiz: 11. Because we have a two-sidedalternative hypothesis (Ha: µ 6= 7.0), our P-value is theprobability that z is greater than 2.50 PLUS the probabil-ity that z is less than -2.50. Because of the symmetry ofthe normal distribution, we need only calculate one of theareas (say, the probability that z > 2.50) and then doublethat area to obtain the P-value. The area of the normaldistribution to the right of z = 2.50 is 1 - 0.9938 = 0.0062.So, our P-value is 2(0.0062) = 0.0124.

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Solutions to Quizzes 26

Solution to Quiz: 12. No additional information.End Quiz

Solutions to Quizzes 27

Solution to Quiz: 13. No additional information.End Quiz