Exercises 1–5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 “Body Data” in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit

0

1

2

3

4

5

6

7

8

9

Frequency

30

35

24

25

35

36

37

27

27

24

Given that the P-value for the hypothesis test is 0.501, what do you conclude? Does it appear that the heights were obtained through measurement or that the subjects reported their heights?

Short Answer

Expert verified

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to conclude that the last digits of heights do not occur equally frequently.

Also, it appears that the heights were measured rather than reported because if the heights were reported, the frequencies corresponding to the last digits of 0 and 5 would be significantly greater than the rest of the digits.

Step by step solution

01

Given information

The last digits of the heights of a sample of people are tabulated along with their respective frequencies.

02

Conclusion of the test

The null hypothesis and the alternative hypothesis is written as follows:

\[{H_0}:\]The last digits of the heights of people are equally likely to occur.

\[{H_1}:\]The last digits of the heights of people are not equally likely to occur.

The p-value is equal to 0.501.

The level of significance is equal to 0.05.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

Thus, there is not enough evidence to conclude that the last digits of heights do not occur equally frequently.

03

Reported values vs. measured values

If the heights have been reported, then most of them would have rounded off the values such that a majority of the heights would end in 0 or 5.

Since the frequencies corresponding to the last digits of 0 and 5 are not significantly greater than those of the remaining digits, it can be said that the heights were measured and not reported.

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Most popular questions from this chapter

Benford’s Law. According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

Author’s Computer Files The author recorded the leading digits of the sizes of the electronic document files for the current edition of this book. The leading digits have frequencies of 55, 25, 17, 24, 18, 12, 12, 3, and 4 (corresponding to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively). Using a 0.05 significance level, test for goodness-of-fit with Benford’s law.

The accompanying table is from a study conducted

with the stated objective of addressing cell phone safety by understanding why we use a particular ear for cell phone use. (See “Hemispheric Dominance and Cell Phone Use,” by Seidman, Siegel, Shah, and Bowyer, JAMA Otolaryngology—Head & Neck Surgery,Vol. 139, No. 5.)

The goal was to determine whether the ear choice is associated with auditory or language brain hemispheric dominance. Assume that we want to test the claim that handedness and cell phone ear preference are independent of each other.

a. Use the data in the table to find the expected value for the cell that has an observed frequency of 3. Round the result to three decimal places.

b. What does the expected value indicate about the requirements for the hypothesis test?

Right Ear

Left Ear

No Preference

Right-Handed

436

166

40

Left-Handed

16

50

3

Weather-Related Deaths Review Exercise 5 involved weather-related U.S. deaths. Among the 450 deaths included in that exercise, 320 are males. Use a 0.05 significance level to test the claim that among those who die in weather-related deaths, the percentage of males is equal to 50%. Provide an explanation for the results.

Exercises 1–5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 “Body Data” in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit

0

1

2

3

4

5

6

7

8

9

Frequency

30

35

24

25

35

36

37

27

27

24

What are the null and alternative hypotheses corresponding to the stated claim?

Benford’s Law. According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

Tax Cheating? Frequencies of leading digits from IRS tax files are 152, 89, 63, 48, 39, 40, 28, 25, and 27 (corresponding to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively, based on data from Mark Nigrini, who provides software for Benford data analysis). Using a 0.05 significance level, test for goodness-of-fit with Benford’s law. Does it appear that the tax entries are legitimate?

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