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Exercises 1鈥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 鈥淏ody Data鈥 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

Critical Thinking: Was Allstate wrong? The Allstate insurance company once issued a press release listing zodiac signs along with the corresponding numbers of automobile crashes, as shown in the first and last columns in the table below. In the original press release, Allstate included comments such as one stating that Virgos are worried and shy, and they were involved in 211,650 accidents, making them the worst offenders. Allstate quickly issued an apology and retraction. In a press release, Allstate included this: 鈥淎strological signs have absolutely no role in how we base coverage and set rates. Rating by astrology would not be actuarially sound.鈥

Analyzing the Results The original Allstate press release did not include the lengths (days) of the different zodiac signs. The preceding table lists those lengths in the third column. A reasonable explanation for the different numbers of crashes is that they should be proportional to the lengths of the zodiac signs. For example, people are born under the Capricorn sign on 29 days out of the 365 days in the year, so they are expected to have 29/365 of the total number of crashes. Use the methods of this chapter to determine whether this appears to explain the results in the table. Write a brief report of your findings.

Zodiac sign

Dates

Length(days)

Crashes

Capricorn

Jan.18-Feb. 15

29

128,005

Aquarius

Feb.16-March 11

24

106,878

Pisces

March 12-April 16

36

172,030

Aries

April 17-May 13

27

112,402

Taurus

May 14-June 19

37

177,503

Gemini

June 20-July 20

31

136,904

Cancer

July21-Aug.9

20

101,539

Leo

Aug.10-Sep.15

37

179,657

Virgo

Sep.16-Oct.30

45

211,650

Libra

Oct.31-Nov 22

23

110,592

Scorpio

Nov. 23-Nov. 28

6

26,833

Ophiuchus

Nov.29-Dec.17

19

83,234

Sagittarius

Dec.18-Jan.17

31

154,477

The accompanying table lists results of overtime football

games before and after the overtime rule was changed in the National Football League in 2011. Use a 0.05 significance level to test the claim of independence between winning an overtime game and whether playing under the old rule or the new rule. What do the results suggest about

the effectiveness of the rule change?

Before Rule Change

After Rule Change

Overtime Coin Toss Winner Won the Game

252

24

Overtime Coin Toss Winner Lost the Game

208

23

Flat Tire and Missed Class A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn鈥檛 have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author鈥檚 claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn鈥檛 have a flat tire?

Tire

Left Front

Right Front

Left Rear

Right Rear

Number Selected

11

15

8

16

Winning team data were collected for teams in different sports, with the results given in the table on the top of the next page (based on data from 鈥淧redicting Professional Sports Game Outcomes fromIntermediateGame Scores,鈥 by Copper, DeNeve, and Mosteller, Chance,Vol. 5, No. 3鈥4). Use a 0.10significance level to test the claim that home/visitor wins are independent of the sport. Given that among the four sports included here, baseball is the only sport in which the home team canmodify field dimensions to favor its own players, does it appear that baseball teams are effective in using this advantage?

Basketball

Baseball

Hockey

Football

Home Team Wins

127

53

50

57

Visiting Team Wins

71

47

43

42

Loaded Die The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 27, 31, 42, 40, 28, and 32. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die?

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