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In Exercises 5鈥20, conduct the hypothesis test and provide the test statistic and the P-value and , or critical value, and state the conclusion.

Testing a Slot Machine The author purchased a slot machine (Bally Model 809) and tested it by playing it 1197 times. There are 10 different categories of outcomes, including no win, win jackpot, win with three bells, and so on. When testing the claim that the observed outcomes agree with the expected frequencies, the author obtained a test statistic of\({\chi ^2} = 8.185\). Use a 0.05 significance level to test the claim that the actual outcomes agree with the expected frequencies. Does the slot machine appear to be functioning as expected?

Short Answer

Expert verified

There is not enough evidence to support the statement that the observed frequencies differ from the expected frequency.

Thus, the slot machine is operating as expected.

Step by step solution

01

Given information

A slot machine is tested to examine whether the observed outcomes agree with the expected frequencies. The machine is run 1197 times to test the 10 different categories. The test statistic \({\chi ^2} = 8.185\).

02

State the hypotheses

Assume that the samples are randomly selected, where the data includes a frequency counts. Also, it is assumed that the expected frequency is lesser than 5.

The hypotheses for conducting the given test is as follows:

\({H_0}:{p_0} = {p_1} = ... = {p_{10}}\)

\({H_a}:\)at least one of the proportions measure is different from others.

Where,\({p_0},{p_1},...,{p_{10}}\)are the proportions corresponding to different categories.

The test is right-tailed.

03

Determine the test statistic

The value of the test statistic is given to be equal to 8.185.

Let k be the number of categories.

The degrees of freedom for computing the critical value are:

\(\begin{aligned}{c}df = k - 1\\ = 10 - 1\\ = 9\end{aligned}\)

Thus, the critical value of\({\chi ^2}\)corresponding to\(\alpha = 0.05\)and degrees of freedom equal to 9 is equal to 16.919.

The p-value is,

\(\begin{aligned}{c}p - value = P\left( {{\chi ^2} > 8.815} \right)\\ = 0.516\end{aligned}\)

Since the test statistic value is less than the critical value along with the p-value which is greater than 0.05, then the null hypothesis is failed to be rejected.

04

State the conclusion

There is not enough evidence to support the statement that the observed frequencies differ from the expected frequency.

Thus, the slot machine appears to be working as expected.

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Most popular questions from this chapter

Benford鈥檚 Law. According to Benford鈥檚 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鈥24, test for goodness-of-fit with the distribution described by Benford鈥檚 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鈥檚 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鈥檚 law.

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

Cybersecurity The accompanying Statdisk results shown in the margin are obtained from the data given in Exercise 1. What should be concluded when testing the claim that the leading digits have a distribution that fits well with Benford鈥檚 law?

Questions 6鈥10 refer to the sample data in the following table, which describes the fate of the passengers and crew aboard the Titanic when it sank on April 15, 1912. Assume that the data are a sample from a large population and we want to use a 0.05 significance level to test the claim that surviving is independent of whether the person is a man, woman, boy, or girl.


Men

Women

Boys

Girls

Survived

332

318

29

27

Died

1360

104

35

18

Given that the P-value for the hypothesis test is 0.000 when rounded to three decimal places, what do you conclude? What do the results indicate about the rule that women and children should be the first to be saved?

Do World War II Bomb Hits Fit a Poisson Distribution? In analyzing hits by V-1 buzz bombs in World War II, South London was subdivided into regions, each with an area of 0.25\(k{m^2}\). Shown below is a table of actual frequencies of hits and the frequencies expected with the Poisson distribution. (The Poisson distribution is described in Section 5-3.) Use the values listed and a 0.05 significance level to test the claim that the actual frequencies fit a Poisson distribution. Does the result prove that the data conform to the Poisson distribution?

Number of Bomb Hits

0

1

2

3

4

Actual Number of Regions

229

211

93

35

8

Expected Number of Regions

(from Poisson Distribution)

227.5

211.4

97.9

30.5

8.7

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