In Exercises 5–20, 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

Bias in Clinical Trials? Researchers investigated the issue of race and equality of access to clinical trials. The following table shows the population distribution and the numbers of participants in clinical trials involving lung cancer (based on data from “Participation in Cancer Clinical Trials,” by Murthy, Krumholz, and Gross, Journal of the American Medical Association, Vol. 291, No. 22). Use a 0.01 significance level to test the claim that the distribution of clinical trial participants fits well with the population distribution. Is there a race/ethnic group that appears to be very underrepresented?

Race/ethnicity

White

non-Hispanic

Hispanic

Black

Asian/

Pacific

Islander

American Indian/

Alaskan Native

Distribution of

Population

75.6%

9.1%

10.8%

3.8%

0.7%

Number in Lung

Cancer Clinical Trials

3855

60

316

54

12

Car Repair Costs Listed below are repair costs (in dollars) for cars crashed at 6 mi/h in full-front crash tests and the same cars crashed at 6 mi/h in full-rear crash tests (based on data from the Insurance Institute for Highway Safety). The cars are the Toyota Camry, Mazda 6, Volvo S40, Saturn Aura, Subaru Legacy, Hyundai Sonata, and Honda Accord. Is there sufficient evidence to conclude that there is a linear correlation between the repair costs from full-front crashes and full-rear crashes?

Front

936

978

2252

1032

3911

4312

3469

Rear

1480

1202

802

3191

1122

739

2767

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?

Using Yates’s Correction for Continuity The chi-square distribution is continuous, whereas the test statistic used in this section is discrete. Some statisticians use Yates’s correction for continuity in cells with an expected frequency of less than 10 or in all cells of a contingency table with two rows and two columns. With Yates’s correction, we replace

\(\sum \frac{{{{\left( {O - E} \right)}^2}}}{E}\)with \(\sum \frac{{{{\left( {\left| {O - E} \right| - 0.5} \right)}^2}}}{E}\)

Given the contingency table in Exercise 9 “Four Quarters the Same as $1?” find the value of the test \({\chi ^2}\)statistic using Yates’s correction in all cells. What effect does Yates’s correction have?

The table below shows results since 2006 of challenged referee calls in the U.S. Open. Use a 0.05 significance level to test the claim that the gender of the tennis player is independent of whether the call is overturned. Do players of either gender appear to be better at challenging calls?

Was the Challenge to the Call Successful?


Yes

No

Men

161

376

Women

68

152

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