Fundamentals of Statistics (5th Edition)
Fundamentals of Statistics (5th Edition)
5th Edition
ISBN: 9780134508306
Author: Michael Sullivan III
Publisher: PEARSON
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Chapter 12, Problem 1RE

Roulette Wheel A pit boss suspects that a roulette wheel is out of balance. A roulette wheel has 18 black slots, 18 red slots, and 2 green slots. The pit boss spins the wheel 500 times and records the following frequencies:

Outcome Frequency
Black 233
Red 237
Green 30

Is the wheel out of balance? Use the α = 0.05 level of significance.

Expert Solution & Answer
Check Mark
To determine

To test: Whether the wheel is balanced at α=0.05 level of significance.

Answer to Problem 1RE

There is not enough evidence to conclude that the wheel is balanced at α=0.05 level of significance.

Explanation of Solution

Given info:

The data represent the frequencies recorded when a pit boss spins the wheel 500 times.

Calculation:

The test hypotheses are given below:

Null hypothesis:

H0: The wheel is balanced.

Alternative hypothesis:

H1: The wheel is not balanced.

Decision rule:

If χ02>χα2 , then reject the null hypothesis.

The requirements that must be satisfied to perform a goodness-of-fit test are as follows:

  • All the expected frequencies are greater than or equal to 1. That is, all Ei1 where Ei represents the expected counts of the category i.
  • Not more than 20% of the expected frequencies are less than 5.

The roulette wheel has 18 black slots, 18 red slots, and 2 green slots. The proportion of each slot is obtained by dividing each of the number of slots with the total number of slots.

The total number of slots is 38. The expected proportion of black is 1838=919 . The expected proportion of red is 1838=919 . The expected proportion of green is 238=119 .

The expected count for each category is obtained by multiplying the obtained proportion for each color with the total number of spins.

The expected count of black is as follows:

Ei=500(919)=236.842

The expected count of red is as follows:

Ei=500(919)=236.842

The expected count of green is as follows:

Ei=500(119)=26.316

Since all Ei1 , the requirements for the goodness-of-fit test are satisfied.

The test statistic for the goodness-of-fit test is given as follows:

χ02=(OiEi)2Ei

The test statistic can be obtained as follows:

Outcome Oi Ei (OiEi)2 (OiEi)2Ei
Red 233 236.842 14.76096 0.062324
Black 237 236.842 0.024964 0.000105
Green 30 26.316 13.57186 0.515726

Therefore, the test statistic is as follows:

χ02=0.062324+0.000105+0.515726=0.578156=0.578

Thus, the value of the test statistic is 0.578.

Degrees of freedom:

Here, there are three categories.

The degree of freedom is as follows:

k1=31=2

Thus, the degree of freedom for the test statistic is 2.

Critical value:

The level of significance is α=0.05 .

Therefore, the critical value is obtained from Table VIII of ‘Chi-square Distribution Area to the Right of critical value’.

Procedure:

  • Locate 2 in the column of degrees of freedom in Table VIII.
  • Take the value corresponding to α=0.05 .

From the table, the critical value is 5.991.

Thus, the critical value using α=0.05 is 5.991.

Conclusion:

Here, the chi-square test statistic is 0.578, and the critical value is 5.991.

The chi-square test statistic is less than the critical value.

That is χ02(=0.578)<χα2(=5.991) .

Therefore, the null hypothesis is not rejected.

Thus, there is not enough evidence to conclude that the wheel is balanced at α=0.05 level of significance.

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Chapter 12 Solutions

Fundamentals of Statistics (5th Edition)

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