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Chapter 11: Problem 2

Roulette Casinos are required to verify that their games operate as advertised. American roulette wheels have 38 slots -18 red, 18 black, and 2 green. In one casino, managers record data from a random sample of 200 spins of one of their American roulette wheels. The one-way table below displays the results. $$ \begin{array}{lccc} \hline \text { Color: } & \text { Red } & \text { Black } & \text { Green } \\\ \text { Count: } & 85 & 99 & 16 \\ \hline \end{array} $$ (a) State appropriate hypotheses for testing whether these data give convincing evidence that the distribution of outcomes on this wheel is not what it should be. (b) Calculate the expected counts for each color. Show your work.

Short Answer

Expert verified
Set hypotheses: H_0: Outcomes match expected distribution. Calculate expected counts: Red 95, Black 95, Green 11.

Step by step solution

01

Understand the Expected Distribution

In an ideal American roulette wheel, there are 38 slots with equal chances of landing on each slot. This means that the probability of each color occurring should be divided as follows: 18 red, 18 black, and 2 green out of 38 slots. Thus, the expected distribution should be equivalent to the expected probabilities of the wheel's outcomes.
02

Set Up the Hypotheses

The null hypothesis ( H_0 ) states that the distribution of outcomes matches the expected distribution of an American roulette wheel (18 red, 18 black, 2 green). The alternative hypothesis ( H_a ) states that the distribution of outcomes is different from what is expected.
03

Calculate Expected Counts for Each Color

The expected count for each color can be found by multiplying the total number of spins by the probability of landing on each color.- For Red: \[ E_{Red} = 200 \times \left( \frac{18}{38} \right) \approx 94.7368 \]- For Black: \[ E_{Black} = 200 \times \left( \frac{18}{38} \right) \approx 94.7368 \]- For Green: \[ E_{Green} = 200 \times \left( \frac{2}{38} \right) \approx 10.5263 \]Round these to reasonable whole numbers based on context and use them for further analysis.
04

Interpret the Results

Compare the observed counts (85 Red, 99 Black, 16 Green) against the expected counts (approximately 95 Red, 95 Black, 11 Green). You would typically conduct a chi-square goodness-of-fit test to determine the likelihood of the observed distribution arising if the null hypothesis were true. However, further calculations are needed for testing significance and concluding whether the hypothesis should be rejected or not.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

American roulette
In American roulette, a distinct wheel is used that presents players with 38 numbered slots. This version of roulette slightly differs from its European cousin, which contains only 37 slots. In American roulette, the numbers are divided into three colors: 18 slots are marked red, 18 are marked black, and two additional slots are marked green. These green slots are labeled 0 and 00.
This unique setup impacts the game's odds and probabilities. Consequently, calculating probabilities and outcomes in American roulette involves considering these 38 slots rather than just alternate red and black options. Understanding this distinction is essential for interpreting game outcomes and probabilities, especially when analyzing a series of spins in a casino context.
Statistical hypothesis testing
Statistical hypothesis testing provides a framework for determining if observed data significantly deviates from what we expect under a given hypothesis. In the context of the American roulette wheel,
we craft two hypotheses to analyze:
  • Null hypothesis ( $H_0$): The distribution of outcomes on the roulette wheel precisely fits the expected probabilities (18 red, 18 black, 2 green).
  • Alternative hypothesis ( $H_a$): The distribution of outcomes does not match the expected probabilities.
These hypotheses form a foundation for performing a chi-square goodness-of-fit test, which will quantify any discrepancies between observed and expected data.
By comparing observed counts to what we theorize should happen under the null hypothesis, we can conclude if the differences are due to natural variation or if the spinning wheel isn't operating as expected, suggesting a deviation from fair play.
Expected distribution
The expected distribution is a critical concept in probability and statistical analysis. It outlines what should theoretically occur based on known conditions. In our American roulette example,
we determine the expected outcomes based on the wheel鈥檚 uniform slot distribution. With 18 red, 18 black, and 2 green slots, each spin should yield:
  • 18/38 chance of landing on red or black (approximately 47.37% each)
  • 2/38 probability of landing on green (approximately 5.26%)
To find the expected counts over 200 spins, multiply these probabilities by the total spins:
For red and black: \(E_{color} = 200 \times (\frac{18}{38}) \approx 94.74\)
For green: \(E_{green} = 200 \times (\frac{2}{38}) \approx 10.53\)
These values guide hypotheses testing and indicate whether real-world data aligns with theoretical expectations.
Probability calculation
Understanding probability calculation is necessary in interpreting the outcome of a roulette wheel spin. Probability quantifies the likelihood of an event occurring and is calculated as the ratio of favorable outcomes to total outcomes. In American roulette,
the probability of landing on a red or black slot is \(P_{Red, Black} = \frac{18}{38}\)
and for a green slot, it's \(P_{Green} = \frac{2}{38}\).
When conducting 200 spins, to compute expected counts for each color:
  • Multiply the total spins by each color's probability (e.g., Red: \(200 \times \frac{18}{38} \approx 94.74\)
    • li>Round these results when necessary for further analysis.
The calculations offer insights and establish a baseline to compare observed outcomes, essential for executing a chi-square goodness-of-fit test to statistically infer on fair play.

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Most popular questions from this chapter

Skittles Statistics teacher Jason Molesky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: "The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor." (a) State appropriate hypotheses for a significance test of the company's claim. (b) Find the expected counts for a bag of Skittles with 60 candies. (c) How large a \(\chi^{2}\) statistic would you need to have significant evidence against the company's claim at the \(\alpha=0.05\) level? At the \(\alpha=0.01\) level? (d) Create a set of observed counts for a bag with 60 candies that gives a \(P\) -value between 0.01 and \(0.05 .\) Show the calculation of your chi-square statistic.

Benford's law Faked numbers in tax returns, invoices, or expense account claims often display patterns that aren't present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford's law. \({ }^{3}\) Call the first digit of a randomly chosen record \(X\) for short. Benford's law gives this probability model for \(X\) (note that a first digit can't be 0 ): $$ \begin{array}{lccccccccc} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Probability: } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\ \hline \end{array} $$ A forensic accountant who is familiar with Benford's law inspects a random sample of 250 invoices from a company that is accused of committing fraud. The table below displays the sample data. $$ \begin{array}{lcrrrrrrrr} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Count: } & 61 & 50 & 43 & 34 & 25 & 16 & 7 & 8 & 6 \\ \hline \end{array} $$ (a) Are these data inconsistent with Benford's law? Carry out an appropriate test at the \(\alpha=0.05\) level to support your answer. If you find a significant result, perform a follow-up analysis. (b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious?

Students and catalog shopping What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university: \(^{26}\) $$ \begin{array}{lcc} \hline & \text { American } & \text { Asian } \\ \text { Save time } & 29 & 10 \\ \text { Easy } & 28 & 11 \\ \text { Low price } & 17 & 34 \\ \text { Live far from stores } & 11 & 4 \\ \text { No pressure to buy } & 10 & 3 \\ \hline \end{array} $$ (a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?

Aw, nuts! A company claims that each batch of its deluxe mixed nuts contains \(52 \%\) cashews, \(27 \%\) almonds, \(13 \%\) macadamia nuts, and \(8 \%\) brazil nuts. To test this claim, a quality-control inspector takes a random sample of 150 nuts from the latest batch. The one-way table below displays the sample data. $$ \begin{array}{lcccc} \hline \text { Nut: } & \text { Cashew } & \text { Almond } & \text { Macadamia } & \text { Brazil } \\ \text { Count: } & 83 & 29 & 20 & 18 \\ \hline \end{array} $$ (a) State appropriate hypotheses for performing a test of the company's claim. (b) Calculate the expected counts for each type of nut. Show your work.

Exercises 51 to 55 refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ Which of the following would be the most appropriate type of graph for these data? (a) A bar chart showing the marginal distribution of opinion about marriage (b) A bar chart showing the marginal distribution of gender (c) A bar chart showing the conditional distribution of gender for each opinion about marriage (d) A bar chart showing the conditional distribution of opinion about marriage for each gender (e) Dotplots that display the number in each opinion category for each gender
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