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See exercise \(10.21\) for an explanation of playing with the dreidel. This time the family used a plastic dreidel and got the following outcomes. The four outcomes are believed to be equally likely (that is, has a uniform probability distribution). Determine whether the plastic dreidel does not follow the uniform distribution using a significance level of \(0.05\). $$\begin{array}{cccc}\text { gimmel } & \text { hey } & \text { nun } & \text { shin } \\ 11 & 9 & 11 & 9\end{array}$$

Short Answer

Expert verified
There is not enough evidence to support the claim that the plastic dreidel is not fair and does not follow a uniform distribution at the \(0.05\) level of significance.

Step by step solution

01

State The Null and Alternative Hypotheses

The null hypothesis \(H_0\) is that the outcomes are uniformly distributed (each has an equal chance of occurrence). The alternative hypothesis \(H_1\) is that the outcomes are not uniformly distributed.
02

Calculate The Expected Frequencies Under \(H_0\)

Under \(H_0\), each of the four outcomes (gimmel, hey, nun, and shin) should occur with equal probability, hence the expected frequency for each outcome is the total number of spins divided by the number of outcomes. The total number of spins is given by sum of the observed frequencies for the four outcomes; \(11 + 9 + 11 + 9 = 40\). Hence, the expected frequency for each outcome is \(40/4 = 10\)
03

Calculate The Test Statistic

The test statistic for Chi-Squared Goodness of Fit test is given by \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\), where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. Substituting the observed and expected frequencies for each outcome into this formula gives \(\chi^2 = \frac{(11 - 10)^2}{10} + \frac{(9 - 10)^2}{10} + \frac{(11 - 10)^2}{10} + \frac{(9 - 10)^2}{10} = 0.4\)
04

Determine The Critical Value

The critical value of \(\chi^2\) for a significance level of \(0.05\) and degrees of freedom given by number of outcomes minus 1 (4 - 1 = 3) can be obtained from standard statistical tables or online calculators. The critical value is \(7.815\)
05

Make A Decision

Since the computed test statistic \(\chi^2 = 0.4\) is less than the critical value of \(7.815\), the decision is to not reject the null hypothesis \(H_0\). This means that the evidence does not support the claim that the plastic dreidel does not follow a uniform distribution at the \(0.05\) level of significance.

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Key Concepts

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

Uniform Distribution
A uniform distribution is a type of probability distribution where all outcomes are equally likely. Think of it like a perfectly balanced die - each side has an equal chance of landing face up. In the context of the dreidel game, a uniform distribution means each outcome (gimmel, hey, nun, and shin) should ideally appear the same number of times over many spins. When you are told something has a uniform distribution, it means that there are no biases or preferences towards any outcome. Every possibility is on a level playing field. This idea is central when we perform tests like the Chi-Squared Goodness of Fit, as we assume uniformity to test for deviations.
Null and Alternative Hypotheses
The null hypothesis ( \( H_0 \)) and the alternative hypothesis ( \( H_1 \)) are essential components of hypothesis testing.
  • **Null Hypothesis ( \( H_0 \)):** This is a statement asserting that there is no effect or no difference. For the dreidel, it states that the outcomes are uniformly distributed.
  • **Alternative Hypothesis ( \( H_1 \)):** This posits that there is an effect or a difference. In our case, it suggests the outcomes are not uniformly distributed.
When conducting a test, you assume the null hypothesis is true until evidence suggests otherwise. You only consider switching to the alternative hypothesis if the data strongly indicates a deviation from the null hypothesis claims.
Significance Level
The significance level often denoted by \( \alpha \), is the threshold for determining whether to reject the null hypothesis. It's typically set at values like 0.05, which is used in our dreidel exercise.This number represents the risk you are willing to take in rejecting the null hypothesis when it is actually true. A 0.05 significance level means there's a 5% chance of making such a mistake. It acts like a balance point - the lower it is, the stricter you are in assuming changes are not due to random chance.With the dreidel, if our computed test statistic exceeds the critical value corresponding to this significance level, we would reject the null hypothesis, concluding the outcomes aren't as uniform as expected.
Expected Frequency
Expected frequency is a cornerstone of the Chi-Squared test. It is the number of occurrences you would expect to see for each outcome if the null hypothesis were correct. To calculate expected frequency, you divide the total number of observations by the number of possible outcomes. In the dreidel exercise, there are 40 spins divided by 4 outcomes, yielding an expected frequency of 10 for each outcome. Comparing these expected frequencies against what was actually observed (observed frequencies) allows us to compute the Chi-Squared statistic, which helps in determining if the deviations from expected frequencies can be attributed to chance or not.

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Most popular questions from this chapter

Refer to the description in exercise 10.71. There were 22 trials with only cockroaches (no robots) that went under one shelter. In 16 of these 22 trials, the group chose the darker shelter, and in 6 of the 22 the group chose the lighter shelter. There were 28 trials with a mixture of real cockroaches and robots that all went under one shelter. In 11 of these trials, the group chose the darker shelter, and in 17 the group chose the lighter shelter. The robot cockroaches were programmed to choose the lighter shelter (as well as preferring groups; Halloy et al. 2007 ) Is the introduction of robot cockroaches associated with the type of shelter when the group went under one shelter? Assume cockroaches were randomly sampled from some meaningful population of cockroaches. a. Use the chi-square test to see whether the presence or absence of robots is associated with whether they went under the darker or the brighter shelter. Use a significance level of \(0.05\) b. Do Fisher's Exact Test with the data. If your software does not do Fisher's Exact Test, search the Internet for a Fisher's Exact Test calculator and use it. Report the p-value and your conclusion. c. Compare the p-values for parts a and b. Which do you think is the more accurate procedure? The p-values that result from the two methods in this question are closer than the p-values in the previous question. Why do you think that is?

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