/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 In Exercise 4.40 on page 236 , w... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercise 4.40 on page 236 , we introduce a series of Euchre games played between two teams: Team A and Team B. After 40 games, Team A has won 16 times and Team \(\mathrm{B}\) has won 24 times. Can we conclude that one team is better than the other? Use the normal approximation to test this. Clearly state the null and alternative hypotheses, calculate the test statistic and p-value, and interpret the result.

Short Answer

Expert verified
Based on the calculated p-value of 0.2076, which is greater than the significance level of 0.05, we cannot conclude that one team is better than the other. The wins of Team A and Team B can be attributed to chance, and not necessarily skill.

Step by step solution

01

State the hypothesis

The null hypothesis is that the teams are equally good, i.e., Team A and Team B have the same winning probability. This means that the probability of Team A winning a game, \(P(A)\), is 0.5. The alternative hypothesis is that one team is better than the other, i.e., \(P(A) \neq 0.5\).
02

Calculate the test statistic

We will use the z-test to calculate the test statistic. The sample proportion of wins for Team A, \(\hat{p}\), is 16/40 = 0.4. The standard deviation of this binomial distribution is \(\sqrt{P(A) \cdot P(B) / n}\) = \(\sqrt{0.5 \cdot 0.5 / 40}\) = 0.079. The z-score is then \((\hat{p} - P(A)) / \sigma\), therefore \((0.4 - 0.5) / 0.079 = -1.26.
03

Calculate the p-value

The p-value associated with a z-score of -1.26 using the standard normal table is 0.1038. This is a two-tailed test since the alternative hypothesis is that \(P(A) \neq 0.5\), and so we multiply this value by 2 to get a final p-value of 0.2076.
04

Interpret the result

Since the p-value (0.2076) is greater than the commonly used significance level of 0.05, we do not reject the null hypothesis. This means we do not have enough evidence to conclude that one team is better than the other.

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Key Concepts

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

Null Hypothesis
In any statistical test, the null hypothesis ( H_0 ) serves as the foundation of the experiment. It represents a statement of no effect or no difference. In the context of our Euchre game problem, the null hypothesis states that both teams, Team A and Team B, have equal skill levels.
This implies that both teams have the same probability of winning any given game. For this exercise, we can set "the probability of Team A winning a game" as 0.5. Why a probability of 0.5? This is because we assume the teams to be of equal strength, meaning each has an equal chance of winning. The null hypothesis is often assumed to be true at the beginning of the testing process. It's only when the data shows otherwise that we have reason to abandon it.
Alternative Hypothesis
The alternative hypothesis ( H_1 ) is what you seek to prove. It's contrary to the null hypothesis and generally represents the argument that there is a significant effect or difference. In Euche game terms, the alternative hypothesis suggests that one team is actually better than the other.
Mathematically, this means that the probability of Team A winning a game ( P(A) ) is not equal to 0.5. This hypothesis usually captures the research question at hand. In a two-tailed test such as this one, you're interested in changes in either direction — that is, whether Team A is better or worse than expected under the null hypothesis.
p-value
Understanding the p-value is key to deciding whether to reject the null hypothesis. The p-value indicates the likelihood of observing the test results, or something more extreme, assuming the null hypothesis is true. In our Euchre exercise, a p-value of 0.2076 was calculated for the test statistic. What does this mean? A high p-value (like 0.2076) suggests the data is consistent with the null hypothesis — making it likely that any observed effect is due to random chance. When a p-value is greater than the significance level (commonly set at 0.05), we do not reject H_0 . Thus, in this context, there's insufficient evidence to support that one team is better than the other.
z-test
The z-test is a statistical method used primarily for hypothesis testing in cases where sample sizes are large or the data is normally distributed. In this exercise, we apply a z-test to check if the winning probabilities of Team A and Team B differ significantly from each other.
Here’s a breakdown of using the z-test:
  • Calculate the sample proportion. For Team A's wins: \(\hat{p} = \frac{16}{40} = 0.4\).
  • Compute the standard deviation of the sample distribution, \(\sigma = \sqrt{P(A) \cdot P(B) / n} = 0.079\).
  • Find the z-score: \(z = \frac{(\hat{p} - P(A))}{\sigma} = \frac{(0.4 - 0.5)}{0.079} = -1.26\).
The z-score represents how many standard deviations the observed data (or sample mean) is from the expected mean under the null hypothesis. The further from zero the score is, the more it indicates the sample does not fit the null hypothesis.

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