Question

In Exercise 3.129 on page \(254,\) we see that the home team was victorious in 70 games out of a sample of 120 games in the FA premier league, a football (soccer) league in Great Britain. We wish to investigate the proportion \(p\) of all games won by the home team in this league. (a) Use StatKeyor other technology to find and interpret a \(90 \%\) confidence interval for the proportion of games won by the home team. (b) State the null and alternative hypotheses for a test to see if there is evidence that the proportion is different from 0.5 . (c) Use the confidence interval from part (a) to make a conclusion in the test from part (b). State the confidence level used. (d) Use StatKey or other technology to create a randomization distribution and find the p-value for the test in part (b). (e) Clearly interpret the result of the test using the p-value and using a \(10 \%\) significance level. Does your answer match your answer from part (c)? (f) What information does the confidence interval give that the p-value doesn't? What information does the p-value give that the confidence interval doesn't? (g) What's the main difference between the bootstrap distribution of part (a) and the randomization distribution of part (d)?

Short answer

The 90% confidence interval does not contain \(0.5\), therefore, there's evidence to suggest that the proportion of victories is not equal to 0.5 under the significance level of 10%. Both bootstrap and randomization distributions are used to make statistical inferences, the difference lies in their applications. Bootstrap is used for estimating sampling distribution of a statistic, while a randomization distribution is for assessing possible outcomes under a null hypothesis.

Step-by-step solution

Computing the 90% Confidence Interval

First, calculate the sample proportion \(\hat{p}\) by dividing the number of successes (games won, here 70) by the number of trials (total games, here 120). This gives \( \hat{p} = \frac{70}{120} = 0.583 \). Next, the 90% confidence interval can be calculated using the formula \(\hat{p} \pm z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\), where \(z*\) value corresponds to the z score of a 90% confidence interval, which is 1.645. Inputting the values, we get the confidence interval.

Formulate Hypotheses

The null hypothesis \(H_0\) is that the true proportion is \(0.5\) while the alternative hypothesis \(H_A\) is that the true proportion is not \(0.5\). This can be formally stated as: \(H_0: p = 0.5\) and \(H_A: p ≠ 0.5\)

Conclusion Based On Confidence Interval

We use the 90% confidence interval computed from step 1 to draw a conclusion about our hypothesis. If the value stated in the null hypothesis is within the confidence interval, then it is reasonable to state that there's no evidence against \(H_0\). If, however, it isn't included in the interval, then \(H_0\) is in doubt.

Find P-value By Using Randomization Distribution

Now, use StatKey or other technology to create a randomization distribution. The p-value of the test can be found by looking at the proportion of randomized samples that are as extreme or more extreme than the observed statistics. If the p-value is less than the significance level (\(10\%\)), reject \(H0\), otherwise, do not reject.

Interpreting the Test Result

The interpretation of the p-value depends on its size. A smaller p-value indicates stronger evidence against the null hypothesis. If the p-value is less than the pre-determined significance level of \(10\%\), we reject the null hypothesis. This means that the proportion of home team wins is significantly different from \(0.5\). If the p-value is greater than \(10\%\), we fail to reject the null hypothesis, indicating less evidence that the true proportion is different from \(0.5\).

Distinguishing Between Confidence Interval and P-Value

The confidence interval gives an estimated range of values which is likely to include an unknown population parameter. The p-value, on the other hand, gives us the strength of evidence in support of a statistical hypothesis and helps us determine the significance of our results.

Difference Between Bootstrap Distribution and Randomization Distribution

Bootstrap distribution is used when estimating the sampling distribution of a statistic, while a randomization sampling distribution is used to determine the possible outcomes under a null hypothesis.

Key concepts

Statistical Hypothesis Testing

In statistics, hypothesis testing is a systematic method used to evaluate assumptions called hypotheses about a parameter in a population. At the core of hypothesis testing are two hypotheses: the null hypothesis \(H_0\) and the alternative hypothesis \(H_A\). The null hypothesis represents a statement of no effect or no difference and serves as a starting point for the statistical test. In contrast, the alternative hypothesis suggests that there is an effect or difference.

In the provided exercise, the null hypothesis posits that the true proportion of games won by the home team \(p\) is 0.5, and the alternative hypothesis contends that the proportion is not 0.5. Hypothesis testing involves using sample data to determine whether there is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. The outcome of this testing process is often informed by the calculation of a p-value, which depends on the chosen significance level and the test statistic computed from the sample data.

Process of Hypothesis Testing

Typically, the steps in hypothesis testing include:

• Formulating the null and alternative hypotheses.
• Choosing a significance level (alpha), which is the threshold for deciding when to reject the null hypothesis.
• Calculating the test statistic from the sample data.
• Comparing the test statistic to a critical value or calculating a p-value.
• Making a decision to either reject or fail to reject the null hypothesis based on the test statistic or the p-value.

P-value Interpretation

The p-value is a probability that measures the evidence against the null hypothesis. It is calculated under the assumption that the null hypothesis is true. Specifically, the p-value is the probability of observing a test statistic at least as extreme as the value observed in the sample, if the null hypothesis were correct.

If the p-value is small, it suggests that the observed data is unlikely under the null hypothesis, and thus provides evidence against it. Conversely, a large p-value indicates that the observed data is more consistent with the null hypothesis, and hence does not provide strong evidence against it.

Interpreting the P-value

What distinguishes p-value interpretation from other statistical measures is its direct relationship with the significance level, \(\alpha\). If the p-value is less than or equal to the chosen significance level, it suggests statistical significance, prompting the rejection of the null hypothesis. For the exercise in question, a significance level of 10% \(\alpha = 0.10\) was chosen. This means that if the computed p-value from the randomization distribution is below 0.10, the evidence suggests that the proportion of games won by the home team is significantly different from 0.5.

Randomization Distribution

Randomization distribution is a concept related to permutation tests or randomization tests, which are non-parametric methods for hypothesis testing. It is the distribution of a test statistic that would be observed if the null hypothesis were true and the observed data were due to random variation. This distribution is created by randomly rearranging the data points and recalculating the test statistic multiple times. By generating a series of randomized samples, we can observe variations in the test statistic purely by chance.

In our exercise, creating a randomization distribution allows us to simulate the statistical behavior under the null hypothesis, assisting us in understanding the probability of observing a test statistic as extreme as the one calculated from the actual data. This contributes to the calculation of a p-value, which is a probability indicating how often a test statistic as extreme as the observed one occurs when the null hypothesis is true.

Application in the Exercise

The creation of the randomization distribution involves the use of statistical software to assess the proportion of randomized samples with outcomes as or more extreme than the home team's observed winning proportion. This simulative approach not only supports the calculation of a p-value but also enhances the student's grasp of the underlying principles of statistical hypothesis testing and significance testing.

Statistical Significance

The concept of statistical significance helps determine whether the results of a study or an experiment reflect a true effect or are merely due to random chance. Statistical significance is typically asserted when the p-value falls below a predetermined threshold, known as the significance level \(\alpha\). The smaller the p-value, the stronger the evidence against the null hypothesis.

Significance levels are often chosen conventionally at 5% (0.05), 1% (0.01), or 10% (0.10), though these values are not fixed and can vary based on the context of the study. Declaring statistical significance implies a certain degree of confidence that the results are real and not a fluke, prompting researchers to reject the null hypothesis in such cases.

Understanding Statistical Significance in Practice

For the exercise regarding the FA premier league games, we are concerned with a 10% significance level. This means we consider results to be statistically significant if the p-value is 0.10 or less. However, it's crucial not to equate statistical significance with importance or practical significance. Statistical significance mainly addresses the reliability of results, while practical significance considers their real-world relevance. Thus, education on this topic emphasizes the correct interpretation of statistical tests and the limitations of the p-value in representing the practical implications of the findings.