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Deflated Footballs? Patriots In the 2015 AFC Championship game, there was a charge the New England Patriots deflated their footballs for an advantage. The balls should be inflated to between \(12.5\) and \(13.5\) pounds per square inch. The measurements were \(11.50,10.85,11.15,10.70,11.10,11.60,11.85,11.10,10.95\) \(10.50\), and \(10.90\) psi (pounds per square inch). (Source: http:// online.wsj.com/public/resources/documents/Deflategate.pdf) a. Test the hypothesis that the population mean is less than \(12.5\) psi using a significance level of \(0.05\). State clearly whether the Patriots' balls are deflated or not. Assume the conditions for a hypothesis test are satisfied. b. Use the data to construct a \(95 \%\) confidence interval for the mean psi for the Patriots' footballs. How does this confidence interval support your conclusion in part a?

Short Answer

Expert verified
Given the conducted test statistics and the confidence interval, verify whether the Patriots' footballs were deflated. If the p-value is \(< 0.05\) and 12.5 psi is not included in the confidence interval, it can be concluded that the balls were deflated. Otherwise, there's not enough evidence to support the claim of deflation. Note that only a statistical approach has been applied here, which provides evidence but not absolute proof.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis \(H_0\) is that the population mean is \(12.5\) psi, which is the standard pressure requirement. The alternative hypothesis \(H_1\) is that the population mean pressure of the Patriots' footballs is less than \(12.5\) psi. Therefore, \(H_0: \mu = 12.5\) and \(H_1: \mu < 12.5\).
02

Calculate Mean and Standard Deviation for Sample

Use the provided measurements to calculate the sample mean \(\overline{x}\) and the sample standard deviation \(s\). The mean of the sample is the sum of all data points divided by the number of data points. Standard deviation gauges the spread of data points from the mean.
03

Conduct a One-Sample t-test

Use these values along with the sample size \(n\) to calculate the t-score using the formula: \(t = \frac{\overline{x} - \mu_0}{s / \sqrt{n}}\), where \(\mu_0\) is the value from the null hypothesis, in this case \(12.5\). After, calculate the p-value associated with the observed t-score for a one-tailed test with degrees of freedom \(n-1\).
04

Decision Making and Conclusion

Make decision by comparing the p-value to the given significance level \(\alpha = 0.05\). If p-value \(< \alpha\), reject \(H_0\) in favor of \(H1\), concluding that the footballs are under-inflated. If p-value \(> \alpha\), do not reject \(H_0\), and the result is inconclusive about the Patriots' football deflation.
05

Construct a 95% Confidence Interval

For part b, use the sample mean, standard deviation, and the t value associated with a two-tailed 95% confidence interval and degrees of freedom \(n-1\) to construct a confidence interval for the mean psi of the Patriots' footballs using the formula: \(\overline{x} \pm t^*(s/\sqrt{n})\).
06

Interpret the Confidence Interval

If 12.5 psi, belonging to desired pressure, is not in this interval, this supports the conclusion from the t-test that the balls are deflated, and vice versa.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values, derived from sample data, that is likely to contain the value of an unknown population parameter. More specifically, a 95% confidence interval suggests that if we were to take 100 different samples and compute a confidence interval for each sample, then approximately 95 of the 100 confidence intervals will contain the population mean.

In the context of the deflated footballs, constructing a 95% confidence interval for the mean psi of the Patriots' footballs gives us a range of values that we are 95% confident contains the true average psi of all their footballs. If this confidence interval does not contain the league's standard pressure requirement of 12.5 psi, it provides statistical support towards the claim that the Patriots' footballs were indeed deflated.
One-sample t-test
The one-sample t-test is used to determine whether there is a statistically significant difference between the mean of a single sample and a known population mean. This test is particularly useful when the population standard deviation is unknown and the sample size is relatively small, which often necessitates the use of the t-distribution.

In our exercise, the one-sample t-test compares the average psi of the sample of footballs to the standard psi of 12.5. The formula for the t-test is given by \[t = \frac{\overline{x} - \mu_0}{s / \sqrt{n}}\], where \overline{x} is the sample mean, \mu_0 is the population mean asserted by the null hypothesis (in this case, 12.5 psi), s is the sample standard deviation, and n is the sample size. The resulting t-score and corresponding p-value will help us determine whether the observed average psi is statistically significantly different from the expected 12.5 psi.
Statistical Significance
Statistical significance is a determination that the observed difference between groups or the relationship between variables is unlikely to have occurred purely by chance, given the null hypothesis is true. Significance is usually determined by the p-value, which, in simplest terms, is the probability of observing your data or something more extreme under the assumption that the null hypothesis is true.

In the case of the deflated footballs problem, a p-value that is less than the chosen significance level of 0.05 (or 5%) would indicate that there is statistically significant evidence to reject the null hypothesis. This would mean there is less than a 5% chance of obtaining a sample mean of 11.5 psi (or lower) if the true population mean is the required 12.5 psi. Therefore, a p-value less than 0.05 would suggest that the footballs were statistically significantly deflated with respect to the league standard.

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Most popular questions from this chapter

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