91Ó°ÊÓ

An extensive study of the cost of health care in the United States presented data showing that the mean spending per Medicare enrollee in 2003 was \(\$ 6883\) (Money, Fall 2003 ). To investigate differences across the country, a researcher took a sample of 40 Medicare enrollees in Indianapolis. For the Indianapolis sample, the mean 2003 Medicare spending was \(\$ 5980\) and the standard deviation was \(\$ 2518\) a. State the hypotheses that should be used if we would like to determine whether the mean annual Medicare spending in Indianapolis is lower than the national mean. b. Use the preceding sample results to compute the test statistic and the \(p\) -value. c. Use \(\alpha=.05 .\) What is your conclusion? d. Repeat the hypothesis test using the critical value approach.

Step by step solution

01

State the Hypotheses

To determine if the mean annual Medicare spending in Indianapolis is lower than the national mean, we will set up the null and alternative hypotheses.- Null hypothesis \((H_0)\): The mean spending in Indianapolis is equal to the national mean, \(\mu = 6883\).- Alternative hypothesis \((H_a)\): The mean spending in Indianapolis is less than the national mean, \(\mu < 6883\).

02

Compute the Test Statistic

The test statistic can be calculated using the formula for the t-statistic for a sample: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Where:\(\bar{x} = 5980\) is the sample mean spending in Indianapolis.\(\mu = 6883\) is the national mean spending.\(s = 2518\) is the sample standard deviation.\(n = 40\) is the sample size.Substituting these values, we have:\[ t = \frac{5980 - 6883}{\frac{2518}{\sqrt{40}}} \approx \frac{-903}{398.21} \approx -2.27 \]

03

Compute the P-value

The p-value is calculated to determine the probability of observing a test statistic as extreme as the calculated \( t \) value, assuming the null hypothesis is true. Using a t-distribution with \( n-1 = 39 \) degrees of freedom, we find that the p-value corresponding to \( t = -2.27 \) is approximately \( 0.014 \) (using a t-table or statistical software).

04

Conclusion using P-value

With \( \alpha = 0.05 \), we reject the null hypothesis if the p-value is less than \( \alpha \). Since the p-value \( 0.014 \) is less than \( 0.05 \), we reject the null hypothesis. There is sufficient evidence to conclude that the mean annual Medicare spending in Indianapolis is lower than the national mean.

05

Critical Value Approach

In the critical value approach, we compare the test statistic to the critical t-value for the given significance level \( \alpha = 0.05 \) and \( 39 \) degrees of freedom. For a one-tailed test with \( \alpha = 0.05 \), the critical t-value is approximately \( -1.685 \). Since the test statistic \( t = -2.27 \) is less than \( -1.685 \), we reject the null hypothesis.

06

Conclusion using Critical Value

The conclusion derived from the critical value approach aligns with the p-value approach. Since \( t = -2.27 \) is more extreme than \( -1.685 \), we again conclude that the mean annual Medicare spending in Indianapolis is significantly lower than the national mean.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

t-test

The t-test is a statistical test used to compare the mean of a sample to a known value or to another sample's mean. In our exercise, we use the t-test to determine whether the mean Medicare spending in Indianapolis is different from the national average. This helps us understand whether the observed difference in spending is due to chance or if it's a significant deviation.

To compute the t-test, we use the formula: \[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] Where:

• \(\bar{x}\) is the sample mean
• \(\mu\) is the population mean
• \(s\) is the sample standard deviation
• \(n\) is the sample size

The purpose of the t-test in this context is to gather evidence on whether Indianapolis's spending is statistically lower than the national average.

p-value

The p-value is a critical component of hypothesis testing. It tells us the probability of obtaining a test statistic as extreme as the one calculated, assuming the null hypothesis is true. In simpler terms, it's the likelihood of observing what we did, or something more extreme, purely by chance.

In our exercise, a p-value of 0.014 was found. Since this value is small, it suggests that such a result is improbable under the assumption that the null hypothesis is true. By comparing the p-value to our significance level \(\alpha\), we decide whether to reject the null hypothesis. A common threshold is \(\alpha = 0.05\). If the p-value is less than \(\alpha\), as it is in this case, we reject the null hypothesis.

statistical significance

Statistical significance helps researchers decide whether their findings are unlikely to have occurred by chance. It gives weight to the results, suggesting they reflect true differences or effects in the population.

In this exercise, the finding that Medicare spending in Indianapolis is lower than the national mean is statistically significant. This conclusion is supported by both the p-value and the critical value approaches. With a p-value of 0.014 being less than 0.05, this result is less likely to be due to random variation. Hence, it's statistically significant. This means that policy makers and stakeholders can be confident in these findings.

null hypothesis

The null hypothesis is a fundamental part of hypothesis testing. It's a statement that there's no effect or no difference. In our scenario, the null hypothesis \((H_0)\) posits that the mean Medicare spending in Indianapolis is equal to the national mean of \(\$6883\).

The null hypothesis serves as the baseline comparison. Researchers use it to test against the alternative. If the data provides strong enough evidence, we can reject the null hypothesis. The process of testing a null hypothesis involves calculating statistical measures, like the t-test and p-value, to decide if the observed data significantly deviates from this baseline. Rejection of the null hypothesis indicates that the sample data shows a statistical difference from the comparison group.

alternative hypothesis

The alternative hypothesis represents what researchers aim to support or demonstrate through their analysis. It suggests there is a meaningful effect or difference. In our exercise, the alternative hypothesis \((H_a)\) is that the mean annual Medicare spending in Indianapolis is less than the national mean, i.e., \(\mu < 6883\).

Formulating an alternative hypothesis is crucial, as it provides direction to the research. By accepting the alternative hypothesis, researchers claim that the differences observed in the data are significant and not due to random chance. The p-value and the critical value test both support the acceptance of this hypothesis in our exercise, leading to the conclusion that Medicare spending in Indianapolis is indeed lower than the national mean.