91Ó°ÊÓ

Arnold Palmer and Tiger Woods are two of the best golfers to ever play the game. To show how these two golfers would compare if both were playing at the top of their game, the following sample data provide the results of 18 -hole scores during a PGA tournament competition. Palmer's scores are from his 1960 season, while Woods' scores are from his 1999 season (Golf Magazine, February 2000). $$Palmer, 1960 \quad Woods, 1999$$ $$ \begin{array}{ll} n_{1}=112 & n_{2}=84 \\ \bar{x}_{1}=69.95 & \bar{x}_{2}=69.56 \end{array} $$ Use the sample results to test the hypothesis of no difference between the population mean 18-hole scores for the two golfers. a. Assume a population standard deviation of 2.5 for both golfers. What is the value of the test statistic? b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.01,\) what is your conclusion?

Step by step solution

01

State the Hypotheses

We need to test the null hypothesis that there is no difference between the population mean scores of Palmer and Woods. The null hypothesis and the alternative hypothesis are \[ H_0: \mu_1 - \mu_2 = 0 \] \[ H_a: \mu_1 - \mu_2 eq 0 \] where \( \mu_1 \) and \( \mu_2 \) are the population mean scores for Palmer and Woods, respectively.

02

Calculate the Test Statistic

We use the formula for the test statistic for two means with known population standard deviations:\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}}}\]where \( \bar{x}_1 = 69.95 \), \( \bar{x}_2 = 69.56 \), \( \sigma = 2.5 \), \( n_1 = 112 \), and \( n_2 = 84 \). Inserting these values:\[ z = \frac{69.95 - 69.56}{\sqrt{\frac{2.5^2}{112} + \frac{2.5^2}{84}}} \approx \frac{0.39}{0.3708} \approx 1.052\]

03

Find the p-value

The test statistic found in Step 2 is \( z = 1.052 \). Using a standard normal distribution table, we find the area to the right of 1.052. Since it is a two-tailed test, multiply this area by 2:\[\text{p-value} = 2 \times P(Z > 1.052) \approx 2 \times 0.1469 = 0.2938\]

04

Compare p-value with α

Given \( \alpha = 0.01 \), compare it with the p-value from Step 3. Since \( 0.2938 > 0.01 \), we do not reject the null hypothesis \( H_0 \).

05

Draw Conclusion

Since we do not reject the null hypothesis, we conclude there is not enough statistical evidence at the \( \alpha = 0.01 \) level to argue that there is a significant difference in the mean 18-hole scores between Palmer and Woods at their peak performance.

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Key Concepts

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

Population Mean

Understanding the concept of population mean is crucial in statistical analysis. The population mean is the average score of an entire group or population. In hypothesis testing, it helps us compare different groups to determine if there is a significant difference.

For instance, in the exercise comparing Arnold Palmer's and Tiger Woods' golf scores, the population means are the average 18-hole scores for both golfers during their peak playing years. Identifying these means enables us to determine if any difference truly exists between the golfers' performances or if they performed similarly across their respective sample sizes.

• The mean for Palmer in 1960 was 69.95,
• while for Woods in 1999, it was 69.56.

These averages, while close, provide the basis for running statistical tests to confirm any significant differences.

Test Statistic

The test statistic is a critical component in hypothesis testing. It indicates how far your sample statistic is from the null hypothesis' expected value.

To calculate it, we use specific formulas that help quantify differences. For comparing two means with known population standard deviations, the formula involves subtracting the means and dividing by the standard error of the difference between them.The formula used in this exercise was:
\[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}}} \]
where:

• \( \bar{x}_1 = 69.95 \) is Palmer's mean score,
• \( \bar{x}_2 = 69.56 \) is Woods' mean score,
• \( \sigma = 2.5 \) is the standard deviation,
• \( n_1 = 112 \) and \( n_2 = 84 \) are the sample sizes.

The test statistic calculated came out to be approximately 1.052, which is essential for finding the p-value.

P-value

The p-value helps us determine the significance of the test statistic obtained. It quantifies the probability of observing the test results under the null hypothesis.

A smaller p-value indicates stronger evidence against the null hypothesis. Typically, a p-value less than 0.05 suggests a rejection of the null hypothesis for a standard significance level. However, the problem at hand used a confidence level \(\alpha = 0.01\).
In the given problem:

• The computed test statistic was 1.052.
• The p-value calculated was approximately 0.2938.
• This indicates a 29.38% chance of observing such an extreme result when the null hypothesis is true.

Thus, the high p-value means there is insufficient evidence to show significant differences between the two population means.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is an initial claim we test using data. It generally asserts that no effect or no difference exists. In the exercise at hand:

\[ H_0: \mu_1 - \mu_2 = 0 \]
This states that there is no difference in the population means of the 18-hole golf scores for Palmer and Woods.
Testing the null hypothesis involves comparing the p-value with the predetermined significance level (\(\alpha\)). In this exercise:

• The significance level was set at \(\alpha = 0.01\).
• Given a p-value of 0.2938, which is greater than \(\alpha\),
• we do not reject the null hypothesis.

This conclusion indicates that, statistically, we don't have enough proof to say Palmer and Woods' performances differed significantly at their peaks. Hypothesis testing uses such methods to make informed decisions based on statistical evidence.