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Two random samples of earnings of professional golfers were selected. One sample was taken from the Professional Golfers Association, and the other was taken from the Ladies Professional Golfers Association. At \(\alpha=0.05\), is there a difference in the means? The data are in thousands of dollars. $$\begin{array}{rrrrr}\text { PGA } & & & & \\\\\hline 446 & 1147 & 1344 & 9188 & 5687 \\\10,508 & 4910 & 8553 & 7573 & 375 \\\\\text { LPGA } & & & & \\\\\hline 48 & 76 & 122 & 466 & 863 \\\100 & 1876 & 2029 & 4364 & 2921\end{array}$$

Short Answer

Expert verified
Conduct a two-sample t-test; decide using the calculated t-statistic and critical value at \(\alpha=0.05\).

Step by step solution

01

Define Hypotheses

We need to set up the null and alternative hypotheses. - Null hypothesis, \(H_0\): There is no difference in the means of the earnings of the PGA and LPGA golfers, \(\mu_1 = \mu_2\).- Alternative hypothesis, \(H_1\): There is a difference in the means, \(\mu_1 eq \mu_2\).
02

Gather and Describe Data

Let's list the earnings data for both groups. For PGA: \(446, 1147, 1344, 9188, 5687, 10508, 4910, 8553, 7573, 375\). For LPGA: \(48, 76, 122, 466, 863, 100, 1876, 2029, 4364, 2921\). Each group has 10 data points.
03

Calculate Sample Means

- Calculate the mean of the PGA earnings.\[ \bar{x}_1 = \frac{446 + 1147 + 1344 + 9188 + 5687 + 10508 + 4910 + 8553 + 7573 + 375}{10} \]- Calculate the mean of the LPGA earnings.\[ \bar{x}_2 = \frac{48 + 76 + 122 + 466 + 863 + 100 + 1876 + 2029 + 4364 + 2921}{10} \]
04

Calculate Sample Standard Deviations

- Use the formula for sample standard deviation:\[ s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \]- Compute for PGA and LPGA separately using their respective means from Step 3.
05

Calculate Test Statistic

We will use a two-sample t-test for independent samples. The test statistic \(t\) is given by:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes for PGA and LPGA, respectively.
06

Determine Critical Value and Decision

Using the level of significance \(\alpha = 0.05\) for a two-tailed test and degrees of freedom approximated by the smallest sample size minus one, check the critical \(t\) value from the \(t\)-distribution table. Compare the calculated \(t\) statistic with the critical value to decide whether to reject \(H_0\) or not.
07

Conclusion

Based on the comparison: - If \(|t|\) is greater than the critical value, reject the null hypothesis, suggesting a significant difference in means.- If \(|t|\) is less than or equal to the critical value, fail to reject the null hypothesis, indicating no significant difference in means.

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Key Concepts

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

Two-Sample T-Test
The two-sample t-test is a statistical method used to compare the means of two independent groups to determine if there is a statistically significant difference between them. In our exercise, we are comparing the earnings of professional golfers from the PGA and LPGA.
The primary goal of the two-sample t-test is to ascertain whether the average earnings of players in these associations differ significantly. To do this, we analyze the data, calculate sample statistics such as means and standard deviations, and compute a test statistic to decide whether any observed difference is likely due to chance.
The test assumes that the data in each group is normally distributed and that the variances are equal. When performed, the t-test calculates a t-score that compares your sample data against the null hypothesis, which, in this case, states there is no difference between the two groups' means.
Sample Mean
The sample mean is an essential statistic that represents the average value of a dataset. In our context, it involves calculating the average earnings from the given set of data for each group of golfers.
For the PGA earnings, the sample mean is derived by summing up all individual earnings and dividing by the number of observations (which is 10). The formula used is:
  • \( \bar{x}_1 = \frac{\sum x}{n} \)

The LPGA sample mean is calculated similarly. These means give us a central value around which the earnings data clusters, and serve as a pivotal component in conducting further analysis, including the computation of the test statistic for the t-test.
Standard Deviation
Standard deviation is a measure of the variance or spread within a set of data. It quantifies how much the earnings of golfers vary from their respective means for the PGA and LPGA groups.
To calculate the standard deviation, we use:
  • \( s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \)

It requires us to compute the difference between each data point and the sample mean, square these differences, sum them, and divide by the number of observations minus one. Finally, taking the square root gives us the standard deviation.
The standard deviation is crucial in determining how varied the earnings are within each group and plays a critical role in calculating the t-test statistic, which assesses the credibility of our null hypothesis.
Null Hypothesis
In statistical hypothesis testing, the null hypothesis (7_0) proposes that there is no effect or no difference in the context being studied. For our exercise, the null hypothesis is that the average earnings of PGA and LPGA golfers are the same.
Formally, this can be expressed as:
  • \( H_0: \mu_1 = \mu_2 \)

The purpose of setting a null hypothesis is to establish a baseline assumption so that the analysis can either provide support for or challenge this assumption. If the data suggests rejecting the null hypothesis, it indicates a potential significant difference between the two groups, leading us to consider the alternative hypothesis.
Alternative Hypothesis
The alternative hypothesis (7_1) contradicts the null hypothesis and suggests that there is an effect or a difference. For this study, the alternative hypothesis posits that the average earnings of PGA and LPGA golfers are not the same.
Mathematically, it can be written as:
  • \( H_1: \mu_1 eq \mu_2 \)

The alternative hypothesis is what we aim to support through our statistical tests. In performing a two-sample t-test, we investigate whether the data provides adequate evidence to reject the null hypothesis in favor of this alternative.
If our test results in a t-value that indicates significance, we may conclude that the earnings differ between the PGA and LPGA groups, thus supporting the alternative hypothesis.

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Most popular questions from this chapter

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A recent random survey of 100 individuals in Michigan found that 80 drove to work alone. A similar survey of 120 commuters in New York found that 62 drivers drove alone to work. Find the \(95 \%\) confidence interval for the difference in proportions.

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if the variance of the areas in square miles for counties in Indiana is less than the variance of the areas for counties in Iowa. A random sample of counties is selected, and the data are shown. At \(\alpha=0.01,\) can it be concluded that the variance of the areas for counties in Indiana is less than the variance of the areas for counties in Iowa? $$ \begin{array}{llll|llll} &&&{\text { Indiana }} &{\text { Iowa }} \\ \hline 406 & 393 & 396 & 485 & 640 & 580 & 431 & 416 \\ 431 & 430 & 369 & 408 & 443 & 569 & 779 & 381 \\ 305 & 215 & 489 & 293 & 717 & 568 & 714 & 731 \\ 373 & 148 & 306 & 509 & 571 & 577 & 503 & 501 \\ 560 & 384 & 320 & 407 & 568 & 434 & 615 & 402 \end{array} $$

The mean age of a random sample of 25 people who were playing the slot machines is 48.7 years, and the standard deviation is 6.8 years. The mean age of a random sample of 35 people who were playing roulette is 55.3 with a standard deviation of 3.2 years. Can it be concluded at \(\alpha=0.05\) that the mean age of those playing the slot machines is less than those playing roulette?

A company owned two small Bath and Body Goods stores in different cities. It was desired to see if there was a difference in their mean daily sales. The following results were obtained from a random sample of daily sales over a six-week period. At \(\alpha=0.01,\) can a difference in sales be concluded? Use the \(P\) -value method. $$ \begin{array}{llcc} & & \text { Population } & \\ & & \text { standard } & \text { Sample } \\ \text { Store } & \text { Mean } & \text { deviation } & \text { size } \\ \hline \text { A } & \$ 995 & \$ 120 & 30 \\ \text { B } & 1120 & 250 & 30 \end{array} $$

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A random survey of 1000 students nationwide showed a mean ACT score of 21.4. Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is 3 , can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

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