91Ó°ÊÓ

Perform each of these steps. Assume that all variables are normally or approximately normally distributed 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. PGA Golf Scores At a recent PGA tournament (the Honda Classic at Palm Beach Gardens, Florida) the following scores were posted for eight randomly selected golfers for two consecutive days. At \(\alpha=0.05,\) is there evidence of a difference in mean scores for the two days? $$ \begin{array}{l|rrrrrrrr} \text { Golfer } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text { Thursday } & 67 & 65 & 68 & 68 & 68 & 70 & 69 & 70 \\ \hline \text { Friday } & 68 & 70 & 69 & 71 & 72 & 69 & 70 & 70 \end{array} $$

Step by step solution

01

State the Hypotheses and Identify the Claim

We need to determine whether there is a significant difference in mean scores between Thursday and Friday. Thus, our null hypothesis \( H_0 \) is that there is no difference in the mean scores: \( H_0: \mu_{Thursday} = \mu_{Friday} \). The alternative hypothesis \( H_1 \) is that there is a difference in the mean scores: \( H_1: \mu_{Thursday} eq \mu_{Friday} \). The claim is represented by the alternative hypothesis.

02

Find the Critical Values

Given that \( \alpha = 0.05 \) and we are conducting a two-tailed test for the difference in means, we will use the t-distribution. For \( n = 8 \), degrees of freedom \( df = n - 1 = 7 \). Looking at a t-distribution table, the critical t-value for \( \alpha = 0.05 \) with 7 degrees of freedom is \( \pm 2.3646 \).

03

Compute the Test Value

First, calculate the differences in scores for each golfer: \([-1, -5, -1, -3, -4, 1, -1, 0]\). The mean difference \( \bar{d} \) is \(-1.75\), and the standard deviation of the differences \( s_d \) is approximately \( 1.8325 \). Calculate the test statistic using the formula: \( t = \frac{\bar{d}}{s_d/\sqrt{n}} = \frac{-1.75}{1.8325/\sqrt{8}} \approx -2.70 \).

04

Make the Decision

Compare the calculated t-value (\(-2.70\)) with the critical t-value (\(-2.3646\)). Since \(-2.70 < -2.3646\), we reject the null hypothesis \( H_0 \).

05

Summarize the Results

At a significance level of \( \alpha = 0.05 \), there is sufficient evidence to support the claim that there is a difference in mean scores between Thursday and Friday for the golfers.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a fundamental part of statistics that helps us make decisions about population parameters based on sample data. In simple terms, it is like a way to test if our data shows a significant effect or change. Here's how it works:

• Null Hypothesis (\( H_0 \)): This is a default statement that there is no effect or no change. We assume it to be true until evidence suggests otherwise. For example, we hypothesized there was no difference in mean scores for two consecutive days of a tournament.
• Alternative Hypothesis (\( H_1 \)): This hypothesis suggests there is an effect or difference. It's what we test against the null hypothesis. In our exercise, this claims there is a difference in the golfers' scores.

We gather data and perform calculations to determine whether to reject or fail to reject the null hypothesis. This decision is based on a significance level (\( \alpha \)), often set at 0.05, which measures the probability of making a Type I error, that is, incorrectly rejecting the null hypothesis when it's true.

Critical Value

The critical value serves as a threshold in hypothesis testing. It helps determine the cut-off points at which the null hypothesis is rejected. When conducting a t-test, this value is dictated by the chosen significance level (\( \alpha \)) and the degrees of freedom from the sample data.

When conducting a two-tailed test, as in our exercise, the critical value defines the boundary for extreme values on both sides of a distribution curve. Here's what to keep in mind:

• The critical value varies depending on whether the test is one-tailed or two-tailed.
• For a significance level of 0.05 with 7 degrees of freedom (from 8 golfers), the critical value was found to be \( \pm 2.3646 \).
• If the test statistic falls beyond this critical value, we reject the null hypothesis.

Understanding the critical value is essential because it allows us to visualize and make decisions based on how extreme the test statistic is relative to the expected range under the null hypothesis.

Test Statistic

The test statistic is a standardized value calculated from sample data. It helps determine whether to support or reject the null hypothesis. In the context of the t-test, it reflects how far away the sample mean is from the null hypothesis mean, measured in units of standard error.

Our example involved a paired t-test, looking at the differences in scores between two days for the same golfers. Here’s how we arrived at the test statistic:

• Calculate each golfer's score difference from Thursday to Friday.
• Find the average of these differences (\( \bar{d} \)) which was -1.75.
• Compute the standard deviation of the differences, noted as \( s_d \).
• Use the formula for the test statistic: \( t = \frac{\bar{d}}{s_d/\sqrt{n}} \).

In our calculation, this resulted in a t-score of approximately -2.70, signifying the observed difference was notably different from the null hypothesis in terms of standard errors.

P-value

The p-value is a practical tool in hypothesis testing that quantifies the evidence against the null hypothesis. It represents the probability of obtaining a result as extreme as the one observed, assuming the null hypothesis is true. Here's how it interacts with \( \alpha \):

• A low p-value (typically ≤ 0.05) indicates strong evidence against the null hypothesis, supporting the alternative hypothesis.
• Conversely, a high p-value (> 0.05) suggests the data do not provide sufficient evidence to reject the null hypothesis.

In the given exercise, although a p-value is not directly calculated, it would correspond to the probability of observing the calculated test statistic (-2.70) or more extreme, under the null hypothesis. Since the test statistic was beyond the critical value, the implied p-value was less than 0.05, leading to rejection of the null hypothesis. Thus, the p-value helps confirm the decision inferred from the critical value.