Question

For Exercises 2 through \(12,\) 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? $$ \begin{array}{l|cccccccc}{\text { Golfer }} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} \\ \hline \text { Thursday } & {67} & {65} & {68} & {68} & {68} & {70} & {69} & {70} \\ \hline \text { Friday } & {68} & {70} & {69} & {71} & {72} & {69} & {70} & {70}\end{array} $$

Short answer

At \( \alpha = 0.05 \), there is no significant difference in mean scores.

Step-by-step solution

State the Hypotheses

We are conducting a two-tailed test to determine if there is a significant difference in the mean scores for Thursday and Friday.Null Hypothesis \(H_0: \mu_1 = \mu_2\), which states that there is no difference in the mean scores.Alternative Hypothesis \(H_a: \mu_1 eq \mu_2\), which indicates there is a difference in the mean scores.The claim is represented by the alternative hypothesis \(H_a\).

Find the Critical Values

Since we are performing a two-tailed test with a significance level \( \alpha = 0.05 \), we need to find the critical value from the t-distribution table for \(n-1\) degrees of freedom, where \(n\) is the number of paired differences.In this case, \( n = 8 \) (the number of golfers). Thus, the degrees of freedom \(df = n-1 = 7\). For \( \alpha/2 = 0.025 \) on each tail, the critical value \(t\) is approximately \(\pm 2.364\).

Compute the Test Value

First, calculate the differences between the paired scores:\[D = [-1, -5, -1, -3, -4, 1, -1, 0]\]Next, compute the mean of the differences \(\bar{D}\):\[\bar{D} = \frac{-1 - 5 - 1 - 3 - 4 + 1 - 1 + 0}{8} = \frac{-14}{8} = -1.75\]Then, find the standard deviation of the differences \(s_D\):\[s_D = \sqrt{\frac{\sum (D - \bar{D})^2}{n-1}} = \sqrt{\frac{43.5}{7}} = 2.495\]Compute the t-value:\[t = \frac{\bar{D} - 0}{s_D/\sqrt{n}} = \frac{-1.75}{2.495/\sqrt{8}} \approx -1.983\]

Make the Decision

Compare the test value to the critical value.The test value \(-1.983\) falls between the critical values \(-2.364\) and \(2.364\). Since it does not exceed the critical values, we fail to reject the null hypothesis.

Summarize the Results

Based on the results of the hypothesis test, there is not enough evidence at \(\alpha = 0.05\) to suggest a significant difference in mean golf scores between Thursday and Friday for the sampled golfers.

Key concepts

t-test

The t-test is a common method used in statistics to determine whether there is a significant difference between the means of two groups. In this scenario, we're using a paired t-test because we're comparing the golf scores of the same golfers on two consecutive days. This type of t-test is ideal for situations where measurements are taken on the same subjects under different conditions.
A paired t-test considers the differences between paired observations. It checks if the mean of these differences is significantly different from zero. In our exercise, the scores from Thursday and Friday are treated as paired data. Each golfer's score difference is computed and analyzed. The test helps us understand if the change in scores from one day to the next is statistically significant or just due to random variations.

• Use when comparing two sets of related data
• Requires normal distribution of differences
• Ideal for repeated measurements of the same group

null hypothesis

The null hypothesis (\(H_0\)) is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference. In this context, the null hypothesis states that there is no difference in the mean golf scores for Thursday and Friday. Establishing a null hypothesis offers a baseline that can be tested statistically. For our exercise, it is formulated as \(H_0: \mu_1 = \mu_2\), indicating no difference in the average scores across the two days. The goal of the test is to determine whether we can reject this null hypothesis based on the data available. It's crucial to remember that we never prove a null hypothesis. Instead, we assess whether there is enough statistical evidence to reject it.

• Represents no change or effect
• The foundation for all hypothesis tests
• Always assumes that observed differences are due to random chance

critical value

In hypothesis testing, the critical value is a threshold that determines the boundary for rejecting or failing to reject the null hypothesis. For a paired t-test, the critical value refers to the values beyond which we would consider the results statistically significant.
To find the critical value, we use the t-distribution table, taking into account our significance level (\(\alpha = 0.05\)) and degrees of freedom. In our case, with a two-tailed test and 7 degrees of freedom (\(n-1 = 8-1\)), the critical values are approximately \(\pm 2.364\). This means our test statistic must fall outside this range to reject the null hypothesis.Understanding critical values is vital as they help determine the test's outcome based on the specified risk level (\(\alpha\)).

• Defines rejection region for null hypothesis
• Dependent on significance level and degrees of freedom
• Provides a clear boundary for statistical decision-making

paired differences

Paired differences refer to the computation of differences between two related sets of data. When conducting a paired t-test, these differences form the basis of analysis. Our exercise involved calculating the difference in scores for each golfer between Thursday and Friday.
The formula for paired differences \(D_i = X_{1i} - X_{2i}\) calculates each individual difference. Once all differences are calculated, we analyze the average and variance of these differences to determine if there is enough evidence to claim a significant effect. Analyzing paired differences enhances the accuracy of results by accounting for individual variations between paired observations.

• Calculated as the difference between two related observations
• Offers insight into changes at an individual level
• Serves as the foundation for paired t-test calculations