A Golf Instructor Is Interested In Determining If Her New To

Question

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective A golf instructor wishes to evaluate the effectiveness of her new teaching technique on players’ scores. She records their 18-hole scores before undertaking the new method and then again after completing her instruction. The data collected from four students are as follows: | Player | Score Before | Score After | |---------|----------------|--------------| | 1 | 85 | 78 | | 2 | 90 | 83 | | 3 | 88 | 80 | | 4 | 92 | 85 | The objective is to determine whether the new technique significantly improves players’ scores. To do this, a hypothesis test will be conducted to compare the before and after scores for the same players. The null hypothesis (H₀) posits that the new technique has no effect—that is, there is no difference in scores before and after the instruction. The alternative hypothesis (H₁) suggests that the new technique results in a significant improvement, meaning the scores after are lower than before. Given the paired data, the appropriate statistical test is the paired sample t-test, which compares the means of two related groups. This test accounts for the fact that the same subjects are measured twice, ensuring the differences are assessed accurately. It assumes that the differences are approximately normally distributed, which is reasonable given the small sample size but could be further assessed with normality tests. Calculating the differences for each player: - Player 1: 85 - 78 = 7 - Player 2: 90 - 83 = 7 - Player 3: 88 - 80 = 8 - Player 4: 92 - 85 = 7 The mean difference (d̄) is (7 + 7 + 8 + 7)/4 = 7.25. Next, the standard deviation of the differences (sd) is calculated: Differences: 7, 7, 8, 7 Mean difference: 7.25 - (7 - 7.25)² = 0.0625 - (7 - 7.25)² = 0.0625 - (8 - 7.25)² = 0.5625 - (7 - 7.25)² = 0.0625 Sum of squared deviations = 0.0625 + 0.0625 + 0.5625 + 0.0625 = 0.75 Variance of differences = 0.75 / (4 - 1) = 0.25 Standard devi

Dr. Jack HW Helper · Accepted Answer

Introduction Assessing the efficacy of instructional techniques in sports, particularly golf, involves rigorous statistical analysis to determine if observed improvements are statistically significant or could have occurred by chance. Evaluating the data from four students undergoing a new golf instruction technique provides insights into its effectiveness. Data and Methodology The data set consists of pre- and post-instruction scores for four players: - Player 1: Before 85, After 78 - Player 2: Before 90, After 83 - Player 3: Before 88, After 80 - Player 4: Before 92, After 85 Since the same players are measured twice, a paired sample t-test is appropriate. This test compares the mean difference between paired observations, accounting for the dependence between measurements. The null hypothesis (H₀) states that the mean difference in scores before and after the instruction is zero, implying no effect. The alternative hypothesis (H₁) suggests the scores improved after instruction, with a mean difference greater than zero. Calculations Differences for each player: - Player 1: 7 - Player 2: 7 - Player 3: 8 - Player 4: 7 Mean difference (d̄) = (7+7+8+7)/4 = 7.25 Standard deviation of differences: - Deviations squared: 0.0625, 0.0625, 0.5625, 0.0625 - Sum: 0.75 - Variance: 0.75 / 3 = 0.25 - Standard deviation (sd): √0.25 = 0.5 Standard error: SE = sd / √n = 0.5 / 2 = 0.25 t-statistic: t = d̄ / SE = 7.25 / 0.25 = 29 Degrees of freedom = 3 Results and Interpretation The critical t-value at α = 0.05 (one-tailed) for df=3 is approximately 2.353. The calculated t-value (29) far exceeds this critical value, leading to the rejection of H₀. This indicates a statistically significant improvement in scores following the new instructional technique. Conclusion The statistical analysis demonstrates that the new golf teaching technique significantly enhances players’ scores, with the data strongly supporting its effectiveness. Coaches and instructors may consider adopting this metho

A Golf Instructor Is Interested In Determining If Her New To

A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective

Sample Paper For Above instruction

Introduction

Data and Methodology

Calculations

Results and Interpretation

Conclusion

References