91Ó°ÊÓ

The "spring-like effect" in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the clubhead). Twelve randomly selected drivers produced by two clubmakers are tested and the coefficient of restitution measured. The data follow: $$\begin{aligned}\text { Club } 1: & 0.8406,0.8104,0.8234,0.8198,0.8235,0.8562 \\\& 0.8123,0.7976,0.8184,0.8265,0.7773,0.7871 \\\\\text { Club 2: } & 0.8305,0.7905,0.8352,0.8380,0.8145,0.8465 \\\& 0.8244,0.8014,0.8309,0.8405,0.8256,0.8476\end{aligned}$$ a. Is there evidence that coefficient of restitution is approximately normally distributed? Is an assumption of equal variances justified? b. Test the hypothesis that both brands of clubs have equal mean coefficient of restitution. Use \(\alpha=0.05 .\) What is the P-value of the test? c Construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean difference in coefficient of restitution for the two brands of golf clubs. d. What is the power of the statistical test in part (b) to detect a true difference in mean coefficient of restitution of \(0.2 ?\) e. What sample size would be required to detect a true difference in mean coefficient of restitution of 0.1 with power of approximately \(0.8 ?\)

Step by step solution

01

Test for Normality

To check if the data is normally distributed, we can use normality tests such as the Shapiro-Wilk test. For Club 1 and Club 2, input data into a statistical software to obtain p-values. If p-values are greater than 0.05, we assume normality.

02

Test for Equal Variances

Use an F-test to check for equal variances between the two data sets. Calculate the F-statistic using the ratio of sample variances: \[F = \frac{s_1^2}{s_2^2}\]where \(s_1\) and \(s_2\) are the standard deviations of Club 1 and Club 2, respectively. If the p-value is greater than 0.05, variances are considered equal.

03

Perform Two-sample t-test

Perform a two-sample t-test (with pooled or unpooled variances depending on results from Step 2) to test the hypothesis:\[H_0: \mu_1 = \mu_2\]\[H_a: \mu_1 eq \mu_2\]Calculate the test statistic and find the p-value. If the p-value is less than 0.05, reject the null hypothesis.

04

Construct a 95% Confidence Interval

Using the results of the t-test, calculate the confidence interval for the mean difference:\[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]where \(\bar{x}_1\) and \(\bar{x}_2\) are sample means, \( n_1 \) and \( n_2 \) are sample sizes.

05

Calculate Power of the Test

The power of the test can be calculated using power analysis tools or software that considers effect size, sample sizes, alpha level, and variance. A common method is to use statistical software with input parameters from the study.

06

Determine Required Sample Size

Utilize a sample size calculator or statistical software given the desired power (0.8), significance level (0.05), and effect size (0.1) to find the sample size needed to detect the mean difference.

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Key Concepts

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

Coefficient of Restitution

In the context of golf clubs, the coefficient of restitution (COR) measures the "spring-like effect" of the club when it hits a golf ball. Formally, it is defined as the ratio of the outbound velocity of the ball to the inbound velocity.
High COR values suggest that the clubface is efficiently transferring energy to the ball, resulting in greater distances.
For engineers and manufacturers, understanding the COR is vital for designing clubs that optimize performance within regulatory limits.

• Materials used in the clubface can affect COR.
• The COR also influences the ball's flight trajectory.

Normality Test

A normality test assesses whether a data set is approximately normally distributed. This property is critical as many statistical tests, such as the t-test, assume normal distribution of the data.
In the example of analyzing golf club data, a normality test like the Shapiro-Wilk test can be employed.
If the p-value from the normality test is greater than 0.05, we typically conclude that the data are approximately normally distributed.
Employing these tests ensures that statistical analysis results are valid and reliable.

F-test for Equal Variances

The F-test for equal variances helps determine whether two data sets have the same variance. It is essential when deciding on pooled variance in a t-test.
For golf club attributes, you'd compare variances of restitution coefficients from two brands.

• Compute the F-statistic: \[ F = \frac{s_1^2}{s_2^2} \]
• If p-value > 0.05, assume equal variances.

Understanding variances helps ensure the proper application of subsequent statistical tests, reducing the risk of incorrect conclusions.

Two-sample t-test

A two-sample t-test evaluates if the means of two independent groups are significantly different. It combines each group's data to produce a single test statistic.
When comparing the mean COR of two brands of golf clubs, you'd test:\[ H_0: \mu_1 = \mu_2 \]\[ H_a: \mu_1 eq \mu_2 \] The conclusion drawn depends on the p-value. If the p-value is less than 0.05, reject the null hypothesis, suggesting a significant difference between the brands.

Confidence Interval (CI)

A confidence interval provides a range of values likely to contain the true mean difference between two groups. For the golf club example, a 95% CI on the mean COR difference gives insight into the direction and magnitude of differences. Calculate using:\[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \] Interpreting the CI involves recognizing that if a CI excludes zero, it indicates a significant difference between the group means.

Power of a Test

The power of a test is the probability that it correctly rejects a false null hypothesis. Higher power means a higher likelihood of detecting a true effect or difference.
91Ó°ÊÓ like statistical software can help compute this by considering sample size, variance, and significance level.

• Power is often targeted at 0.8, indicating an 80% chance of detecting an actual effect.
• Factors like sample size greatly influence the power.

Understanding test power highlights the adequacy of the study design in detecting differences.

Sample Size Determination

Sample size determination is crucial for designing experiments with sufficient power to detect meaningful differences.
In the context of golf clubs, determining the required sample size to detect a COR difference of 0.1 with a power of 0.8 and a significance level of 0.05 involves calculations that account for expected effect size and variability.

• Larger sample sizes can increase test power and accuracy.
• Software tools are available to compute these values efficiently.

Ensuring adequate sample size mitigates the risk of type II errors where true differences go undetected.