Step-by-Step Solution

Part 1: Hypotheses

• The null hypothesis ($H_0$): The mean diameter of the golf balls conforms to the association's standard of $1.68$ inches. $H_0: \mu = 1.68$

• The alternative hypothesis ($H_1$): The mean diameter of the golf balls does not conform to the standard. $H_1: \mu \neq 1.68$

This is a two-tailed test since we are testing for "not equal to."

Part 2: Test Statistic

The formula for the test statistic in a one-sample $t$-test is: $t = \frac{\bar{x} - \mu}{s / \sqrt{n}}$ Where:

• $\bar{x}$: Sample mean
• $\mu$: Hypothesized mean ($1.68$)
• $s$: Sample standard deviation
• $n$: Sample size

Step 1: Calculate the sample mean ($\bar{x}$) and standard deviation ($s$): The sample data is: $1.683, 1.685, 1.684, 1.685, 1.677, 1.678, 1.684, 1.682, 1.682, 1.685, 1.673, 1.674$

• $n = 12$

Using formulas for mean and standard deviation: $\bar{x} = \frac{\sum x_i}{n}, \quad s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

1. Sum of the data: $\sum x_i = 20.218$ $\bar{x} = \frac{20.218}{12} = 1.684833 \approx 1.6848$

2. Sum of squared deviations: $\sum (x_i - \bar{x})^2 = 0.000944$ $s = \sqrt{\frac{0.000944}{11}} = 0.003 \approx 0.0030$

Step 2: Compute the $t$-statistic: $t = \frac{1.6848 - 1.68}{0.003 / \sqrt{12}} = \frac{0.0048}{0.000866} \approx 5.54$

Part 3: $P$-Value

For a two-tailed test with $t = 5.54$ and $df = 11$, we use a $t$-distribution table or software. The $P$-value for $|t| > 5.54$ is very small, less than $0.001$.

Part 4: Conclusion

• At $\alpha = 0.05$: Since the $P$-value $(< 0.001)$ is less than $0.05$, we reject the null hypothesis.

• Conclusion: B. Reject $H_0$: There is sufficient evidence to conclude that the golf balls do not conform to the association's standards at the $\alpha = 0.05$ level of significance.

Summary of Results:

1. Hypotheses: $H_0: \mu = 1.68, \quad H_1: \mu \neq 1.68$
2. Test statistic: $t \approx 5.54$
3. $P$-value: $< 0.001$
4. Conclusion: Reject $H_0$

Follow-Up Questions:

1. How is the test statistic affected if the sample size increases?
2. What are the implications of using a smaller significance level (e.g., $\alpha = 0.01$)?
3. How can the association ensure compliance with their diameter standards in the future?
4. Would a one-tailed test have been more appropriate for this scenario?
5. What are the potential sources of error in this hypothesis test?

Tip: Always ensure your data meets the assumptions of the test being applied, such as normality and independence.