 golf club manufacturer claims that golfers can lower their scores by using the
manufacturer’s new designed golf clubs. Eight golfers are randomly selected, and each is
asked to give his or her most recent score. After using the new clubs for one month, the
golfers are again asked to give their most recent score. The scores for each golfer are in
the table below. Test the manufacturer’s claim to the 0.10 significance level.
Golfer 1 2 3 4 5 6 7 8
Before
clubs
89 84 96 82 74 92 85 91
After clubs 83 83 92 84 76 87 84 91


To test the manufacturer's claim that golfers can lower their scores by using the new golf clubs, we can perform a **paired t-test**. This test is appropriate because we are comparing two sets of related scores: the scores before and after using the new clubs for each golfer.

### Steps to Perform the Paired t-Test

1. **State the Hypotheses**:
   - Null Hypothesis (\(H_0\)): There is no significant difference in the golfers' scores before and after using the new clubs. (\(\mu_d = 0\))
   - Alternative Hypothesis (\(H_1\)): There is a significant reduction in the golfers' scores after using the new clubs. (\(\mu_d > 0\))

   Here, \(\mu_d\) represents the mean difference between the scores before and after using the clubs.

2. **Calculate the Differences**:
   For each golfer, calculate the difference between the scores before and after using the new clubs (\(d_i = \text{Before}_i - \text{After}_i\)).

   | Golfer | Before | After | Difference (\(d_i\)) |
   |--------|--------|-------|---------------------|
   | 1      | 89     | 83    | 6                   |
   | 2      | 84     | 83    | 1                   |
   | 3      | 96     | 92    | 4                   |
   | 4      | 82     | 84    | -2                  |
   | 5      | 74     | 76    | -2                  |
   | 6      | 92     | 87    | 5                   |
   | 7      | 85     | 84    | 1                   |
   | 8      | 91     | 91    | 0                   |

3. **Calculate the Mean and Standard Deviation of the Differences**:
   - Mean of differences (\(\bar{d}\)): 
   - Standard deviation of differences (\(s_d\)): 

4. **Calculate the Test Statistic**:
   The t-statistic is calculated using the formula:
   \[
   t = \frac{\bar{d}}{s_d/\sqrt{n}}
   \]
   where:
   - \(\bar{d}\) is the mean of the differences,
   - \(s_d\) is the standard deviation of the differences,
   - \(n\) is the number of paired samples (in this case, \(n = 8\)).

5. **Determine the Critical Value**:
   For a one-tailed test at the 0.10 significance level with \(n - 1 = 7\) degrees of freedom, find the critical value \(t_{0.10, 7}\) from the t-distribution table.

6. **Compare the Test Statistic with the Critical Value**:
   - If \(t\) is greater than the critical value, reject the null hypothesis (\(H_0\)).
   - If \(t\) is less than or equal to the critical value, fail to reject the null hypothesis.

7. **Conclusion**:
   Based on the comparison, draw a conclusion about whether the golfers' scores have significantly decreased after using the new clubs.

I'll calculate the necessary values now.Here are the results of the paired t-test:

1. **Mean of Differences** (\(\bar{d}\)): 1.625
2. **Standard Deviation of Differences** (\(s_d\)): 3.068
3. **Calculated t-Statistic**: 1.498
4. **Critical t-Value** (for \(df = 7\) and \(\alpha = 0.10\)): 1.415

### Conclusion:
- Since the calculated t-statistic (1.498) is greater than the critical t-value (1.415), we **reject the null hypothesis**.

This means there is statistically significant evidence at the 0.10 significance level to support the manufacturer's claim that the golfers' scores decreased after using the new golf clubs.

Would you like more details on this process or have any other questions?

Here are some related questions you might consider:

1. What if the significance level was 0.05 instead of 0.10?
2. How does the sample size affect the results of a paired t-test?
3. What assumptions must be met to use a paired t-test?
4. Can you explain the concept of a p-value in hypothesis testing?
5. How would the results change if one golfer's scores had a large difference?
6. What if the differences in scores were not normally distributed?
7. How would we conduct a two-tailed test instead of a one-tailed test?
8. What are some other statistical tests that could be used in this scenario?

**Tip:** In hypothesis testing, it's important to ensure that the assumptions of the test are met, such as normality of the differences when using a paired t-test.

Learn how to test a manufacturer's claim that golfers can lower their scores using new golf clubs with a paired t-test. Includes step-by-step calculation of mean differences, standard deviation, and critical values. Results show statistically significant evidence supporting the claim at the 0.10 significance level.

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