Question

Miniature Golf Scores A large group of friends went miniature golfing together at a par 54 course and decided to play on two teams. A random sample of scores from each of the two teams is shown. At \(\alpha=0.05\) is there a difference in mean scores between the two teams? Use the \(P\) -value method. $$ \begin{array}{l|lllllll}{\text { Team } 1} & {61} & {44} & {52} & {47} & {56} & {63} & {62} & {55} \\ \hline \text { Team 2} & {56} & {40} & {42} & {58} & {48} & {52} & {51}\end{array} $$

Short answer

No significant difference in mean scores between teams at \(\alpha=0.05\).

Step-by-step solution

Define Hypotheses

Begin by stating the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\). For this problem: \(H_0\): There is no difference in mean scores between the two teams (\(\mu_1 = \mu_2\)). \(H_a\): There is a difference in mean scores (\(\mu_1 eq \mu_2\)).

Gather Sample Statistics

Calculate the sample means and standard deviations for both teams. Team 1 scores: 61, 44, 52, 47, 56, 63, 62, 55. Team 2 scores: 56, 40, 42, 58, 48, 52, 51. Calculate the means: \(\bar{x}_1 = 55\), \(\bar{x}_2 = 49\). Calculate the standard deviations: \(s_1 \approx 7.23\), \(s_2 \approx 7.08\).

Calculate Test Statistic

Use a two-sample t-test to calculate the test statistic. The formula is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \(n_1 = 8\) and \(n_2 = 7\). The calculation results in \(t \approx 1.59\).

Determine Degrees of Freedom

Calculate the degrees of freedom for the two-sample t-test using the formula: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]Substitute the values to get \( df \approx 13.85 \), which can be rounded to \( df = 14 \) for practical purposes.

Find P-value

Using a t-distribution table or calculator, find the P-value for the calculated t-statistic \(t \approx 1.59\) with \(df = 14\). The P-value is approximately 0.134.

Make a Decision

Compare the P-value to the significance level \(\alpha = 0.05\). Since the P-value 0.134 is greater than 0.05, we fail to reject the null hypothesis \(H_0\).

Conclusion

There is not enough statistical evidence at the \(\alpha=0.05\) level to conclude that there is a difference in mean scores between the two teams.

Key concepts

Two-sample t-test

A key tool in hypothesis testing, the two-sample t-test helps us determine if there is a statistically significant difference between the means of two independent groups. In our miniature golf exercise, we are comparing the mean scores of two teams.

The formula used for a two-sample t-test is:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Where:

• \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means for Team 1 and Team 2, respectively.
• \( s_1 \) and \( s_2 \) are the standard deviations of the two samples.
• \( n_1 \) and \( n_2 \) are the sample sizes.

In our case, the test statistic calculated was approximately 1.59, indicating the difference in means needs to be further evaluated against the null hypothesis.

P-value method

The P-value method in hypothesis testing involves comparing the P-value to a predetermined significance level, known as alpha (\( \alpha \)). This method helps us decide if we should reject the null hypothesis.

Here's how it works:

• The P-value represents the probability of observing a test statistic as extreme as, or more extreme than the one we calculated, assuming the null hypothesis is true.
• We compare this P-value to our alpha level, in this case, 0.05.

For the miniature golf problem, the calculated P-value was about 0.134. Because 0.134 is greater than 0.05, we do not reject the null hypothesis. In simpler terms, there's not enough evidence to say the team's scores are significantly different.

Degrees of Freedom

Degrees of freedom are a crucial component in statistical tests, including the two-sample t-test. They are an indication of the number of values in a calculation that are free to vary.

For the two-sample t-test, degrees of freedom are determined using:\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]This formula helps us find the degrees of freedom, which we then use to identify the appropriate distribution for finding the critical value or P-value.

In our exercise, the degrees of freedom calculated were approximately 13.85. For practical reasons, this was rounded to 14, assisting us in using standard statistical tables or calculators effectively.

Null and Alternative Hypotheses

Formulating null and alternative hypotheses is the first step in any hypothesis test. It's the foundation that dictates how we interpret our results.

- **Null Hypothesis (\( H_0 \))**: This is a statement suggesting no effect or no difference. For our golfing example, \( H_0 \) was that there's no difference in the mean scores of the two teams (\( \mu_1 = \mu_2 \)).- **Alternative Hypothesis (\( H_a \))**: This proposes that there is an effect or a difference. In the exercise, it states that there is a difference in the mean scores (\( \mu_1 eq \mu_2 \)).Hypotheses guide statistical tests and decision-making.

We test whether to reject the null hypothesis based on our calculations and the results provided, like the P-value. In this particular problem, the result was in favor of not rejecting the null hypothesis since evidence did not show a significant difference between the two teams' mean scores.