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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

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

Step by step solution

01

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 (\(\mu_1 eq \mu_2\)).
02

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: \(s_1 \approx 7.23\), \(s_2 \approx 7.08\).
03

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 \(t \approx 1.59\).
04

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.
05

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.
06

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 \(H_0\).
07

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. School Teachers' Salaries A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salaries of secondary school teachers. A random sample of the salaries of 30 elementary school teachers has a variance of \(8324,\) and a random sample of the salaries of 30 secondary school teachers has a variance of \(2862 .\) At \(\alpha=0.05\) can the researcher conclude that the variation in the elementary school teachers' salaries is greater than the variation in the secondary school teachers' salaries? Use the \(P\) -value method.

Noise Levels in Hospitals The mean noise level of 20 randomly selected areas designated as "casualty doors", was \(63.1 \mathrm{dBA}\), and the sample standard deviation is \(4.1 \mathrm{dBA}\). The mean noise level for 24 randomly selected areas designated as operating theaters was \(56.3 \mathrm{dBA}\), and the sample standard deviation was \(7.5 \mathrm{dBA}\). At \(\alpha=0.05,\) can it be concluded that there is a difference in the means?

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Hypertension It has been found that \(26 \%\) of men 20 years and older suffer from hypertension (high blood pressure) and \(31.5 \%\) of women are hypertensive. A random sample of 150 of women are hypertensive. from recent hospital records, and the following results were obtained. Can you conclude that a higher percentage of women have high blood pressure? Use \(\alpha=0.05 .\) Men 43 patients had high blood pressure Women 52 patients had high blood pressure

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Wolf Pack Pups Does the variance in the number of pups per pack differ between Montana and Idaho wolf packs? Random samples of packs were selected for each area, and the numbers of pups per pack were recorded. At the 0.05 level of significance, can a difference in variances be concluded? $$ \begin{array}{l|cccccccc}{\text { Montana }} & {4} & {3} & {5} & {6} & {1} & {2} & {8} & {2} \\ \hline \text { wolf packs } & {3} & {1} & {7} & {6} & {5} & {} & {} \\ \hline \text { Idaho } & {2} & {4} & {5} & {4} & {2} & {4} & {6} & {3} \\ \hline \text { wolf packs } & {1} & {4} & {2} & {1} & {}\end{array} $$

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. High School Graduation Rates The overall U.S. public high school graduation rate is \(73.4 \% .\) For Pennsylvania it is \(83.5 \%\) and for Idaho \(80.5 \%-\mathrm{a}\) difference of \(3 \%\). Random samples of 1200 students from each state indicated that 980 graduated in Pennsylvania and 940 graduated in Idaho. At the 0.05 level of significance, can it be concluded that there is a difference in the proportions of graduating students between the states?

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