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Students in Greek organizations at schools throughout the country sent volunteers to a yearly blood drive. The numbers from each randomly selected participating school are listed. Test the claim that there is no difference in the number of students participating from fraternities and sororities at \(\alpha=0.10\). $$\begin{array}{l|l}\text { Fraternities } & 4,5,10,7,7,15,12,11,13,15,12,12 \\\\\hline \text { Sororities } & 3,5,6,7,4,7,10,9,9,14\end{array}$$

Short Answer

Expert verified
There is insufficient evidence to reject the null hypothesis at \( \alpha = 0.10 \); no significant difference in participation.

Step by step solution

01

Define the Hypotheses

The null hypothesis (H0) states that there is no difference in the number of students participating from fraternities and sororities: \( H_0: \mu_1 = \mu_2 \). The alternative hypothesis (H1) says there is a difference: \( H_1: \mu_1 eq \mu_2 \).
02

Calculate Sample Means and Standard Deviations

Calculate the means for each group: Fraternity mean \( \bar{x_1} \) and Sorority mean \( \bar{x_2} \), and their sample standard deviations \( s_1 \) and \( s_2 \).
03

Use the Two-Sample t-Test Formula

Use the formula for the two-sample t-test: \[ t = \frac{\bar{x_1} - \bar{x_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \( n_1 \) and \( n_2 \) are the number of samples in each group.
04

Determine the Degrees of Freedom and Critical t-Value

Calculate the degrees of freedom using the formula: \[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1-1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2-1}} \]. Then, retrieve the critical t-value from t-distribution table at \( \alpha = 0.10 \) for a two-tailed test.
05

Compare and Conclude

Compare the calculated t-value with the critical t-value. If the calculated t-value is greater, reject the null hypothesis; otherwise, fail to reject it.

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Key Concepts

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

Two-Sample t-Test
In the world of statistics, a two-sample t-test is a method used to determine if there are significant differences between the means of two independent groups. This test is particularly useful when comparing data like the number of fraternity and sorority participants in the exercise example. The core idea is to see if there is evidence that the two groups have different average values or if any observed differences happened by chance.
The two-sample t-test answers the question: Are the observed differences in sample means reliable enough to reflect a real difference in the populations from which the samples were drawn? Here's how you tackle it:
  • Gather the data from the two groups - in this case, the number of participants from fraternities and sororities.
  • Calculate the sample means: \( \bar{x_1} \) for fraternities and \( \bar{x_2} \) for sororities.
  • Use the standard deviations of both samples, \( s_1 \) for fraternities and \( s_2 \) for sororities, in the test formula.
  • Apply the two-sample t-test formula to compute the t-score.
This t-score helps us understand how the samples relate to the hypothesized population means and if group differences are statistically significant.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It assumes that there is no significant difference between the specified populations. In the given exercise, the null hypothesis \( H_0 \) is expressed as \( \mu_1 = \mu_2 \), suggesting that fraternities and sororities have the same average number of participants attending the blood drive.
Formulating this hypothesis involves setting a baseline that any observed differences in sample data can be attributed to random variation, rather than a real effect. The strength of hypothesis testing lies in its ability to reject the null hypothesis, therefore suggesting that differences do exist.
Whenever conducting a hypothesis test:
  • Step 1: Clearly state the null hypothesis \( H_0 \).
  • Step 2: State the alternative hypothesis \( H_1 \), which indicates that a difference does exist.
These steps set the stage to utilize statistical calculations, compare results, and make informed conclusions.
Degrees of Freedom
Degrees of freedom are a critical part of determining the credibility of test statistics in hypothesis testing. It reflects the number of independent values or quantities which can vary in a data set. The formula for degrees of freedom when comparing two independent sample means is rather complex:
\[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1}} \]
  • This formula helps account for the variability and size of the two groups being compared.
  • Larger sample sizes typically increase the degrees of freedom, which can lead to more reliable results.
Knowing the degrees of freedom allows us to reference the correct t-distribution table to find the critical t-value. Hence, it plays a vital role when verifying if results are significant enough to reject the null hypothesis.
Critical t-Value
The critical t-value is part of the two-sample t-test and serves as a threshold to determine whether the observed test statistic (t-value) is extreme enough to reject the null hypothesis. This value depends on the degrees of freedom and the level of significance.
In the current exercise:
  • A single critical t-value is extracted from the t-distribution table.
  • The level of significance is given as \( \alpha = 0.10 \), indicating a 10% risk of concluding that a difference exists when there is none (Type I error).
  • For a two-tailed test like ours, both upper and lower critical t-values are considered.
Once you have the critical t-value, compare it to the computed t-value from your test:
  • If the absolute t-value calculated from your data exceeds the critical t-value, the null hypothesis is rejected.
  • If not, there's insufficient evidence to claim a significant difference.
Understanding the relationship between the observed and critical t-values guides researchers to make sound decisions in hypothesis testing, highlighting real differences in populations.

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Most popular questions from this chapter

Daily Lottery Numbers Listed below are the daily numbers (daytime drawing) for the Pennsylvania State Lottery for February 2007. Using O for odd and E for even, test for randomness at \(\alpha=0.05\). $$\begin{array}{lllllll}270 & 054 & 373 & 204 & 908 & 121 & 121 \\ 804 & 116 & 467 & 357 & 926 & 626 & 247 \\\ 783 & 554 & 406 & 272 & 508 & 764 & 890 \\ 441 & 964 & 606 & 568 & 039 & 370 & 583\end{array}$$

For Exercises 5 through \(20,\) 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. The U.S. median annual income for men in 2014 (in constant dollars) was \(\$ 35,642 .\) A random sample of recent male college graduates indicated the following incomes. At the 0.05 level of significance, test the claim that the median is more than \(\$ 35,642\) $$ \begin{array}{lllll} 35,000 & 37,682 & 39,800 & 32,500 & 30,000 \\ 41,050 & 36,198 & 31,500 & 29,650 & 35,800 \\ 34,500 & 38,850 & 39,750 & & \end{array} $$

For Exercises 5 through \(20,\) 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. The median age at first marriage in 2014 for women was 27 years - the highest it has ever been. A random sample of women's ages (in years) from recently applied for marriage licenses resulted in the following set of ages. At \(\alpha=0.05,\) is there sufficient evidence that the median is not 27 years? $$ \begin{array}{lllll} 34.6 & 31.2 & 28.9 & 28.4 & 24.3 \\ 29.8 & 25.9 & 21.4 & 25.1 & 26.2 \\ 28.3 & 30.6 & 35.6 & 34.2 & 34.1 \end{array} $$

Use the Kruskal-Wallis test and perform these steps. 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. Maximum Speeds of Animals A human is said to be able to reach a maximum speed of 27.89 miles per hour. The maximum speeds of various randomly selected types of other animals are listed below. Based on these particular groupings, is there evidence of a difference in speeds? Use the 0.05 level of significance. $$ \begin{array}{ccc} \begin{array}{c} \text { Predatory } \\ \text { mammals } \end{array} & \begin{array}{c} \text { Deerlike } \\ \text { animals } \end{array} & \begin{array}{c} \text { Domestic } \\ \text { animals } \end{array} \\ \hline 70 & 50 & 47.5 \\ 50 & 35 & 39.35 \\ 43 & 32 & 35 \\ 42 & 30 & 30 \\ 40 & 61 & 11 \end{array} $$

A group of compulsive gamblers was selected. The amounts (in dollars) they spent on lottery tickets for one week are shown. Then they were required to complete a workshop showing that the chances of winning were not in their favor. After they complete the workshop, test the claim that, at \(\alpha=0.05,\) the workshop was effective in reducing the weekly amount spent on lottery tickets. $$ \begin{array}{l|cccccccc} \text { Subject } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } \\ \hline \text { Before } & 86 & 150 & 161 & 197 & 98 & 56 & 122 & 76 \\ \hline \text { After } & 72 & 143 & 123 & 186 & 102 & 53 & 125 & 72 \end{array} $$

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