Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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. Speaking Confidence Fear of public speaking is a common problem for many individuals. A researcher wishes to see if educating individuals on the aspects of public speaking will help people be more confident when they speak in public. She designs three programs for individuals to complete. Group A studies the aspects of writing a good speech. Group \(\mathrm{B}\) is given instruction on delivering a speech. Group \(\mathrm{C}\) is given practice and evaluation sessions on presenting a speech. Then each group is given a questionnaire on selfconfidence. The scores are shown. At \(\alpha=0.05\), is there a difference in the scores on the tests? $$ \begin{array}{ccc} \text { Group A } & \text { Group B } & \text { Group C } \\ \hline 22 & 18 & 16 \\ 25 & 24 & 17 \\ 27 & 25 & 19 \\ 26 & 27 & 23 \\ 33 & 29 & 18 \\ 35 & 31 & 31 \\ 30 & 17 & 15 \\ 36 & 15 & 36 \end{array} $$

Short Answer

Expert verified
Perform Kruskal-Wallis test: reject null if H > 5.991 at \(\alpha=0.05\).

Step by step solution

01

State Hypotheses and Identify the Claim

The null hypothesis (H_0a): There is no difference in the scores among the three groups. The alternative hypothesis (H_1): There is a difference in the scores among the three groups. Our claim is that there is a difference in the confidence scores after undergoing different programs.
02

Find the Critical Value

The critical value for the Kruskal-Wallis test can be found using a chi-square distribution table. With 3 groups, the degrees of freedom (df) is calculated as the number of groups minus one, df = 3 - 1 = 2. For a significance level  = 0.05, the critical value of chi-square ( 5) at df = 2 is approximately 5.991.
03

Compute the Test Value

First, list all the scores from the three groups and rank them from lowest to highest. Calculate the sum of ranks for each group: Group A (R_1), Group B (R_2), and Group C (R_3). Use the formula for the test statistic, \[H = \frac{12}{N(N+1)} \sum \frac{R_i^2}{n_i} - 3(N+1)\]where \(N\) is the total number of observations across all groups, \(n_i\) is the number of observations in group \(i\), and \(R_i\) is the sum of ranks for group \(i\). Calculate \(H\) and compare it with the critical value from Step 2.
04

Make the Decision

If the computed test statistic \(H\) is greater than the critical value from Step 2 (5.991), we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.
05

Summarize the Results

Based on our decision from Step 4, we conclude whether there is sufficient evidence to support the claim that there is a significant difference in the confidence scores of individuals subjected to different public speaking programs.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to determine if there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. It involves making an assumption called the "null hypothesis" and then testing the validity of this assumption based on the data collected. Usually, a "null hypothesis" indicates that there is no effect or no difference in the given situation.

Here, in our Kruskal-Wallis test scenario, the null hypothesis is that there is no difference in public speaking confidence scores across the three different training groups (Group A, Group B, and Group C). This means that if the null hypothesis holds true, the different educational programs do not influence confidence levels.
  • The alternate hypothesis, on the other hand, proposes that there is a difference among the groups' scores.
  • The goal of hypothesis testing is to determine which hypothesis is supported by the evidence collected from the samples.
At the end of the hypothesis testing process, we either reject the null hypothesis or fail to reject it, making a decision based on the statistical evidence available.
Public Speaking Confidence
Public speaking is often intimidating for a lot of people. Sometimes, fear arises due to a lack of confidence. Increasing confidence in public speaking situations can potentially alleviate some of this fear.

In the given study, the researcher creates educational programs intended to boost public speaking confidence. These are:
  • Group A: Focuses on writing a good speech.
  • Group B: Emphasizes delivering a speech effectively.
  • Group C: Involves practice and receiving evaluations on speech presentations.
Each participant's confidence is then measured through a questionnaire, and the scores obtained are analyzed statistically. By examining the scores across these groups using the Kruskal-Wallis test, the researcher aims to understand whether these programs have significantly different effects on confidence levels.

This analysis helps identify which aspect of public speaking training contributes more to increasing confidence and can guide future interventions or educational approaches.
Critical Value
In statistical hypothesis testing, the critical value plays a crucial role. It is the threshold that the test statistic must exceed to reject the null hypothesis.

In the context of the Kruskal-Wallis test, which is a type of non-parametric statistical test used to compare three or more independent groups, the critical value is derived from the chi-square distribution corresponding to the specified significance level (often denoted as \( \alpha \)).
  • The degrees of freedom (df) are calculated by subtracting one from the number of groups being compared. In this example, with three groups, df is 2.
  • At a significance level \( \alpha = 0.05 \), the critical value from the chi-square distribution table with df = 2 is 5.991.
This value indicates how extreme the sample statistic must be for us to reject the null hypothesis. If our computed test statistic \( H \) is greater than 5.991, then this threshold is exceeded, suggesting the findings are statistically significant; hence, we reject the null hypothesis.

Understanding the critical value is key to making informed decisions in hypothesis testing, as it confirms whether the differences observed in the data are likely due to the effect being investigated or by random chance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Accidents or Illnesses The people who went to the emergency room at a local hospital were treated for an accident (A) or illness (I). Test the claim \(\alpha=0.10\) that the reason given occurred at random. $$ \begin{array}{llllllllll} \text { I } & \text { A } & \text { I } & \text { A } & \text { A } & \text { A } & \text { A } & \text { A } & \text { A } & \text { I } \\ \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { A } \\ \text { A } & \text { I } & \text { A } & \text { I } & \text { A } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } \\ \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { I } & \text { A } & \text { I } & \text { A } & \text { I } \\ \text { A } & \text { I } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { A } & \text { I } \\ \text { I } & \text { A } & \text { I } & \text { A } & \text { A } & \text { I } & \text { I } & \text { A } & \text { A } & \text { A } \end{array} $$

Perform these steps. a. Find the Spearman rank correlation coefficient. b. State the hypotheses. c. Find the critical value. Use \(\alpha=0.05\). d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Motion Picture Releases and Gross Revenue In Chapter 10 it was demonstrated that there was a significant linear relationship between the numbers of releases that a motion picture studio put out and its gross receipts for the year. Is there a relationship between the two at the 0.05 level of significance? $$ \begin{array}{l|ccccccccc} \begin{array}{l} \text { No. of } \\ \text { releases } \end{array} & 361 & 270 & 306 & 22 & 35 & 10 & 8 & 12 & 21 \\ \hline \text { Receipts } & 2844 & 1967 & 1371 & 1064 & 667 & 241 & 188 & 154 & 125 \end{array} $$

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. A researcher wishes to test the effects of a pill on a person's appetite. Twelve randomly selected subjects are allowed to eat a meal of their choice, and their caloric intake is measured. The next day, the same subjects take the pill and eat a meal of their choice. The caloric intake of the second meal is measured. The data are shown here. At \(\alpha=0.02,\) can the researcher conclude that the pill had an effect on a person's appetite? $$ \begin{array}{l|ccccccc} \text { Subject } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Meal 1 } & 856 & 732 & 900 & 1321 & 843 & 642 & 738 \\ \hline \text { Meal 2 } & 843 & 721 & 872 & 1341 & 805 & 531 & 740 \end{array} $$ $$ \begin{array}{l|rrrrr} \text { Subject } & 8 & 9 & 10 & 11 & 12 \\ \hline \text { Meal 1 } & 1005 & 888 & 756 & 911 & 998 \\ \hline \text { Meal 2 } & 900 & 805 & 695 & 878 & 914 \end{array} $$

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. Sodium Content of Microwave Dinners Three brands of microwave dinners were advertised as low in sodium. Random samples of the three different brands show the following milligrams of sodium. At \(\alpha=0.05,\) is there a difference in the amount of sodium among the brands? $$ \begin{array}{ccc} \text { Brand A } & \text { Brand B } & \text { Brand C } \\ \hline 810 & 917 & 893 \\ 702 & 912 & 790 \\ 853 & 952 & 603 \\ 703 & 958 & 744 \\ 892 & 893 & 623 \\ 732 & & 743 \\ 713 & & 609 \\ 613 & & \end{array} $$

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. Expenditures for Pupils The expenditures in dollars per pupil for randomly selected states in three sections of the country are listed below. At \(\alpha=0.05,\) can it be concluded that there is a difference in spending between regions? $$ \begin{array}{ccc} \text { Eastern third } & \text { Middle third } & \text { Western third } \\ \hline 6701 & 9854 & 7584 \\ 6708 & 8414 & 5474 \\ 9186 & 7279 & 6622 \\ 6786 & 7311 & 9673 \\ 9261 & 6947 & 7353 \end{array} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free