/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 True or False: The expected freq... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 1

True or False: The expected frequencies in a chi-square test for independence are found using the formula Expected frequency \(=\frac{(\text { row total })(\text { column total })}{\text { table total }}\)

Short Answer

Expert verified
True

Step by step solution

01

Understanding the Formula

The formula given in the exercise is \( \text{Expected frequency} = \frac{(\text{row total})(\text{column total})}{\text{table total}} \). This formula helps in calculating the expected frequency for each cell in a contingency table, assuming the null hypothesis of independence is true.
02

Row Total

Identify the row total, which is the sum of frequencies in that particular row of the contingency table.
03

Column Total

Identify the column total, which is the sum of frequencies in that particular column of the contingency table.
04

Table Total

Determine the table total by summing all the frequencies in the entire contingency table.
05

Applying the Formula

Substitute the row total, column total, and table total into the formula: \( \text{Expected frequency} = \frac{(\text{row total})(\text{column total})}{\text{table total}} \) to find the expected frequency for each cell.
06

Conclusion

Given the steps and the correct application of the formula, the statement provided in the exercise is 'True'. This formula is indeed used for finding the expected frequencies in a chi-square test for independence.

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 91影视!

Key Concepts

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

Contingency Table
A contingency table is a type of table that displays the frequency distribution of variables. It's a way to understand the relationship between two categorical variables. For example, you could have a table showing the number of male and female students who prefer different school subjects. Each cell in the table shows the frequency count of the variables. To perform a chi-square test for independence, we first need to organize our data into a contingency table. This helps us see how often combinations of variables occur together.
Expected Frequency
Expected frequency is a crucial concept in the chi-square test for independence. It represents the frequency we would expect in each cell of a contingency table if there was no association between the two categorical variables. To calculate expected frequency, use the formula: \( \text{Expected frequency} = \frac{(\text{row total})(\text{column total})}{\text{table total}} \). This formula involves:
- Determining the row total, which is the sum of all frequencies in a specific row.
- Determining the column total, the sum of all frequencies in a specific column.
- Finding the table total, which is the sum of all frequencies in the entire table.
Once you have these values, plug them into the formula to get the expected frequency for each cell.
Null Hypothesis
In a chi-square test for independence, the null hypothesis is a statement that assumes there is no association between the two categorical variables being tested. For example, if you are investigating the relationship between gender (male/female) and preference for a subject (math/science), the null hypothesis would state that gender does not affect subject preference. It's important to remember:
- The null hypothesis is what you are testing against.
- You use the expected frequencies, based on the null hypothesis, to compare with the observed frequencies in your table.
If the observed frequencies are significantly different from the expected frequencies, you would reject the null hypothesis, indicating that there is a relationship between the variables.
Statistical Formulas
Several statistical formulas are used in the chi-square test for independence, and understanding them helps in performing the test correctly. Here are the key formulas:
1. **Expected Frequency:** \( \text{Expected frequency} = \frac{(\text{row total})(\text{column total})}{\text{table total}} \).
2. **Chi-square Statistic:** The formula for the chi-square statistic is: \( \text{Chi-square} = \frac{( \text{Observed frequency} - \text{Expected frequency})^2}{\text{Expected frequency}} \).
Summed over all cells in the contingency table.

These formulas enable you to calculate the chi-square statistic, which you then compare to a critical value from the chi-square distribution table (based on degrees of freedom and significance level). If your chi-square statistic is greater than the critical value, you reject the null hypothesis.

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

The following table contains observed values and expected values in parentheses for two categorical variables, \(X\) and \(Y\), where variable \(X\) has three categories and variable \(Y\) has two categories: $$ \begin{array}{cccc} & \boldsymbol{X}_{\mathbf{1}} & \boldsymbol{X}_{\mathbf{2}} & \boldsymbol{X}_{3} \\ \hline \boldsymbol{Y}_{\mathbf{1}} & 34(36.26) & 43(44.63) & 52(48.11) \\ \hline \boldsymbol{Y}_{\mathbf{2}} & 18(15.74) & 21(19.37) & 17(20.89) \\ \hline \end{array} $$ (a) Compute the value of the chi-square test statistic. (b) Test the hypothesis that \(X\) and \(Y\) are independent at the \(\alpha=0.05\) level of significance.

In a survey of 3029 adult Americans, the Harris Poll asked people whether they smoked cigarettes and whether they always wear a seat belt in a car. The table shows the results of the survey. For each activity, we define a success as finding an individual who participates in the hazardous activity. $$ \begin{array}{lcc} & \begin{array}{c} \text { No Seat Belt } \\ \text { (success) } \end{array} & \begin{array}{c} \text { Seat Belt } \\ \text { (failure) } \end{array} \\ \hline \text { Smoke (success) } & 67 & 448 \\ \hline \text { Do not smoke (failure) } & 327 & 2187 \\ \hline \end{array} $$ (a) Why is this a dependent sample? (b) Is there a significant difference in the proportion of individuals who smoke and the proportion of individuals who do not wear a seat belt? In other words, is there a significant difference between the proportion of individuals who engage in hazardous activities? Use the \(\alpha=0.05\) level of significance.

The following table contains the number of successes and failures for three categories of a variable. $$ \begin{array}{lccc} & \text { Category } 1 & \text { Category } 2 & \text { Category } 3 \\ \hline \text { Success } & 76 & 84 & 69 \\ \hline \text { Failure } & 44 & 41 & 49 \\ \hline \end{array} $$ Test whether the proportions are equal for each category at the \(\alpha=0.01\) level of significance.

The General Social Survey asked a random sample of adult Americans two questions: (1) Would you favor or oppose a law which would require a person to obtain a police permit before he or she could buy a gun? (2) Do you favor or oppose the death penalty for persons convicted of murder? Results of the survey are below. Does the sample evidence suggest the proportion of adult Americans who favor permits for guns is different from the proportion of adult Americans who favor the death penalty for murder convictions? Use the \(\alpha=0.05\) level of significance. $$ \begin{array}{llcc} & & {\text { Death Penalty }} \\ & & \text { Favor } & \text { Oppose } \\ \hline{\text { Gun Laws }} & \text { Favor } & 176 & 69 \\ & \text { Oppose } & 58 & 17 \\ \hline \end{array} $$

In Section 10.2, we tested hypotheses regarding a population proportion using a z-test. However, we can also use the chi-square goodness-of-fit test to test hypotheses with \(k=2\) possible outcomes. In Problems 25 and \(26,\) we test hypotheses with the use of both methods. According to the U.S. Census Bureau, \(7.1 \%\) of all babies born are of low birth weight \((<5 \mathrm{lb}, 8 \mathrm{oz})\) An obstetrician wanted to know whether mothers between the ages of 35 and 39 years give birth to a higher percentage of low-birth-weight babies. She randomly selected 240 births for which the mother was 35 to 39 years old and found 22 low-birth-weight babies. (a) If the proportion of low-birth-weight babies for mothers in this age group is \(0.071,\) compute the expected number of low-birth-weight births to 35 - to 39 -year-old mothers. What is the expected number of births to mothers 35 to 39 years old that are not low birth weight? (b) Answer the obstetrician's question at the \(\alpha=0.05\) level of significance using the chi-square goodness-of-fit test. (c) Answer the question by using the approach presented in Section 10.2
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation
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.