/*! 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 32 Give a two-way table and specify... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give a two-way table and specify a particular cell for that table. In each case find the expected count for that cell and the contribution to the chi- square statistic for that cell. \((\mathrm{B}, \mathrm{E})\) cell $$ \begin{array}{l|rrrr|r} \hline & \text { D } & \text { E } & \text { F } & \text { G } & \text { Total } \\ \hline \text { A } & 39 & 34 & 43 & 34 & 150 \\ \text { B } & 78 & 89 & 70 & 93 & 330 \\ \text { C } & 23 & 37 & 27 & 33 & 120 \\ \hline \text { Total } & 140 & 160 & 140 & 160 & 600 \\ \hline \end{array} $$

Short Answer

Expert verified
The expected count for cell (B, E) is 88 and its contribution to the chi-square statistic is 0.0114.

Step by step solution

01

Calculate the Expected Count

To calculate the expected count for cell (B, E), multiply the total for row B (330) by the total for column E (160), then divide by the overall total (600). So, the expected count \(E(B,E) = \frac{{330 \times 160}}{{600}} = 88\).
02

Calculate the Contribution to the Chi-Square Statistic

To calculate the contribution to the chi-square statistic for cell (B, E), first find the difference between the observed count (O = 89) and the expected count (E = 88). Then square this difference and divide by the expected count. So, the Chi-square contribution \(\chi^2(B,E) = \frac{{(O - E)^2}}{{E}} = \frac{{(89 - 88)^2}}{{88}} = 0.0114\).

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 91Ó°ÊÓ!

Key Concepts

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

Expected Count
In statistical analysis, the expected count is a fundamental step when working with a two-way table to understand how the data should be distributed if there are no associations between variables. To find the expected count for a specific cell in the table, it is necessary to use the formula:
  • Expected Count = \( \frac{{\text{Row Total} \times \text{Column Total}}}{{\text{Overall Total}}} \)
The expected count tells us what the data would look like if the two variables were completely independent of each other.
In the provided exercise, for instance, the expected count for the cell (B, E) was calculated based on:
  • The total for row B: 330
  • The total for column E: 160
  • The overall total of all observations: 600
By plugging these values into the formula, you find the expected count, which helps in the next step of assessing the relationship between variables by calculating the chi-square statistic.
Two-Way Table
A two-way table, also known as a contingency table, is a matrix that displays the frequency distribution of two categorical variables. Each cell within the table shows the count of occurrences for a specific combination of these variables. The arrangement in rows and columns makes it easier to identify potential relationships or patterns among the variables.
For example, in the exercise, a two-way table is presented with three row categories \( A, B, C \) and four column categories \( D, E, F, G \). The intersection of rows and columns, such as cell (B, E), provides specific data points which are used in further analysis like calculating the expected count and contributions to chi-square.
This table format not only allows data organization but also aids in statistical assessments like the chi-square test, which determines if there's a significant association between the two categorical variables being analyzed.
Contribution to Chi-Square
The chi-square test statistic is a way to measure how observed counts deviate from expected counts if two categorical variables are independent. In other words, it helps determine if there is a significant association between the two variables.
For each cell of the two-way table, you can calculate its contribution to the chi-square statistic using the formula:
  • Contribution = \( \frac{{(\text{Observed Count} - \text{Expected Count})^2}}{{\text{Expected Count}}} \)
This formula helps break down the overall chi-square value into individual components.
In the exercise's example, the contribution of cell (B, E) to the chi-square statistic was calculated as 0.0114. This was done by taking the difference between the observed count \( O = 89 \) and the expected count \( E = 88 \), squaring this difference, and dividing by the expected count. Such individual contributions add up to form the total chi-square statistic, which can then be compared to a critical value from the chi-square distribution to assess statistical significance.

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

Age and Frequency of Status Updates on Facebook Exercise 7.48 introduces a study about users of social networking sites such as Facebook. Table 7.35 shows the self-reported frequency of status updates on Facebook by age groups. (a) Based on the totals, if age and frequency of status updates are really unrelated, how many of the 156 users who are 18 to 22 years olds should we expect to update their status every day? (b) Since there are 20 cells in this table, we'll save some time and tell you that the chi-square statistic for this table is \(210.9 .\) What should we conclude about a relationship (if any) between age and frequency of status updates? $$ \begin{array}{l|rrrr|r} \hline \downarrow \text { Status/Age } \rightarrow & 18-22 & 23-35 & 36-49 & 50+ & \text { Total } \\ \hline \text { Every day } & 47 & 59 & 23 & 7 & 136 \\ \text { 3-5 days/week } & 33 & 47 & 30 & 7 & 117 \\ \text { 1-2 days/week } & 32 & 69 & 35 & 25 & 161 \\ \text { Every few weeks } & 23 & 65 & 47 & 34 & 169 \\ \text { Less often } & 21 & 74 & 99 & 170 & 364 \\ \hline \text { Total } & 156 & 314 & 234 & 243 & 947 \\ \hline \end{array} $$

Metal Tags on Penguins In Exercise 6.148 on page 445 we perform a test for the difference in the proportion of penguins who survive over a ten-year period, between penguins tagged with metal tags and those tagged with electronic tags. We are interested in testing whether the type of tag has an effect on penguin survival rate, this time using a chi-square test. In the study, 10 of the 50 metal-tagged penguins survived while 18 of the 50 electronic-tagged penguins survived. (a) Create a two-way table from the information given. (b) State the null and alternative hypotheses. (c) Give a table with the expected counts for each of the four categories. (d) Calculate the chi-square test statistic. (e) Determine the p-value and state the conclusion using a \(5 \%\) significance level.

In Exercises 7.5 to 7.8 , the categories of a categorical variable are given along with the observed counts from a sample. The expected counts from a null hypothesis are given in parentheses. Compute the \(\chi^{2}\) -test statistic, and use the \(\chi^{2}\) -distribution to find the p-value of the test. $$ \begin{array}{lccc} \hline \text { Category } & \text { A } & \text { B } & \text { C } \\ \text { Observed } & 35(40) & 32(40) & 53(40) \\ \text { (Expected) } & & & \\ \hline \end{array} $$

Which Is More Important: Grades, Sports, or Popularity? 478 middle school (grades 4 to 6 ) students from three school districts in Michigan were asked whether good grades, athletic ability, or popularity was most important to them. \({ }^{18}\) The results are shown below, broken down by gender: $$ \begin{array}{lccc} \hline & \text { Grades } & \text { Sports } & \text { Popular } \\ \hline \text { Boy } & 117 & 60 & 50 \\ \text { Girl } & 130 & 30 & 91 \end{array} $$ (a) Do these data provide evidence that grades, sports, and popularity are not equally valued among middle school students in these school districts? State the null and alternative hypotheses, calculate a test statistic, find a p-value, and answer the question. (b) Do middle school boys and girls have different priorities regarding grades, sports, and popularity? State the null and alternative hypotheses, calculate a test statistic, find a p-value, and answer the question.

Binge Drinking The American College Health Association - National College Health Assessment survey \(,{ }^{17}\) introduced on page 60 , was administered at 44 colleges and universities in Fall 2011 with more than 27,000 students participating in the survey. Students in the ACHA-NCHA survey were asked "Within the last two weeks, how many times have you had five or more drinks of alcohol at a sitting?" The results are given in Table 7.31 . Is there a significant difference in drinking habits depending on gender? Show all details of the test. If there is an association, use the observed and expected counts to give an informative conclusion in context. $$ \begin{array}{c|rr|r} \hline & \text { Male } & \text { Female } & \text { Total } \\ \hline 0 & 5,402 & 13,310 & 18,712 \\ 1-2 & 2,147 & 3,678 & 5,825 \\ 3-4 & 912 & 966 & 1,878 \\ 5+ & 495 & 358 & 853 \\ \hline \text { Total } & 8,956 & 18,312 & 27,268 \\ \hline \end{array} $$

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.