/*! 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 14 Not all ethnic groups have the s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 14

Not all ethnic groups have the same mix of blood types and \(\mathrm{Rh}\) factors. For example, Latino-Americans have a high number of Os while Asians have a high number of Bs. \({ }^{12}\) A tabulation of blood types including Rh factors for 300 people in each of these ethnic groups is given below. $$ \begin{array}{lllllllll} \hline \text { Type } & \text { O+ } & \text { O? } & \text { A+ } & \text { A- } & \text { B+ } & \text { B- } & \text { AB+ } & \text { AB- } \\ \hline \text { Latino- } & & & & & & & & \\ \text { American } & 161 & 10 & 88 & 6 & 21 & 5 & 6 & 3 \\ \text { Asian } & 115 & 4 & 79 & 4 & 72 & 3 & 19 & 4 \end{array} $$ Do these data provide evidence to conclude that the proportions of people in the various blood groups differ for these two ethnic groups? Use \(\alpha=.01\)

Short Answer

Expert verified
Answer: [Based on Step 6 conclusion: Either "Yes, the proportions of people in the various blood groups differ for these two ethnic groups." OR "No, we cannot conclude that the proportions of people in the blood groups differ for these two ethnic groups."]

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) states that there is no association between ethnic groups and blood types. The alternative hypothesis (\(H_1\)) states that there is an association between ethnic groups and blood types.
02

Calculate expected counts

We need to calculate the expected counts for each blood group in both ethnic groups based on the assumption that the null hypothesis is true. The expected count for each cell can be calculated using the formula: \(E_{ij} = \frac{(n_{i.})(n_{.j})}{n}\), where \(E_{ij}\) is the expected count for a particular cell, \(n_{i.}\) is the row total of the \(i^{th}\) row, \(n_{.j}\) is the column total of the \(j^{th}\) column, and \(n\) is the total number of samples. Calculate the row and column totals and the total number of samples: Latino-American total = 300 Asian total = 300 Total samples (n) = 600 Column totals: O+: 276 O-: 14 A+: 167 A-: 10 B+: 93 B-: 8 AB+: 25 AB-: 7 Now, calculate the expected counts for each cell using the above formula.
03

Calculate the chi-square test statistic

For each cell, subtract the expected count value from the observed count value, square the result, and divide by the expected count value. Sum these values across all cells to get the chi-square test statistic (\(\chi^2\)). \(\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}\)
04

Determine the degrees of freedom

In a chi-square test for independence, degrees of freedom for the test statistic are calculated as: df = (number of rows - 1) * (number of columns - 1) In this case, df = (2 - 1) * (8 - 1) = 1 * 7 = 7
05

Find the critical chi-square value and make a decision

We will use the chi-square distribution table to find the critical value at the 0.01 significance level with 7 degrees of freedom. The critical value for our test is 18.475. Now, compare the calculated chi-square test statistic with the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is an association between blood groups and ethnic groups. If the test statistic is less than the critical value, we fail to reject the null hypothesis and cannot conclude that there is an association between blood groups and ethnic groups.
06

Conclusion

Based on the comparison in step 5, either reject the null hypothesis or fail to reject it. If rejected, then the proportions of people in the various blood groups differ for these two ethnic groups. If the null hypothesis is not rejected, we cannot conclude that the proportions of people in the blood groups differ for these two ethnic groups.

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.

Null Hypothesis in Statistics
The null hypothesis, often represented as \(H_0\), is a core concept in inferential statistics. It posits that there is no significant effect or relationship between the variables under investigation. In the context of a chi-square test for independence, the null hypothesis maintains that there is no association between the categorical variables being examined, such as ethnic groups and blood type distribution in our example.

When conducting a chi-square test, if the observed data significantly deviates from what would be expected under 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

To study the effect of worker participation in managerial decision making, 100 workers were interviewed in each of two separate German manufacturing plants. One plant had active worker participation in managerial decision making; the other did not. Each selected worker was asked whether he or she generally approved of the managerial decisions made within the firm. The results of the interviews are shown in the table: $$ \begin{array}{lcc} \hline & \text { Participation } & \text { No Participation } \\ \hline \text { Generally Approve } & 73 & 51 \\ \text { Do Not Approve } & 27 & 49 \\ \hline \end{array} $$ a. Do the data provide sufficient evidence to indicate that approval or disapproval of management's decisions depends on whether workers participate in decision making? Test by using the \(X^{2}\) test statistic. Use \(\alpha=.05 .\) b. Do these data support the hypothesis that workers in a firm with participative decision making more generally approve of the firm's managerial decisions than those employed by firms without participative decision making? Test by using the \(z\) -test presented in Section \(9.5 .\) This problem requires a one-tailed test. Why?

Pfizer Canada Inc. is a pharmaceutical company that makes azithromycin, an antibiotic in a cherry-flavored suspension used to treat bacterial infections in children. To compare the taste of their product with three competing medications, Pfizer tested 50 healthy children and 20 healthy adults. Among other taste-testing measures, they recorded the number of tasters who rated each of the four antibiotic suspensions as the best tasting. \({ }^{13}\) Is there a difference in the perception of the best taste between adults and children? If so, what is the nature of the difference, and why is it of practical importance to the pharmaceutical company? $$ \begin{array}{lcccc} \hline &&{\text { Flavor of Antibiotic }} \\ \hline & \text { Banana } & \text { Cherry* } & \text { Wild Fruit } & \text { Strawberry-Banana } \\ \hline \text { Children } & 14 & 20 & 7 & 9 \\ \text { Adults } & 4 & 14 & 0 & 2 \\ \hline \end{array} $$*Azithromycin produced by Pfizer Canada Inc.

Find the appropriate degrees of freedom for the chisquare test of independence. $$\text { three rows and three columns }$$

A department store manager claims that her store has twice as many customers on Fridays and Saturdays than on any other day of the week (the store is closed on Sundays). That is, the probability that a customer visits the store Friday is \(2 / 8\), the probability that a customer visits the store Saturday is \(2 / 8\), while the probability that a customer visits the store on each of the remaining weekdays is \(1 / 8\). During an average week, the following numbers of customers visited the store: $$ \begin{array}{lr} \hline \text { Day } & \text { Number of Customers } \\ \hline \text { Monday } & 95 \\ \text { Tuesday } & 110 \\ \text { Wednesday } & 125 \\ \text { Thursday } & 75 \\ \text { Friday } & 181 \\ \text { Saturday } & 214 \end{array} $$ Can the manager's claim be refuted at the \(\alpha=.05\) level of significance?

Suppose you wish to test the null hypothesis that three binomial parameters \(p_{A}, p_{B},\) and \(p_{c}\) are equal versus the alternative hypothesis that at least two of the parameters differ. Independent random samples of 100 observations were selected from each of the populations. Use the information in the table to answer the questions in Exercises \(5-7 .\) $$ \begin{array}{lrrrr} \hline & {\text { Population }} & \\ & \text { A } & \text { B } & \text { C } & \text { Total } \\ \hline \text { Successes } & 24 & 19 & 33 & 76 \\ \text { Failures } & 76 & 81 & 67 & 224 \\ \hline \text { Total } & 100 & 100 & 100 & 300 \end{array} $$ Write the null and alternative hypotheses for testing the equality of the three binomial proportions.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.