Logout Open In App
Open In App
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

Chapter 7: Problem 54

Gender and ACTN3 Genotype We see in the previous two exercises that sprinters are more likely to have allele \(R\) and genotype \(R R\) versions of the ACTN3 gene, which makes these versions associated with fast-twitch muscles. Is there an association between genotype and gender? Computer output is shown for this chi-square test, using the control group in the study. In each cell, the top number is the observed count, the middle number is the expected count, and the bottom number is the contribution to the chi-square statistic. What is the p-value? What is the conclusion of the test? Is gender associated with the likelihood of having a "sprinting gene"? \(\begin{array}{lrrrr} & \text { RR } & \text { RX } & \text { XX } & \text { Total } \\ \text { Male } & 40 & 73 & 21 & 134 \\ & 40.26 & 69.20 & 24.54 & \\\ & 0.002 & 0.208 & 0.509 & \\ \text { Female } & 88 & 147 & 57 & 292 \\ & 87.74 & 150.80 & 53.46 & \\ & 0.001 & 0.096 & 0.234 & \\ \text { Total } & 128 & 220 & 78 & 426\end{array}\) \(\mathrm{Chi}-\mathrm{Sq}=1.050, \mathrm{DF}=2, \mathrm{P}\) -Value \(=0.592\)

Short Answer

Expert verified
The p-value is 0.592. The conclusion of the test is that there is not enough evidence to suggest there's an association between gender and the likelihood of having a 'sprinting gene.

Step by step solution

01

Review the Chi-square test results

The Chi-square statistic is 1.050 with degree of freedom (DF) equal to 2. The p value is given as 0.592.
02

Interpret the p-value

The p-value is a probability, with a value ranging from 0 to 1. A smaller the p-value means it is more unlikely we would have seen the data we have if there were no association between the categorical variables. Usually, if p-value is less than 0.05, we reject the null hypothesis of no association and conclude that there is an association. In this case, the p-value is 0.592, which is greater than 0.05.
03

Conclusion of the test

Our p-value of 0.592 is greater than our significance level of 0.05, so we do not reject the null hypothesis. That is, we do not have enough evidence to suggest there is an association between gender and the likelihood of having a 'sprinting gene'.

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.

Start your free trial

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

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

Examining Genetic Alleles in Fast-Twitch Muscles Exercise 7.24 discusses a study investigating the \(A C T N 3\) genotypes \(R R, R X,\) and \(X X .\) The same study also examines the \(A C T N 3\) genetic alleles \(R\) and \(X,\) also associated with fast-twitch muscles. Of the 436 people in this sample, 244 were classified \(R\) and 192 were classified \(X .\) Does the sample provide evidence that the two options are not equally likely? (a) Conduct the test using a chi-square goodnessof-fit test. Include all details of the test. (b) Conduct the test using a test for a proportion, using \(H_{0}: p=0.5\) where \(p\) represents the proportion of the population classified \(R .\) Include all details of the test. (c) Compare the p-values and conclusions of the two methods.

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. (Group 3. Yes) cell $$ \begin{array}{l|rr|r} \hline & \text { Yes } & \text { No } & \text { Total } \\ \hline \text { Group 1 } & 56 & 44 & 100 \\ \text { Group 2 } & 132 & 68 & 200 \\ \text { Group 3 } & 72 & 28 & 100 \\ \hline \text { Total } & 260 & 140 & 400 \\ \hline \end{array} $$

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. $$ \begin{array}{c|rr|r} \hline & \text { Male } & \text { Female } & \text { Total } \\ \hline 0 & 5,402 & 13,310 & 18,712 \\ 1-2 & 2,147 & 3,678 & 5,825 \\ 3-4 & 912 & 966 & 1,878 \\ 5+ & 495 & 358 & 853 \\ \hline \text { Total } & 8,956 & 18,312 & 27,268 \\ \hline \end{array} $$

Exercises 7.9 to 7.12 give a null hypothesis for a goodness-of-fit test and a frequency table from a sample. For each table, find: (a) The expected count for the category labeled B. (b) The contribution to the sum of the chi-square statistic for the category labeled \(\mathrm{B}\). (c) The degrees of freedom for the chi-square distribution for that table. $$ \begin{aligned} &H_{0}: p_{a}=0.2, p_{b}=0.80\\\ &H_{a}: \text { Some } p_{i} \text { is wrong }\\\ &\begin{array}{ll} \mathrm{A} & \mathrm{B} \end{array}\\\ &132 \quad 468 \end{aligned} $$

Testing Genotypes for Fast-Twitch Muscles The study on genetics and fast- twitch muscles includes a sample of elite sprinters, a sample of elite endurance athletes, and a control group of nonathletes. Is there an association between genotype classification \((R R, R X,\) or \(X X)\) and group (sprinter, endurance, control group)? Computer output is shown for this chi- square test. In each cell, the top number is the observed count, the middle number is the expected count, and the bottom number is the contribution to the chi-square statistic. \(\begin{array}{lrrrr} & \text { RR } & \text { RX } & \text { XX } & \text { Total } \\ \text { Control } & 130 & 226 & 80 & 436 \\ & 143.76 & 214.15 & 78.09 & \\ & 1.316 & 0.655 & 0.047 & \\ \text { Sprint } & 53 & 48 & 6 & 107 \\\ & 35.28 & 52.56 & 19.16 & \\ & 8.901 & 0.395 & 9.043 & \\ \text { Endurance } & 60 & 88 & 46 & 194 \\ & 63.96 & 95.29 & 34.75 & \\ & 0.246 & 0.558 & 3.645 & \\ \text { Total } & 243 & 362 & 132 & 737\end{array}\) Chi-Sq \(=24.805, \mathrm{DF}=4, \mathrm{P}\) -Value \(=0.000\) (a) What is the expected count for endurance athletes with the \(X X\) genotype? For this cell, what is the contribution to the chi-square statistic? Verify both values by computing them yourself. (b) What are the degrees of freedom for the test? Verify this value by computing it yourself. (c) What is the chi-square test statistic? What is the p-value? What is the conclusion of the test? (d) Which cell contributes the most to the chisquare statistic? For this cell, is the observed count greater than or less than the expected count? (e) Which genotype is most over-represented in sprinters? Which genotype is most overrepresented in endurance athletes?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

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.

Sign-up for free