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7.53 Testing Genetic Alleles 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 genetic allele classification \((R\) or \(X)\) and group (sprinter, endurance, control)? 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. \(\begin{array}{lrrr} & \text { R } & \text { X } & \text { Total } \\ \text { Control } & 244 & 192 & 436 \\ & 251.42 & 184.58 & \\ & 0.219 & 0.299 & \\\ \text { Sprint } & 77 & 30 & 107 \\ & 61.70 & 45.30 & \\ & 3.792 & 5.166 & \\\ \text { Endurance } & 104 & 90 & 194 \\ & 111.87 & 82.13 & \\ & 0.554 & 0.755 & \\ \text { Total } & 425 & 312 & 737\end{array}\) \(\mathrm{Ch} \mathrm{i}-\mathrm{Sq}=10.785, \mathrm{DF}=2, \mathrm{P}\) -Value \(=0.005\) (a) How many endurance athletes were included in the study? (b) What is the expected count for sprinters with the \(R\) allele? For this cell, what is the contribution to the chi-square statistic? Verify both values by computing them yourself. (c) What are the degrees of freedom for the test? Verify this value by computing it yourself. (d) What is the chi-square test statistic? What is the p-value? What is the conclusion of the test? (e) Which cell contributes the most to the chisquare statistic? For this cell, is the observed count greater than or less than the expected count? (f) Which allele is most over-represented in sprinters? Which allele is most over-represented in endurance athletes?

Short Answer

Expert verified
The study includes 194 endurance athletes. The expected count for sprinters with \(R\) allele is 61.7, with a chi-square contribution of 3.792. The degrees of freedom for the test is 2. The chi-square test statistic and p-value are 10.785 and 0.005 respectively, thus the null hypothesis is rejected, indicating a significant association. The cell with the most contribution is sprinters with \(R\) allele. The \(R\) allele is over-represented in both sprinters and endurance athletes.

Step by step solution

01

Identify endurance athlete sample size

We can observe from the given table that 194 endurance athletes were included in the study.
02

Calculate the expected count and chi-square contribution

For sprinter with \(R\) allele, as shown in the table, the expected count is 61.7 and its contribution to the chi-square statistic is 3.792. These can be verified by calculation. The expected count is calculated as (Row Total * Column Total) / Grand Total = (107 * 425) / 737. The chi-square contribution is calculated as \((Observed - Expected)^2 / Expected = (77 -61.7)^2 / 61.7 \)
03

Determine the degrees of freedom

Degrees of freedom is calculated as (Number of rows - 1) * (Number of columns - 1) = (3-1)*(2-1) = 2. As given, this value has been confirmed with a DF = 2.
04

Understand the chi-square test statistic and p-value

The chi-square test statistic is 10.785 and the p-value is 0.005, which means the probability that any observed difference between the data sets occurred by chance is quite low. For the conclusion, because the p-value is less than, usually, 0.05, the null hypothesis is rejected, suggesting a significant association between genetic allele classification and groups.
05

Identify cell with the most contribution to chi-square statistic

The cell contributes the most to the chi-square statistic is for sprinters with \(R\) allele, with a value of 3.792. The observed count for this cell (77) is greater than the expected count (61.7).
06

Determine overrepresented allele in groups

For sprinters, the over represented allele is \(R\) as it has 77 individuals, higher than expected count. Likewise, for endurance athletes, the allele \(R\) appears more frequently than expected with a count of 104

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Most popular questions from this chapter

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. $$ \begin{aligned} &H_{0}: p_{a}=p_{b}=p_{c}=p_{d}=0.25\\\ &H_{a}:\\\ &\text { Some } p_{i} \neq 0.25\\\ &\begin{array}{lccc|c} \hline \text { A } & \text { B } & \text { C } & \text { D } & \text { Total } \\\ \text { 40 } & 36 & 49 & 35 & 160 \\ \hline \end{array} \end{aligned} $$

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