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Qualifications of male and female head and assistant college athletic coaches were compared in the article "Sex Bias and the Validity of Believed Differences Between Male and Female Interscholastic Athletic Coaches" (Res. Q. Exercise Sport, 1990: 259-267). Each person in random samples of 2225 male coaches and 1141 female coaches was classified according to number of years of coaching experience to obtain the accompanying two-way table. Is there enough evidence to conclude that the proportions falling into the experience categories are different for men and women? Use \(\alpha=.01\). $$ \begin{array}{lccccc} \hline & \multicolumn{5}{c}{\text { Years of Experience }} \\ \cline { 2 - 6 } \text { Gender } & \mathbf{1 - 3} & \mathbf{4 - 6} & \mathbf{7 - 9} & \mathbf{1 0 - 1 2} & \mathbf{1 3 +} \\ \hline \text { Male } & 202 & 369 & 482 & 361 & 811 \\ \text { Female } & 230 & 251 & 238 & 164 & 258 \\ \hline \end{array} $$

Short Answer

Expert verified
Conduct a chi-square test to see if the distributions differ significantly; follow steps with critical value comparison.

Step by step solution

01

Set Up the Hypothesis

First, we establish the null and alternative hypotheses. The null hypothesis \((H_0)\) states that there is no difference in the proportions of male and female coaches in each experience category. The alternative hypothesis \((H_a)\) states that there is a difference.
02

Collect the Data

From the table, we have the data for male and female coaches distributed across five levels of years of experience. This provides us with the observed frequencies of male and female coaches in each category.
03

Calculate Expected Frequencies

We calculate the expected frequency for each cell in the table using the formula: \[ E_{ij} = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}} \]where \(E_{ij}\) is the expected frequency for the cell in the \(i\)-th row and \(j\)-th column.
04

Compute the Chi-Square Statistic

Use the formula: \[ \chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}} \]where \(O_{ij}\) is the observed frequency and \(E_{ij}\) is the expected frequency calculated for each cell.
05

Determine Degrees of Freedom

The degrees of freedom \((df)\) for the chi-square test is calculated by \[ df = (r-1)(c-1) \]where \(r\) is the number of rows and \(c\) is the number of columns. Here, \(df = (2-1)(5-1) = 4\).
06

Find the Critical Value

Using a chi-square distribution table, find the critical value with \(df = 4\) and \(\alpha = 0.01\). The critical value is approximately 13.277.
07

Make the Decision

Compare the computed chi-square statistic to the critical value. If the statistic is greater than the critical value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.
08

Conclusion

Based on the comparison, if the chi-square statistic is greater, we conclude there is enough evidence to suggest the distribution of years of experience is different for male and female coaches. If not, we conclude there is not enough evidence.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about the data. Typically, it starts with two opposing hypotheses: the null hypothesis (denoted as \(H_0\)) and the alternative hypothesis (denoted as \(H_a\)).
- The null hypothesis asserts there is no effect or no difference between groups. In this exercise, \(H_0\) states there is no difference in the distribution of years of experience between male and female coaches.
- The alternative hypothesis suggests the opposite; in this case, \(H_a\) says there is a difference in those distributions.
Selecting an appropriate level of significance, \(\alpha\), is crucial. Here, we use \(\alpha = 0.01\), which means that we accept a 1% risk of incorrectly rejecting the null hypothesis. This step helps control the probability of type I error, where we mistakenly deduce that there's a difference when there isn't one. The goal of hypothesis testing is to use sample data to decide whether to reject the null hypothesis or not. This decision is made by looking at specific test results, like the chi-square statistic, and comparing them to critical values from statistical tables.
Degrees of Freedom
Degrees of freedom (df) are critical to understanding the shape of the chi-square distribution we use. They essentially represent the number of independent values or quantities that can be assigned to a statistical distribution. In the context of the chi-square test, degrees of freedom relate to how many values can vary in the table without breaking constraints.
For our problem, degrees of freedom are determined by the formula:
\[ df = (r - 1)(c - 1) \]
where \(r\) is the number of rows (here representing gender), and \(c\) is the number of columns (the experience categories).So, substituting the given values, we have:
\[ (2 - 1)(5 - 1) = 4 \]
This means we have four degrees of freedom.
The degrees of freedom will inform which chi-square distribution table we reference in later steps, ensuring we use the correct critical value when making a decision about the null hypothesis.
Critical Value
The critical value is a threshold that helps determine whether our test statistic is extreme enough to reject the null hypothesis. For the chi-square test in this exercise, we rely on a chi-square distribution table to find this critical value based on degrees of freedom and the chosen significance level.
Using \(df = 4\) and \(\alpha = 0.01\), we look up the chi-square distribution table. We find that the critical value is approximately 13.277. This point reflects the value beyond which only 1% of the distribution's area lies, given four degrees of freedom.
- If our calculated chi-square statistic is greater than 13.277, we reject \(H_0\). - If it's less, we do not reject \(H_0\).
This comparison guides our final decision in hypothesis testing. A statistic greater than 13.277 signals significant differences in experience distribution between genders.
Observed and Expected Frequencies
In a chi-square test, we deal with both observed and expected frequencies.
- **Observed frequencies (\(O_{ij}\))** are the frequencies recorded directly from the collected data. They represent the actual counts in each category, like the number of male and female coaches with a certain range of experience.
- **Expected frequencies (\(E_{ij}\))** are what we would theoretically expect if there were no relationship between gender and experience. We calculate them using the formula:
\[ E_{ij} = \frac{(\text{Row Total} \times \text{Column Total})}{\text{Grand Total}} \]
The expected frequency helps us understand what the distribution would look like under the null hypothesis. By comparing observed and expected frequencies, we assess whether any observed differences are too large to be due to random chance.
The comparison is quantified using the chi-square statistic, which measures the extent of deviation from what was expected if \(H_0\) were true. This step is key in identifying if there's a significant association between gender and the years of coaching experience.

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Most popular questions from this chapter

A study of sterility in the fruit fly ("Hybrid Dysgenesis in Drosophila melanogaster: The Biology of Female and Male Sterility," Genetics, 1979: 161-174) reports the following data on the number of ovaries developed for each female fly in a sample of size 1,388 . One model for unilateral sterility states that each ovary develops with some probability \(p\) independently of the other ovary. Test the fit of this model using \(\chi^{2}\). \begin{tabular}{l|ccc} \(x=\) Number of Ovaries Developed & 0 & 1 & 2 \\ \hline Observed Count & 1212 & 118 & 58 \end{tabular}

In a study of 2989 cancer deaths, the location of death (home, acute-care hospital, or chronic-care facility) and age at death were recorded, resulting in the given two-way frequency table ("Where Cancer Patients Die," Public Health Rep., 1983: 173). Using a \(.01\) significance level, test the null hypothesis that age at death and location of death are independent. $$ \begin{array}{lccc} \hline & \multicolumn{3}{c}{\text { Location }} \\ \cline { 2 - 4 } \text { Age } & \text { Home } & \text { Acute-Care } & \text { Chronic-Care } \\ \hline \mathbf{1 5 - 5 4} & 94 & 418 & 23 \\ \mathbf{5 5 - 6 4} & 116 & 524 & 34 \\ \mathbf{6 5 - 7 4} & 156 & 581 & 109 \\ \text { Over } \mathbf{7 4} & 138 & 558 & 238 \\ \hline \end{array} $$

A certain type of flashlight is sold with the four batteries included. A random sample of 150 flashlights is obtained, and the number of defective batteries in each is determined, resulting in the following data: \begin{tabular}{l|ccccc} Number Defective & 0 & 1 & 2 & 3 & 4 \\ \hline Frequency & 26 & 51 & 47 & 16 & 10 \end{tabular} Let \(X\) be the number of defective batteries in a randomly selected flashlight. Test the null hypothesis that the distribution of \(X\) is \(\operatorname{Bin}(4, \theta)\). That is, with \(p_{i}=P(i\) defectives), test \(H_{0}: p_{i}=\left(\begin{array}{l}4 \\ i\end{array}\right) \theta^{i}(1-\theta)^{4-i} \quad i=0,1,2,3,4\) [Hint: To obtain the mle of \(\theta\), write the likelihood (the function to be maximized) as \(\theta^{u}(1-\theta)^{v}\), where the exponents \(u\) and \(v\) are linear functions of the cell counts. Then take the natural \(\log\), differentiate with respect to \(\theta\), equate the result to 0 , and solve for \(\hat{\theta}\).]

The article "Susceptibility of Mice to Audiogenic Seizure Is Increased by Handling Their Dams During Gestation" (Science, 1976: 427-428) reports on research into the effect of different injection treatments on the frequencies of audiogenic seizures. \begin{tabular}{lc|c|c|c} & \multicolumn{1}{c}{ No } & \multicolumn{1}{c}{ Wild } & Clonic & Tonic \\ Treatment & Response & \multicolumn{1}{c}{ Running } & \multicolumn{1}{c}{ Seizure } & \multicolumn{1}{c}{ Seizure } \\ \cline { 2 - 5 } Thienylalanine & 21 & 7 & 24 & 44 \\ \cline { 2 - 5 } Solvent & 15 & 14 & 20 & 54 \\ \cline { 2 - 5 } Sham & 23 & 10 & 23 & 48 \\ \cline { 2 - 5 } Unhandled & 47 & 13 & 28 & 32 \\ \cline { 2 - 5 } & & & & \end{tabular} Does the data suggest that the true percentages in the different response categories depend on the nature of the injection treatment? State and test the appropriate hypotheses using \(\alpha=.005\).

The accompanying data on sex combinations of two recombinants resulting from six different male genotypes appears in the article "A New Method for Distinguishing Between Meiotic and Premeiotic Recombinational Events in Drosophila melanogaster" (Genetics, 1979: 543-554). Does the data support the hypothesis that the frequency distribution among the three sex combinations is homogeneous with respect to the different genotypes? Define the parameters of interest, state the appropriate \(H_{0}\) and \(H_{\mathrm{a}}\), and perform the analysis. $$ \begin{array}{llccr} \hline & & \multicolumn{3}{c}{\text { Sex Combination }} \\ \cline { 3 - 5 } & & \text { M/M } & \text { M/F } & \text { F/F } \\ \cline { 3 - 5 } \text { Male } & 1 & 35 & 80 & 39 \\ & 2 & 41 & 84 & 45 \\ \text { Genotype } & 3 & 33 & 87 & 31 \\ & 4 & 8 & 26 & 8 \\ & 5 & 5 & 11 & 6 \\ & 6 & 30 & 65 & 20 \\ \hline \end{array} $$

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