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Chapter 12: Problem 19

In his book Outliers, Malcolm Gladwell claims that more hockey players are born in January through March than in October through December. The following data show the number of players in the National Hockey League in the 2014-2015 season according to their birth month. Is there evidence to suggest that professional hockey players' birth dates are not uniformly distributed throughout the year at the \(\alpha=0.05\) level of significance? $$ \begin{array}{lc} \text { Birth Month } & \text { Frequency } \\ \hline \text { January-March } & 278 \\ \hline \text { April-June } & 246 \\ \hline \text { July-September } & 163 \\ \hline \text { October-December } & 143\\\ \hline \end{array} $$

Short Answer

Expert verified
Reject the null hypothesis; birth dates are not uniformly distributed.

Step by step solution

01

- State the Hypotheses

The null hypothesis (ull hypothesis) states that birthdates are uniformly distributed throughout the year. This can be written as \( H_0: P(Jan-Mar) = P(Apr-Jun) = P(Jul-Sep) = P(Oct-Dec) \). The alternative hypothesis (ull hypothesis) states that birthdates are not uniformly distributed. This can be written as \( H_a: \text{at least one } P(\text{month group}) e \frac{1}{4} \).
02

- Determine the Expected Frequencies

The total number of players is \( N = 278 + 246 + 163 + 143 = 830 \). Since under the null hypothesis, the birth frequencies should be uniformly distributed, the expected frequency for each quarter of the year is \( E_i = \frac{N}{4} = \frac{830}{4} = 207.5 \).
03

- Calculate the Chi-Square Test Statistic

The Chi-Square test statistic is calculated using the formula \[ \ \chi^2 = \sum\frac{ (O_i - E_i)^2 }{ E_i } \ \] where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. Plug in the values: \[ \ \chi^2 = \frac{(278 - 207.5)^2}{207.5} + \frac{(246 - 207.5)^2}{207.5} + \frac{ (163 - 207.5)^2 }{ 207.5 } + \frac{ (143 - 207.5)^2 }{ 207.5 } \ \chi^2 = \frac{(70.5)^2}{207.5} + \frac{(38.5)^2}{207.5} + \frac{(44.5)^2}{207.5} + \frac{(64.5)^2}{207.5} \ \chi^2 \approx 23.975 \]
04

- Determine the Critical Value

The critical value can be found using a Chi-Square distribution table. Here, the degrees of freedom (ull degrees of freedom) is \ df = k - 1 = 4 - 1 = 3 \ where ull hypothesisull hypothesis, the number of categories (ull hypothesis) is 4. The critical value of \( \chi^2 \) for \( df = 3 \) at \( \ull hypothesis..05 \) is 7.815.
05

- Make the Decision

Compare the test statistic to the critical value: \[ \23.975 \gt 7.815 \]. Since the test statistic is greater than the critical value, reject the null hypothesis. There is enough evidence at the \(\0.05 \) level of significance to conclude that professional hockey players' birth dates are not uniformly distributed throughout the year.

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Key Concepts

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

Uniform Distribution
In statistics, a **uniform distribution** is a type of probability distribution in which all outcomes are equally likely. For example, if birthdates of hockey players were uniformly distributed, each quarter of the year (January-March, April-June, July-September, October-December) would see roughly the same number of births. In mathematical terms, if there are 830 total players, we would expect around 207.5 players to be born in each quarter. This idea forms the bedrock of comparing our observed data (actual birth frequencies) with what we would expect under uniform distribution.
Hypothesis Testing
The process of **hypothesis testing** allows us to use statistical methods to determine if there is enough evidence to reject a preconceived notion regarding our data (the null hypothesis). For this particular problem:

The null hypothesis (\(H_0\)) assumes that the birthdates of hockey players are uniformly distributed. In other words, each quarter has an equal probability of 25% of containing a hockey player's birthdate.
The alternative hypothesis (\(H_a\)) suggests that the birthdates are not uniformly distributed.
Hypothesis testing uses data to decide whether to accept or reject this null hypothesis, based on the computed test statistics and associated critical values.
Expected Frequency
The **expected frequency** is what we anticipate observing in each category if the null hypothesis were true. For a uniform distribution in our problem, this value can be calculated by dividing the total number of observations by the number of categories. With 830 players and four quarters:
\(E_i = \frac{830}{4} = 207.5\)
This means we would expect around 207.5 players to be born in each quarter. This expectation is a key part of calculating the chi-square test statistic, as it provides the baseline against which the actual (observed) frequencies are compared.
Test Statistic
The **test statistic** in a chi-square test measures how much the observed data deviate from the expected data. It helps us quantify the discrepancy between what we observed and what was expected under the null hypothesis. The chi-square test statistic is calculated using:

\(\chi^2 = \sum \frac{ (O_i - E_i)^2 }{ E_i }\)
Where:
\(O_i\) are the observed frequencies and \(E_i\) are the expected frequencies.
In our example, the test statistic calculation is:
\[ \chi^2 = \frac{(278-207.5)^2}{207.5} + \frac{(246-207.5)^2}{207.5} + \frac{(163-207.5)^2}{207.5} + \frac{(143-207.5)^2}{207.5} \approx 23.975 \]
This statistic tells us how far our observed data diverge from what we would expect if birthdates were uniformly distributed.
Significance Level
The **significance level** (\(\alpha\)) determines the threshold for rejecting the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true (also known as Type I error). Common significance levels are 0.05 or 0.01. In our problem,
\(\alpha=0.05\)
This means we are willing to tolerate a 5% chance of incorrectly rejecting the null hypothesis.
In the final step of hypothesis testing, we compare our test statistic (23.975) to the critical value from the chi-square distribution table for 3 degrees of freedom at \(\alpha=0.05\), which is 7.815. Since 23.975 > 7.815, we reject the null hypothesis, concluding there is significant evidence to suggest the birthdates of hockey players are not uniformly distributed.

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Most popular questions from this chapter

Determine the expected counts for each outcome. $$ \begin{array}{lllll} \hline \boldsymbol{n}=\mathbf{5 0 0} & & & & \\ \hline p_{i} & 0.2 & 0.1 & 0.45 & 0.25 \\ \hline \text { Expected counts } & & & & \\ \hline \end{array} $$

At Joliet Junior College, the mathematics department decided to offer a redesigned course in Intermediate Algebra, called the Math Redesign Program (MRP). Laura Egner, the coordinator of the program, wanted to determine if the grade distribution in the course differed from that of traditional courses. The following shows the grade distribution of traditional courses based on historical records and the observed grades in three pilot classes in which the MRP program was utilized. $$ \begin{array}{lcccccc} & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} & \mathbf{F} & \mathbf{W} \\\ \hline \begin{array}{l} \text { Traditional } \\ \text { Distribution } \end{array} & 0.133 & 0.191 & 0.246 & 0.104 & 0.114 & 0.212 \\ \hline \begin{array}{l} \text { Observed Counts } \\ \text { in MRP Program } \end{array} & 7 & 16 & 10 & 13 & 6 & 12 \\ \hline \end{array} $$ (a) How many students were enrolled in the MRP program for the three pilot courses? Based on this result, determine the expected number of students for each grade assuming there is no difference in the distribution of MRP student grades and traditional grades. (b) Does the sample evidence suggest that the distribution of grades is different from the traditional classes at the \(\alpha=0.01\) level of significance? (c) Explain why it makes sense to use 0.01 as the level of significance. (d) Suppose the MRP pilot program continues in three more classes with the grades earned for all six pilot courses shown below. Notice that the sample size was simply doubled with the grade distribution remaining unchanged. Does this sample evidence suggest that the distribution of grades is different from the traditional classes at the \(\alpha=0.01\) level of significance? What does this result suggest about the role of sample size in the ability to reject a statement in the null hypothesis? $$ \begin{array}{lcccccc} & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} & \mathbf{F} & \mathbf{W} \\\ \hline \begin{array}{l} \text { Observed Counts } \\ \text { in MRP Program } \end{array} & 14 & 32 & 20 & 26 & 12 & 24 \end{array} $$

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Religion in Congress Is the religious make-up of the United States Congress reflective of that in the general population? The following table shows the religious affiliation of the 535 members of the 114 th Congress along with the religious affiliation of a random sample of 1200 adult Americans. $$ \begin{array}{lcc} \text { Religion } & \begin{array}{c} \text { Number of } \\ \text { Members } \end{array} & \begin{array}{c} \text { Sample of } \\ \text { Residents } \end{array} \\ \hline \text { Protestant } & 306 & 616 \\ \hline \text { Catholic } & 164 & 287 \\ \hline \text { Mormon } & 16 & 20 \\ \hline \text { Orthodox Christian } & 5 & 7 \\ \hline \text { Jewish } & 28 & 20 \\ \hline \text { Buddhist/Muslim/Hindu/Other } & 6 & 57 \\ \hline \text { Unaffiliated/Don't Know/Refused } & 10 & 193 \\ \hline \end{array} $$ (a) Determine the probability distribution for the religious affiliation of the members of the 114 th Congress. (b) Assuming the distribution of the religious affiliation of the adult American population is the same as that of the Congress, determine the number of adult Americans we would expect for each religion from a random sample of 1200 individuals. (c) The data in the third column represent the declared religion of a random sample of 1200 adult Americans (based on data obtained from Pew Research). Do the sample data suggest that the American population has the same distribution of religious affiliation as the 114 th Congress? (d) Explain what the results of your analysis suggest.

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