In: Statistics and Probability
Birth Date and Canadian Ice Hockey
In his book Outliers: The Story of Success (2008), Malcolm
Gladwell speculates that Canadian ice hockey players that are born
early in the year have an advantage. This is because the birthdate
cutoff for different levels of youth hockey leagues in Canada is
January 1st, so youth hockey players who are born in January and
February are slightly older than teammates born later in the year.
Does this slight age advantage in the beginning lead to success
later on? A 2010 study1 examined the birthdate
distribution of players in the Ontario Hockey League (OHL), a
high-level and selective Canadian hockey league (ages 15-20), for
the 2008–2009 season. The number of OHL players born during the 1st
quarter (Jan–Mar), 2nd quarter (Apr–Jun), 3rd quarter (Jul–Sep),
and 4th quarter (Oct–Dec) of the year is shown in the table below.
The overall percentage of live births in Canada (year 1989) are
also provided for each quarter. Is this evidence that the birthdate
distribution for OHL players differs significantly from the
national proportions? Calculate the chi-square statistic, find the
p-value, and state the conclusion in context.
Qtr 1 | Qtr 2 | Qtr 3 | Qtr 4 | |
---|---|---|---|---|
OHL Player | 147 | 110 | 52 | 50 |
% of Canadian births | 23.7% | 25.9% | 25.9% | 24.5% |
Table 1 Birthdates nationally in Canada and for elite hockey
players
1Nolan, J. and Howell, G., "Hockey success and birth
date: The relative age effect revisited," International Review
of Sociology of Sport, 2010; 45(4): 507–512.
Calculate the chi-square test statistic and the p-value.
Round your answer for the chi-square statistic to two decimal
places, and your answer for the p-value to three decimal
places.
χ2= __________________________
p-value = ______________________
Is there evidence that the birthdate distribution for OHL players differs significantly from the national proportions?
Yes or No?
Categories | Observed | Expected | (fo-fe)2/fe |
Qrt1 | 147 | 359*0.237=85.083 | (147-85.083)2/85.083 = 45.059 |
Qtr2 | 110 | 359*0.259=92.981 | (110-92.981)2/92.981 = 3.115 |
Qtr3 | 52 | 359*0.259=92.981 | (52-92.981)2/92.981 = 18.062 |
Qtr4 | 50 | 359*0.245=87.955 | (50-87.955)2/87.955 = 16.379 |
Sum = | 359 | 359 | 82.614 |
The P-Value is < .00001. The result is significant at p < .05.