Lippa had gathered survey data from more than 50 short-haired men and photographed their pates (women were excluded because their hairstyles, even at the pride festival, were too long for simple determination; crewcuts are the ideal Rorschach, he explains). About 23 percent had counterclockwise hair whorls. In the general population, that figure is 8 percent.
Well, just how meaningful is this 23% discrepancy from the norm of 8%? Maybe it’s just randomness, right? Well, try the omitted calculations for yourself. This is an example of a “hypothesis test” in Statistics. The Null Hypothesis (H0) says that there is no difference in the groups. The Alternative Hypothesis (HA) says there is a statistically significant difference in the groups. In a hypothesis test, the essential question is this: What are the odds that a sample varies this much from the expected percentage (proportion) simply due to natural random variation? (For example, if you flip a coin 10 times, you usually get 5 heads. Sometimes, however, you might get 6. In fact, that should happen 26% of the time. Nothing to be alarmed about. However, the odds of getting 8 heads is only about 3%. If you do get 8 heads, that’s rare enough to indicate the coin might be rigged. Odds are you won’t do it again!)
So, for this hair test, we need to ask, “What are the odds of taking a sample of 50 guys and seeing that 23% having a counterclockwise whorl?” We should expect to get 8%, as per the broad population. If it’s very very rare to get 23%, then we might suspect there is a connection, and gay men do have different hair swirls than the broad population. In Statistics, we define “very very rare” as under 5%. In other words, if the odds that 23% of a sample of 50 have a counterclockwise whorl is under 5%, then it is statistically significant.
The calculations:
H0: Gay men have no difference in their hair whorl orientation.
HA: Gay men do have a difference in their hair whorl orientation.
First, take stock of the given information:
\(n = 50 \\\\ p = .08 \\\\ \hat{p}=.23\)
Next, you calculate the standard deviation of samples of this size.
\(SD(\hat{p})= \sqrt{\frac{(.08)(.92)}{50}}=.04\)
To determine how unlikely your sampling result was, you calculate how many standard deviations away from the expected proportion it was (Z-score).
\(Z(\hat{p})= \frac{\hat{p}-p_0}{SD(\hat{p})}=\frac{.23-.08}{.04}=3.75\)
Then, you calculate the odds of getting this Z-score via the normal cumulative distribution function. (What are the odds of this happening randomly?) If it’s under 5%, then you reject the null hypothesis, because it’s unlikely this variation can be attributed to random chance. ie: Odds are, the hair is indeed different.
\(p(Z \ge 3.75) = .000088 = 0\% \)
Conclusion: If the odds of having counterclockwise hair whorl is 8%, then the odds of having 23% of 50 random men exhibit this trait is unlikely. The odds of this happening by chance (random variations) is basically 0%. So, we reject the null hypothesis (H0), and accept the alternative hypothesis (HA)