Tournament | mean | sd | n |
---|---|---|---|
men's | 150.3 | 14.2 | 16 |
women's | 150.3 | 14.2 | 16 |
Neither strong skewness nor extreme outliers are present in the data.
mean(xbar1-xbar2) = mu1 - mu2.
SE(xbar1-xbar2)
= sqrt(sigma12 / n1 + sigma22 / n2)
The shape is exactly normal if the populations are normal and approximately normal for nonnormal populations and sufficiently large sample sizes.
If population standard deviations are known, the test statistic
z = ((xbar1-xbar2) - (mu1-mu2)) / SE
has the standard normal distribution. If the population standard deviations (sigma1 and sigma2) are replaced with sample standard deviations, the test statistic has an approximate t distribution with degrees of freedom that can be estimated from the observed data.
A common approach is to assume that the two population standard deviations are equal, but unknown. In this case, the standard error is
SE = sigma sqrt( 1/n1 + 1/n2)
The common standard deviation sigma may be estimated by pooling the two sample standard deviations. The pooled estimate is the weighted average of the sample standard deviations with the weights proportional to the degrees of freedom.
sp = sqrt( ((n1-1)s12 + (n1-1)s12) / (n1 + n1 - 2 ) )
The test statistic
t = ((xbar1-xbar2) - (mu1-mu2)) / ( sp sqrt( 1/n1 + 1/n2) )
has a t distribution with n1 + n1 - 2 degrees of freedom.
(estimate) ± (multiplier)(standard error)
The standard error is 18.8 sqrt(1/16 + 1/16) = 6.65. A t multiplier for the middle 95% of a t distribution with 30 degrees of freedom is t* = 2.04. The 95% confidence interval is:
12.3 ± 13.6 or -1.3 to 25.9.
We are 95% confident that the difference in population mean total scores for men's tournaments minus women's tournaments is between a 1.3 point higher score for women to a 25.9 higher score for men.
H0: mu(men) = mu(women)
H0: mu(men) not= mu(women)
The test statistic is t = (150.3 - 138.0) / (18.8 sqrt(1/16 + 1/16) ) = 1.85.
The p-value is twice the area to the right of 1.85 under a t distribution with 30 degrees of freedom. This area is between 0.05 and 0.10.
This result is not significant at the 5% level. There is marginal evidence that the mean score in men's championship games differs from women's.
Bret Larget, larget@mathcs.duq.edu