## Two-Sample Test for Proportions

The Null hypothesis of "No Effect" can be re-written as:
##### H0: "Proportion of Colds among Placebo users = Prop. of Colds among Vitamin C users"

If H0 is true we could estimate the common proportion P of colds by

### EP = (total no. of colds) / (total no. of Skiers) = 0.17

The z test statistic is then given by:
### z = (obs. difference)/ SE

Where
### (obs. difference) = 31/140 - 17/139 = 0.1

and the SE for the difference is given by the sqrt-law:
### SQRT( EP*(1-EP)*(1/140 + 1/139) ) = 4.57e-4

Therefore the observed test statistic is:
### z = 0.1 / 4.57e-4 = 2.19

From the normal table we obtain the area between -2.19 and 2.19 to be about
97.2%. This means that the tail area from 2.19 to the right is (100%-97.2%)/2
or about 1.4%
We conclude that there is evidence in the observed data against the NULL
hypothesis that "Vitamin C" has no effect on Colds. The p-value is 1.4%.

## Solution Using the Chi-square Test

Expected table under independence (the null)

Cold No Cold Totals
Placebo 24.1 115.9 140
Vitamin C 23.9 115.1 139
Totals 48 231 279

Therefore: Chi-square = 4.8 , and the p-value from the Chi-square
table with 1 degree of freedom is less than 5% so we reject the
null. We conclude that vitamin C shows a significant effect.