In: Math
In a USA Today article about an experimental vaccine for children the data found in the table below was presented. Using the technique of Chi Square use a .05 significance level to test if there is an association between treatment type and outcome of flu. Does the vaccine appear to be effective? (Is there an association?)
Developed flu?
yes no
vaccine 14 656
placebo 95 137
Claim: There is an association between treatment type and outcome of flu.
The null and alternative hypothesis is
H0: There is no an association between treatment type and outcome of flu.
H1: There is an association between treatment type and outcome of flu.
Level of signfiicance = 0.05
Test statistic is

O: Observed frequency
E: Expected frequency.
E = ( Row total*Column total) / Grand total
| yes | no | Total | |
| Vaccine | 14 | 656 | 670 |
| placebo | 95 | 137 | 232 |
| Total | 109 | 793 | 902 |
| O | E | (O-E) | (O-E)^2 | (O-E)^2/E |
| 14 | 80.96452 | -66.9645 | 4484.247 | 55.38534 |
| 656 | 589.0355 | 66.96452 | 4484.247 | 7.612865 |
| 95 | 28.03548 | 66.96452 | 4484.247 | 159.949 |
| 137 | 203.9645 | -66.9645 | 4484.247 | 21.98543 |
| Total | 244.93 |

Degrees of freedom = ( Number of rows - 1 ) * ( Number of column - 1) = ( 2 - 1) * (2 - 1) = 1 * 1 = 1
Critical value = 3.841
( From chi-square table)
Test statistic > critical value we reject null hypothesis.
Conclusion: There is an association between treatment type and outcome of flu.