GE1707
Princess Claire R. Chan Bsit 111a
Answers
Column Observe Expected (E) O– (O – E) ² (O – E)
Rox (O) E
²/E
1,1 36 (49 × 290)/565 = 25.16 10.84 117.51 4.67
2,1 13 (49 × 275) = 23.84 10.84 117.51 4.93
1,2 254 (516 × 290)/565 = 10.84 117.51 0.44
264.84
2,2 262 (516 × 275)/565 = 10.84 117.51 0.47
251.16
a. Null Hypothesis (H₀):
There is no relationship between the treatment and the side effect of nausea — they are independent.
b. Alternative Hypothesis (H₁):
There is a relationship between the treatment and nausea — they are not independent.
c. Level of Significance (α):
We’re using a significance level of 0.10.
d. Degrees of Freedom (df):
Since it’s a 2x2 table, we calculate:
(2 - 1) × (2 - 1) = 1
e. Test Statistic:
The computed chi-square value is 10.51.
f. Critical Value:
from the chi-square distribution table with df = 1 and α = 0.10, the critical value is 2.71.
g. Decision:
Because the test statistic (10.51) is greater than the critical value (2.71), we reject the null hypothesis.
h. Conclusion There is enough statistical evidence to conclude that the treatment is not independent of
the side effect of nausea. In other words, the type of treatment may influence whether a person experiences
nausea.