In: Statistics and Probability
In a study of treatments for very painful "cluster" headaches,
159159
patients were treated with oxygen and
154154
other patients were given a placebo consisting of ordinary air. Among the
159159
patients in the oxygen treatment group,
111111
were free from headaches 15 minutes after treatment. Among the
154154
patients given the placebo,
3434
were free from headaches 15 minutes after treatment. Use a
0.050.05
significance level to test the claim that the oxygen treatment is effective. Complete parts (a) through(c) below.
p1 = 111/159 = 0.6981 Sample Proportion of
oxygen treatment group free from
headache
n1 = 159 Sample size of oxygen treatment
group
p2 = 34/154 = 0.2208 Sample Proportion of placebo
treatment group free from
headache
n2 = 154 Sample size of placebo treatment
group
the null and alternative hypothesis are
Ho : p1 =
p2
H1 : p1 >
p2
Whether oxygen treatment is effective than placebo
treatment
This is a one sided right tailed
test
Let α =
0.05
Level of
significance
Finding the Standard
error
We find the pooled
proportion
p̂ =
0.4633
Pooled proportion
Standard Error (SE) =
0.056378
where pdiff = 0 since null hypothesis
states p1 = p2
Zcalc =
8.4661
p-value
We find p-value such that
p-value = P(Z >
8.4661)
We find p-value using z-tables or the Excel function
NORM.S.DIST
p-value = 1 - NORM.S.DIST(8.4661,
TRUE)
p-value = 1 -
1
p-value =
0
Decision
0 <
0.05
p-value <
α
Hence we reject the null
hypothesis
Conclusion
Ho is
rejected
Interpretation
:
There exists enough statistical evidence to show that
proportion of patients free from headaches given oxygen treatment
is more than those given
placebo
That is the oxygen treatment is
effective.