Question

In: Statistics and Probability

QUESTION 1 (15 pts) Until 2002, hormone replacement therapy (HRT), estrogen and/or progesterone, was commonly prescribed...

QUESTION 1 (15 pts) Until 2002, hormone replacement therapy (HRT), estrogen and/or progesterone, was commonly prescribed to post-menopausal women. This changed in 2002, when the results of a large clinical trial were published.8506 women were randomized to take HRT, 8102 were randomized to placebo. 166 HRT and 124 placebo women developed invasive breast cancer. Does HRT increase risk of breast cancer? (P value = 0.03) From the information above,

a) (2 pts) Define all the parameters of interest

b) (2 pts) State the null and alternative hypotheses

c) (2 pts) What is informal description of the strength of evidence against H0

d) (2 pts) Are the results statistically significant?

e) (2 pts) State the formal decision about H0, using α = 0.05 f) (2 pts) Conclusion in the context of the question

g) (3 pts )Calculate the statistics of interest

Solutions

Expert Solution

a)
parameter of interest is the difference between the two population proportions p1 - p2

b)
Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

c)
p1cap = X1/N1 = 166/8506 = 0.0195
p1cap = X2/N2 = 124/8102 = 0.0153
pcap = (X1 + X2)/(N1 + N2) = (166+124)/(8506+8102) = 0.0175

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.0195-0.0153)/sqrt(0.0175*(1-0.0175)*(1/8506 + 1/8102))
z = 2.06

P-value = 0.0197

d)
As P-value < 0.05, reject the null hypothesis.

Yes, the results are significant

e)
There is sufficient evidence to conclude that HRT increase risk of breast cancer

g)
p1cap - p2cap = (0.0195-0.0153) = 0.0042


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