Pharmaceutical companies advertise that the birth control pill has an efficacy o

Question

Pharmaceutical companies advertise that the birth control pill has an efficacy of 99.5% in preventing pregnancy. However, under typical use the real efficacy is only 95%. That is, 5% of women taking the pill for a year will experience an unplanned pregnancy that year. A gynecologist looks back at the random sample of medical records of patients who were prescribed the pill one year ago.

a) What are the mean and standard deviation of the distribution of sample proportions of women experiencing unplanned pregnancies?

b) The gynecologist takes an SRS of 200 records and finds that 14 women had become pregnant within one year while taking the pill. How surprising is this finding? Give the probability of finding 7% or more pregnant women in the sample.

c) How surprising would it be if the gynecologist had found 16 pregnant women (8%)? 20 pregnant women (10%)?

Explanation / Answer

Here p = 0.05
a)
mean = np
std. dev. = sqrt(np*(1-p))

b)
pcap = 14/200 = 0.07
SE = sqrt(p*(1-p)/n) = sqrt(0.05*0.95/200) = 0.0154

Test statistics, z = (0.07 - 0.05)/0.0154 = 1.2987

P(z > 1.2987) = 0.097

Hence 0.097 is the probability of finding 7% or more pregnant women in the sample

finding 7% women pregnent is not surprising because p-value calculated does not make us to reject the null hypothesis at 0.05 significance level.

c)
Test statistics, z = (0.08 - 0.05)/0.0154 = 1.9481

P(z > 1.9481) = 0.0257

Hence 0.0257 is the probability of finding 8% or more pregnant women in the sample.

this is surprising at 0.05 significance level because it makes us to reject null hypothesis which means there are significant evidence to conclude that pregnency rate is not 5%

Test statistics, z = (0.1 - 0.05)/0.0154 = 3.2468

P(z > 3.2468) = 0.0006

Hence 0.0006 is the probability of finding 10% or more pregnant women in the sample.

Finding 10% of women pregnent is highly surprising because probability of its occurrance is very rare i.e. 0.06%

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