Question

It is well known that a placebo, a fake medication or treatment, can sometimes have a positive effect just because patients often expect the medication or treatment to be helpful. An article gave examples of a less familiar phenomenon, the tendency for patients informed of possible side effects to actually experience those side effects. The article cited a study in which a group of patients diagnosed with benign prostatic hyperplasia is randomly divided into two subgroups. One subgroup of size 60 received a compound of proven efficacy along with counseling that a potential side effect of the treatment is erectile dysfunction. The other subgroup of size 50 is given the same treatment without counseling. The percentage of the no-counseling subgroup that reported one or more sexual side effects is 18%, whereas 45% of the counseling subgroup reported at least one sexual side effect. State and test the appropriate hypotheses at significance level 0.05 to decide whether the nocebo effect is operating here. (Note: The estimated expected number of "successes" in the no-counseling sample is a bit shy of 10, but not by enough to be of great concern (some sources use a less conservative cutoff of 5 rather than 10).] in USE SALT State the relevant hypotheses. (Use p, for the true proportion of patients experiencing one or more sexual side effects when given no counseling and P2 for the true proportion of patients experiencing one or more sexual side effects when receiving counseling that a potential side effect of the treatment is erectile dysfunction.) OH: P2 - P2 = 0 HP1-P2 > 0 O Ho: P1 - P2 = 0 HP - P2 < 0 Ho: P1 - P2 = 0 H.: P.-P220 O HIP - P2 = 0 H: P1-P2*0 Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value =
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Answer 1

1)

	1st group	2nd group
x=	11	23
n =	60	50
p̂=x/n=	0.1800	0.4500
estimated prop. diff =p̂1-p̂2    =		-0.2700
pooled prop p̂ =(x1+x2)/(n1+n2)=		0.3027
std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) =		0.0880
test stat z=(p̂1-p̂2)/Se =		-3.07
P value   =	0.0021	(from excel:2*normsdist(-3.07)

2)

	YF	OF
sample mean x =	30.80	16.00
standard deviation s=	3.29	4.30
sample size n=	10	5

test statistic f =(s1/s2)^2 =0.59

(please try 1.71 if this comes wrong)