Math 2710 A Second Course in Statistics

Frequentist inference in Example "A clinical trial "

(Example 3.2.10, p. 74 )

In the clinical trial example, the unknkown quantity of interest is the probability of success p of patients receiving the specific treatment. Data consist in a sample of 40 patient, 22 of which had a success.

Bayesian inference: In the Bayesian approach, a set of possible values of p is postulated, namely j 1 p = − , j =1,...,11, 10 and p is assumed to be random with prior probabilities 1/11 for each value. Then, posterior probabilities are calculated on the bais of the observed data.

Frequentist inference: in this more "traditional" type of statistical inference, the quantity of interest p is not assumed to be random. Hypothesis tests and confidence intervals are constructed under the assumption that p is not random, but just unknown.

Let n =40(sample size) and 22 pˆ = =0.55 40 the sample proportion of successes, this estimates the unknown "population proportion" p of successes (in fact the success probability). a) Confidence interval. We obtain a confidence interval of level C

[ˆp m, pˆ + m] − with margin of error pˆ(1 pˆ) m = zα∗ − r n where zα∗ is the upper α-quantile of the standard normal random variable Z

Pr (Z>zα∗ )=α with α =(1 C)/2=0.025 for confidence level C =0.95 (95%). From the normal table − zα∗ =1.96,and 0.55 (1 0.55) m =1.96 · − =0.154 · r 40 so the confidence interval is

[0.55 0.154 , 0.55 + 0.154 ] = [0.396, 0.704 ] . − This interval includes the values 0.4,...,0.7, the values where also the posterior probabilities in the Bayesian model are concentrated (cf Figure 2.3, p. 75 in the text).

1 Here we used the normal approximation to the standardized sample proportion pˆ p − Z (0.1) p (1 p) /n ≈ − for large n. Use of a large samplep sample approximation is not specificforfrequentiest inference; is can also be used in a Bayesian framework. b) Hypothesis test. Suppose we set up a null hypothesis H0 : p = p0 andanalternative Ha : p>p0. Here we are "trying to reject the null hypothesis", that is trying to collect statistically significant evidence for the claim that the treatment is better than a certain threshold. For instance, p0 mightbetheknowneffectiveness of another treatment, or it might be the probability of success (no relapse) with no treatment. Let us consider p0 =0.3. We will be using a one sided Z-test. The test statistic is, with p0 =0.3

pˆ p0 Zˆ = − p0 (1 p0) /n − and we reject at the α =0.05 (5%) signifipcance level if ˆ Z>zα∗ . where Pr (Z>zα∗ )=α =0.05, thus zα∗ =1.645.Wefind

pˆ p0 0.55 0.3 Zˆ = − = − =3.450 . p0 (1 p0) /n (0.3) (1 0.3) /40 − − ˆ p p :ThusZ>zα∗ =1.645 and there enough evidence at the 5% level to reject the null hypothesis. In fact the evidence is strong, since the P-value is very small:

P =Pr Z>Zˆ =Pr(Z>3. 45) = 0.0003. ³ ´ Again we used the normal approximation (0.1) at the value p = p0 =0.3. This should be justified by checking the success-failure condition:

np0 =12> 10,

n(1 p0)=28> 10. −

2