Question

A new drug is being tested in a medical trial for 60 randomly selected patients who suffer from a certain illness. The drug’s manufacturer claims that this drug will help relieve the symptoms in 40% of cases, for people suffering from this illness. If the manufacturer’s claim is correct, then the sampling distribution of the proportion of the 60 patients whose symptoms will be relieved is (to a good approximation):

		t 59
		N(0.4, 0.24)
		N(0.5,0.24)
		N(40, 2.4)
		N(0.4, 0.004)

Answer 1

Solution :

The population proportion of cases in which drug will help relieve the symptoms is p = 0.40

Sample size (n) = 60

If np > 5 and nq > 5 then, sampling distribution of sample proportion follows approximately normal distribution with mean p and variance pq/n.

(Where, p is population proportion, q = 1 - p and n is sample size.)

We have, n = 60, p = 0.40 and q = 1 - 0.40 = 0.60

np = (60 × 0.40) = 24 which is greater than 5.

nq = (60 × 0.60) = 36 which is greater than 5.

Hence, sampling distribution of the proportion of the 60 patients whose symptoms will be relieved is approximately normal distribution with mean p = 0.40 and variance pq/n = (0.40 × 0.60)/60 = 0.004.

i.e. p̂ ~ N(0.40, 0.004)

Sampling distribution of the proportion of the 60 patients whose symptoms will be relieved is N(0.4, 0.004).

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