An urn contains ten chips. An unknown number of chips are white' the others are red. We wish to test:

H0: exactly half the chips are white

vs.

H1: more than half the chips are white

We will draw, without replacement, three chips and reject H0 if two or more are white.

Find α. Also, find β when the urn is (a) 60% white and (b) 70% white.

Let's denote X to be the number of white chips in the sample and E be the event that exactly half of the chips are white. Then,

a) Find α

α = P (reject H0 | H0 is true) = P (X ≥ 2|E)

= P (X = 2|E) + P (X = 3|E),

We took two case, as we can draw only only three chips with two or more white to reject H0, it means we can only take 2 white chips or 3, not more, we get solution

= (5C2 * 5C1) / 10C3 + (5C3 * 5C0) / 10C3

= 0.5

So, α = 0.5

b) Find β

i) Let E1 be the event that the urn contains 6 white and 4 red chips. (As given)

β = P (accept H0 | E1) = P (X ≤ 1|E1)

= (6C0 * 4C3) / 10C3 + (6C1 * 4C2) / 10C3

= 1/3

= 0.333

So, β = 0.333

i) Let E2 be the event that the urn contains 7 white and 3 red chips. (As given)

β = P (accept H0 | E2) = P (X ≤ 1|E2)

= (7C0 * 3C3) / 10C3 + (7C1 * 3C2) / 10C3

= 11/60

= 0.183

So, β = 0.183