Could you have a bit more clarification so there shall be a better understanding toreares your questions
Answer:
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
Answer:
who is Theodore and what even is the question?
Answer:
<u>ACUTE</u><u> </u><u>TRIANGLE</u><u> </u>
Step-by-step explanation:
If the sum of the squares of the two shorter sides of a triangle is greater than the square of the longest side, the triangle is acute. However, if the sum of the squares of the two shorter sides of a triangle is smaller than the square of the longest side, the triangle is obtuse.
Therefore, the triangle with sides 11, 9, and 7 is an acute triangle since 9² + 7² is greater than 11²
