How does an indirect proof different from a regular proof
1 answer:
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Answer:
a) the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548
b)
- the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504
- the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177
- the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281
Step-by-step explanation:
Given the data in the question;
proportion p = 0.6
sample size n = 10
binomial distribution
let x rep number of orders for raw materials arriving late in the sample.
(a) probability of committing a type I error if the true proportion is p = 0.6;
∝ = P( type I error )
= P( reject null hypothesis when p = 0.6 )
= ³∑ b( x, n, p )
= ³∑ b( x, 10, 0.6 )
= ³∑
∝ = 0.0548
Therefore, the probability of committing a type I error if the true proportion is p = 0.6 is 0.0548
b)
the probability of committing a type II error for the alternative hypotheses p = 0.3
β = P( type II error )
= P( accept the null hypothesis when p = 0.3 )
= ¹⁰∑ b( x, n, p )
= ¹⁰∑ b( x, 10, 0.3 )
= ¹⁰∑
= 1 - ³∑
= 1 - 0.6496
= 0.3504
Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.3 is 0.3504
the probability of committing a type II error for the alternative hypotheses p = 0.4
β = P( type II error )
= P( accept the null hypothesis when p = 0.4 )
= ¹⁰∑ b( x, n, p )
= ¹⁰∑ b( x, 10, 0.4 )
= ¹⁰∑
= 1 - ³∑
= 1 - 0.3823
= 0.6177
Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.4 is 0.6177
the probability of committing a type II error for the alternative hypotheses p = 0.5
β = P( type II error )
= P( accept the null hypothesis when p = 0.5 )
= ¹⁰∑ b( x, n, p )
= ¹⁰∑ b( x, 10, 0.5 )
= ¹⁰∑
= 1 - ³∑
= 1 - 0.1719
= 0.8281
Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281