Question

A fabric manufacturer believes that the proportion of orders for raw material arriving late isp= 0.6. If a random sample of 10 orders shows that 3 or fewer arrived late, the hypothesis thatp= 0.6 should be rejected in favor of the alternativep <0.6. Use the binomial distribution.(a) Find the probability of committing a type I error if the true proportion isp= 0.6.(b) Find the probability of committing a type II error for the alternative hypothesesp= 0.3,p= 0.4, andp= 0.5

Answer 1

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∑$_{x=0$ b( x, n, p )

= ³∑$_{x=0$ b( x, 10, 0.6 )

= ³∑$_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.6)^x$$( 1 - 0.6 )^{10-x$

∝ = 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∑$_{x=4$ b( x, n, p )

= ¹⁰∑$_{x=4$ b( x, 10, 0.3 )

= ¹⁰∑$_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.3)^x$$( 1 - 0.3 )^{10-x$

= 1 - ³∑$_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.3)^x$$( 1 - 0.3 )^{10-x$

= 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∑$_{x=4$ b( x, n, p )

= ¹⁰∑$_{x=4$ b( x, 10, 0.4 )

= ¹⁰∑$_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.4)^x$$( 1 - 0.4 )^{10-x$

= 1 - ³∑$_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.4)^x$$( 1 - 0.4 )^{10-x$

= 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∑$_{x=4$ b( x, n, p )

= ¹⁰∑$_{x=4$ b( x, 10, 0.5 )

= ¹⁰∑$_{x=4$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.5)^x$$( 1 - 0.5 )^{10-x$

= 1 - ³∑$_{x=0$ $\left[\begin{array}{ccc}10\\x\\\end{array}\right]$$(0.5)^x$$( 1 - 0.5 )^{10-x$

= 1 - 0.1719

= 0.8281

Therefore, the probability of committing a type II error for the alternative hypotheses p = 0.5 is 0.8281

Answer 2

The probability of committing a type I is 0.0548, type II error for  p = 0.3 is 0.3504, type II error for p = 0.4 is 0.6177, type II error for s p = 0.5 is 0.8281.

The sample size n = 10

What is binomial distribution?

Binomial distribution summarizes the number of trials, or observations when each trial has the same probability of attaining one particular value.

$\rm p= \left[\begin{array}{ccc}n\\x\\\end{array}\right] \times p^{x} \times q^{n-x}$

(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 )

$\rm p= \left[\begin{array}{ccc}10\\x\\\end{array}\right] \times 0.6^{x} \times 0.6^{10-x}$

∝ = 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∑

$\rm p= \left[\begin{array}{ccc}10\\x\\\end{array}\right] \times 0.3^{x} \times 0.3^{10-x}$

= 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 )

$\rm \left[\begin{array}{ccc}10\\x\\\end{array}\right] \times 0.4^{x} \times 0.4^{10-x}$ = 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∑ = $\rm \left[\begin{array}{ccc}10\\x\\\end{array}\right] \times 0.5^{x} \times 0.5^{10-x}$

= 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

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