Question

A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air pollution. If more than 314 voters respond positively, we will conclude that at least 60% of the voters favor the use of these fuels. Round your answers to four decimal places (e.g. 98.7654).

a) Find the probability of type I error if exactly 60% of the voters favor the use of these fuelsb) What is the Type II error probability (Beta) β if 75% of the voters favor this action?

Answer 1

a) 0.0853

b) 0.0000

Explanation:

Parameters given stated that;

H₀ : p = 0.6

H₁ : p = 0.6, this explains the acceptance region as;

p° ≤
`(315)/(500)`=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)

a).

the probability of type I error if exactly 60% is calculated as :

∝ = P (Reject H₀ | H₀ is true)

= P (p°>0.63 | p=0.6)

where p° is represented as pI in the subsequent calculated steps below

= P
`[\frac{p°-p}{\sqrt{(p(1-p))/(n)}} >\frac{0.63-p}{\sqrt{(p(1-p))/(n)}} |p=0.6]`

= P
`[\frac{p°-0.6}{\sqrt{(0.6(1-0.6))/(500)}} >\frac{0.63-0.6}{\sqrt{(0.6(1-0.6))/(500)}} ]`

= P
`[Z>\frac{0.63-0.6}{\sqrt{(0.6(1-0.6))/(500) } } ]`

= P [Z > 1.37]

= 1 - P [Z ≤ 1.37]

= 1 - Ф (1.37)

= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)

≅ 0.0853

b)

The probability of Type II error β is stated as:

β = P (Accept H₀ | H₁ is true)

= P [p° ≤ 0.63 | p = 0.75]

where p° is represented as pI in the subsequent calculated steps below

= P
`[\frac{p°-p} \sqrt{(p(1-p))/(n) } }\leq \frac{0.63-p}{\sqrt{(p(1-p))/(n) } } | p=0.75]`

= P
`[\frac{p°-0.6} \sqrt{(0.75(1-0.75))/(500) } }\leq \frac{0.63-0.75}{\sqrt{(0.75(1-0.75))/(500) } } ]`

= P
`[Z\leq\frac{0.63-0.75}{\sqrt{(0.75(1-0.75))/(500) } } ]`

= P [Z ≤ -6.20]

= Ф (-6.20)

≅ 0.0000 (from Cumulative Standard Normal Distribution Table).