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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 313 voters respond positively, we will conclude that at least 60% of the voters favor the use of these fuels.

Required:
a. Find the probability of type I error if exactly 60% of the voters favor the use of these fuels.
b. What is the type II error probability β if 75% of the voters favor this action?

Respuesta :

okpalawalter8

Answer:

[tex]\alpha=0.11767[/tex]

[tex]\beta=0[/tex]

Step-by-step explanation:

From the question we are told that:

Population size size [tex]N=500[/tex]

Sample mean [tex]n=313[/tex]

Null hypothesis [tex]H_0:P=0.6[/tex]

Alternative hypothesis[tex]H_a:P>0.6[/tex]

Generally the equation for for P value is mathematically given by

[tex]P=\frac{n}{N}[/tex]

[tex]P=\frac{313}{500}[/tex]

[tex]P=0.6[/tex]

Generally the mean [tex]\=x[/tex] is mathematically given by

[tex]\=x=np\\\=x=500*0.6[/tex][tex]\=x=300[/tex]

Generally the standard deviation [tex]\sigma[/tex] is mathematically given by

[tex]\sigma=\sqrt{npq}[/tex]

[tex]\sigma=\sqrt{500*0.6*(1-0.6)}[/tex]

[tex]\sigma=10.95[/tex]

a)

Generally the probability of Type_I error [tex]P(X>313)[/tex] is mathematically given as

[tex]P(X>313)=P(\frac{x-\mu}{\sigma}>\frac{313-300}{10.95} )[/tex]

[tex]P(X>313)=P(z>1.187 )[/tex]

Therefore using z table

[tex]\alpha=0.11767[/tex]

b)

Null hypothesis [tex]H_0:P=0.75[/tex]

Alternative hypothesis[tex]H_a:P>0.75[/tex]

Generally the probability of Type_II error [tex]P(X<313)[/tex] is mathematically given as

[tex]P=0.75,n=500\\q=1-0.75\\q=0.25[/tex]

Generally the mean [tex]\=x'[/tex] is mathematically given by

[tex]\=x'=500*0.75[/tex]

[tex]\=x'=500*0.75[/tex]

[tex]\=x'=375[/tex]

Generally the standard deviation [tex]\sigma[/tex] is mathematically given by

[tex]\sigma=\sqrt{npq}[/tex]

[tex]\sigma=\sqrt{500*0.75*(1-0.75)}[/tex]

[tex]\sigma=9.68[/tex]

Therefore

[tex]P(X<313)=P(\frac{x-\mu}{\sigma}<\frac{313-375}{9.68} )[/tex]

[tex]P(X<313)=P(z<-6.405)[/tex]

Therefore from standard normal table

[tex]P(X<313)=0[/tex]

[tex]\beta=0[/tex]

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