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 a

1 answer
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?

Answers

Answer:a) 0.0853b) 0.0000Step-by-step explanation:Parameters given stated that; H₀ : p = 0.6H₁ : p  = 0.6, this explains the acceptance region as;p° ≤ [tex]\frac{315}{500}[/tex]=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  [tex][\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6][/tex]     = P  [tex][\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ][/tex]     = P   [tex][Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ][/tex]     = P   [Z > 1.37]     = 1 - P   [Z ≤ 1.37]     = 1 - Ф (1.37)     = 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)     ≅ 0.0853b)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 [tex][\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75][/tex]   = P [tex][\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]   = P[tex][Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]   = P [Z ≤ -6.20]   = Ф (-6.20)   ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

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