Arandom 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?

a) 0.0853

b) 0.0000

Step-by-step explanation:

Parameters given stated that;

H₀ : p = 0.6

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

p° ≤ =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  

    = P  

    = P  

    = 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

  = P

  = P

  = P [Z ≤ -6.20]

  = Ф (-6.20)

  ≅ 0.0000 (from Cumulative Standard Normal Distribution Table).

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

could you reply to this with the question so i can ?