Answer:
A) α = 0.04136
B) β = 0.00256
Step-by-step explanation:
We are given;
Sample size; n = 400
Proportion; p = 60% = 0.6
Formula for mean is;
μ = np
μ = 400 × 0.6 = 240
Standard deviation is given by;
σ = √npq
Where q = 1 - p = 1 - 0.6 = 0.4
σ = √(400 × 0.6 × 0.4)
σ = √96
σ = 9.8
A) our null hypothesis is at p = 0.6
Probability of making a type I error means we reject the null hypothesis when it is true.
This can be expressed in reference to the question as;
α = P(x < 220) + P(x > 260) all at p = 0.6
Now,
P(x < 220) = z = (x¯ - μ)/σ = (220 - 240)/9.8 = -2.04
Also;
P(x > 260) = z = (260 - 240)/9.8 = 2.04
Now, from z-distribution table probability of a z-score of -2.04 is 0.02068.
Also, probability of z-score of 2.04 is (1 - P(z < 2.04) = 1 - 0.97932 = 0.02068
Thus;
α = 0.02068 + 0.02068
α = 0.04136
B) Type II error occurs when we fail to reject the null hypothesis even though it's false.
In this case our alternative hypothesis is at p = 48% = 0.48
Thus;
μ = np
μ = 400 × 0.48 = 192
Standard deviation is given by;
σ = √npq
Where q = 1 - p = 1 - 0.48 = 0.52
σ = √(400 × 0.48 × 0.52)
σ = √99.84
σ = 9.992
Type II error would be given by;
β = [((x1¯ - μ)/σ) < z > ((x2¯ - μ)/σ)]
β = [((220 - 192)/9.992) < z > ((260 - 192)/9.992)]
β = (2.8 < z > 6.81)
Rearranging this gives us;
β = P(z < 6.81) - P(z < 2.8)
From z-distribution tables, we have;
β = 1 - 0.99744
β = 0.00256
