Question

The CEO of a large electric company claims that 80 % of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed a random sample of 500 customers. Among the sampled customers, only 380 said they are very satisfied. Is this evidence that the CEOs. claim is false, and less than 80% of customers are very satisfied? Suppose that, in fact, 77% of customers are satisfied. Using a significance level of 0.10 ( α = 0.10 ), find the following: a. Find the probability of committing a Type I error? b. Find the probability of committing a Type II error? c. How can we simultaneously decrease the probability of committing both errors?

Answer 1

a)

Probability of type I error, α = 0.10

b)

p = 0.8, n = 500

Standard error, se = √( p*(1- p )/n) = √(0.8*0.2/500 ) = √(0.8*0.2/500) = 0.0179

z = (p̂ - p) / se

Left tailed critical value, z crit = NORM.S.INV(0.1) = -1.282

Reject if Z ≤ -1.282

p̂ ≤ p + z*se

p̂ ≤ 0.8 + (-1.282) * 0.0179

p̂ ≤ 0.7771

The researchers will reject H₀ if the sample proportion is less than 0.7771

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Standard error, se₁ = √( p₁*(1- p₁ )/n) = √(0.77*0.23/500 ) = 0.0188

If true p₁ = 0.77, then type II error, β =

β = P(p̂ > 0.7771 | p₁ = 0.77)

Z = (p̂ - p₁) / se₁

= P(Z > (0.7771-0.77)/(0.0188) )

= P(Z > 0.3759)

= 1 - P(Z < 0.3759)

= 1 - NORM.S.DIST(0.3759, 1)

= 0.3535

Type II error = 0.3535

c) By decrease the significance level we simultaneously decrease the probability of committing both errors