Question

We are given that P = population proportion of persons 20-30 years old that hold license degrees (unknown) and we are asked to test H0: p≤0.67 versus H1: p>0.67 The observed sample proportion is pˆ = 61/79 = 0.772 from a random sample of size 79. compute the following:

a) P- value of the test

b) Probability of making Type II error and the power of this test at p= 0.87

Answer 1

Answer:

a)

Given,

Null hypothesis

Ho : p <= 0.67

Alternative hypothesis

Ha : p > 0.67

sample proportion p^ = x/n

substitute values

= 61/79

= 0.772

consider,

test statistic z = (p^ - p)/sqrt(pq/n)

substitute values

= (0.772 - 0.67)/sqrt(0.67(1-0.67)/79)

z = 1.931

now corresponding p value is given as follows

i.e.,

P(z < 1.931) = 0.0267415

P value = 0.0267415 [since from z table]

= 0.0027

b)

Given,

p = 0.87

p^ = 0.67

standard error of sampling distribution = sqrt(p^(1-p^)n)

substitute values

= sqrt(0.67(1-0.67)/79)

standard error of sampling distribution = 0.0529

standard error of true proportion = sqrt(pq/n)

substitute values

= sqrt(0.87(1-0.87)/79)

standard error of true proportion = 0.0378

assume , alpha = 0.05

So ,

Z(alpha) = 1.645

let us consider,

p^ <= z*σp^ + p^

substitute values

p^ <= 1.645*0.0529 + 0.67

p^ <= 0.0870 + 0.67

p^ <= 0.7570

Type II error =

= P(p^ <= 0.7570)

where as p = 0.87

= P((p^ - p)/σp < (0.757 - 0.87)/0.0378)

On solving we get

= P(z < - 2.986)

= 0.0014133 [since from z table]

= 0.0014

Type II error = 0.0014