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 pˆp^  = 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

a)

Ho : p = 0.67
H1 : p > 0.67

Number of Items of Interest, x = 61
Sample Size, n = 79
  
Sample Proportion , p̂ = x/n = 0.772
  
Standard Error , SE = √( p(1-p)/n ) = 0.0529
Z Test Statistic = ( p̂-p)/SE = ( 0.7722 - 0.67 ) / 0.0529 = 1.931

  
p-Value = 0.0267

...............

true proportion, p= 0.87
  
hypothesis proportion, po= 0.670
significance level, α = 0.05
sample size, n = 79
  
std error of sampling distribution, σpo = √(po*(1-po)/n) = √ ( 0.670 * 0.330 / 79 ) = 0.0529
std error of true proportion, σp = √(p(1-p)/n) = √ ( 0.87 * 0.13 / 79 ) = 0.0378

Zα = 1.645 (right tailed test)
  
We will fail to reject the null (commit a Type II error) if we get a Z statistic < 1.645
this Z-critical value corresponds to X critical value( X critical), such that
  
(p^ - po)/σpo ≤ Zα
p^ ≤ Zα*σpo + po
p^ ≤ 1.645*0.0529+0.67 = 0.7570
  
now, type II error is ,ß = P( p^ ≤ 0.7570 given that p = 0.87
  
= P ( Z < (p^ - p)/σp )= P(Z < (0.757-0.87) / 0.0378)
= P ( Z < ( -2.986 )
= 0.00141

ß = 0.00141

power =    1 - ß =   0.9986

.................

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