In: Statistics and Probability
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
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 [Excel function
=NORMSDIST(-z)
...................
b)
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
...............
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