Answer:
a) p*n = 46 and q*n = 454 data is enough for inference
b) CI = ( 0,099 ; 0,141)
c) CI we will find with 90% of confidence the value of p in that interval
d) No, we accept H₀ we don´t have evidence to claim that the percentage in the State University is smaller than that of the Department of Education
Step-by-step explanation:
Information from A state´s Department of Education is:
p₀ = 12 % p₀ = 0,12
The State University sample:
Sample size n = 500
Sample proportion 46 out of 500
p = 46/500 p = 0,092 then q = 1- p q = 1 - 0,092 q = 0,908
Both p*n = 46 and q*n = 454
a) Are big enough to approximate the binomial distribution to normal distribution
b) CI = 90% CI = 0,9 significance level α = 0,10 for an CI we need
α/2 = 0,10/2 α/2 = 0,05
z(c) = - 1,64 ( from z-table)
MOE = z(c)*√(p*q)/n
MOE = 1,64 * √ ( 0,092*0,908)/500
MOE = 1,64 *√0,08353/500
MOE = 1,64 * 0,0129
MOE = 0,021
Then CI = 90 % is:
CI = ( 0,12 - 0,021 < P < 0,12 + 0,021 )
CI = ( 0,099 ; 0,141)
We have with 90 % of confidence for finding the value of p in the interval
Test Hypothesis
Null Hypothesis H₀ p = p₀
Alternative Hypothesis Hₐ p < p₀
we have for α = 0,1 z(c) = - 2,32 from z table
And z(s) = ( p - p₀ ) / √ (p*q)/n
z(s) = ( 0,092 - 0,12 ) / √(0,092*0,908)/500
z(s) = - 0,028 / 0,0129
z(s) = - 2,17
Comparing z(s) and z(c) we see that
|z(s)| < |z(c)|
So z(s) is in the acceptance region. We can support the claim that the percentage in the State University is the same than that of the Departament of Education
