Question

A state's Department of Education reports that 12% of the high school students in that state attend private high schools. The State University wonders if the percentage is the same in their applicant pool. Admissions officers plan to check a random sample of the over 10,000 applications on file to estimate the percentage of students applying for admission who attend private schools.
a) They select a random sample of 500 applications, and find that 46 of those students attend private schools. Check the conditions required for inference. (3 marks)
b) Create the 90% confidence interval. (6 marks)
c) Interpret the confidence interval in the context of the problem. (3 marks)
d) Should the admissions officers conclude that the percentage of private school students in their applicant pool is lower than the statewide enrollment rate of 12%? Explain. (2 marks)

Answer 1

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