Question

Data from a recent year showed that 72​% of the tens of thousands of applicants to a certain program were accepted. A company that trains applicants claimed that 216 of the 280 students it trained that year were accepted. Assume these trainees were representative of the population of applicants. Has the company demonstrated a real improvement over the​ average? Find z score

Answer 1

Solution:

Given:

p = proportion of the tens of thousands of applicants to a certain program were accepted = 0.72

n = sample size = 280

x = Number of 0 students company trained that year were accepted = 216

We have to test if the company demonstrated a real improvement over the​ average.

That is we have to test : if p > 0.72 or not.

Thus hypothesis are:

H0: p = 0.72

H1: p > 0.72

Find z score:

where

thus

P-value:

Since this is right tailed test:

P-value = P(Z > z )

P-value = P(Z > 1.92 )

P-value = 1 - P(Z < 1.92 )

Look in z table for z = 1.9 and 0.02 and find corresponding area.

P(Z < 1.92) = 0.9726

thus

P-value = 1 - P(Z < 1.92 )

P-value = 1 - 0.9726

P-value = 0.0274

Decision Rule:
Reject H0, if P-value < 0.05 level of significance, otherwise we fail to reject H0

Since P-value = 0.0274 < 0.05 level of significance, we reject null hypothesis H0.

Thus we conclude that: the company demonstrated a real improvement over the​ average.

That is: The P-value provides strong evidence that the program is successful