Question

According to a Pew Research Center nationwide telephone survey of adults conducted between March 15 and April 24, 2011, 55% of college graduates said that college education prepared them for a job (Time, May 30, 2011). Suppose this result was true of all college graduates at that time. In a recent sample of 2000 college graduates, 59% said that college education prepared them for a job. Is there significant evidence at a 5% significance level to conclude that the current percentage of all college graduates who will say that college education prepared them for a job is different from 55%? Use both the p-value and the critical-value approaches. Round your answers for the observed value of z and the critical value of z to two decimal places, and the p-value to four decimal places. Enter the critical values in increasing order.

z_observed =______

p-value =_____
Critical values:_______and _______

We can conclude that the current percentage of all college graduates who will say that college education prepared them for a job is ______55%

not different from

different from

Answer 1

Solution:

Null and alternative hypothesis are

H0 : p = 0.55

H1 : p   0.55

The test statistic z is

z_observed =   

=  (0.59 - 0.55)/[0.55*(1 - 0.55)/2000]

= 3.60

z_observed = 3.60

Now ,

p value

sign in H1 indicates that the test id "Two Tailed"

p value = P(Z < -z) + P(Z > +z) = P(Z < -3.60) + P(Z > +3.60) = 0.0002+0.0002 = 0.0004

p-value = 0.0004

Now , given , = 5% = 0.05

/2 = 0.025

= 1.96

Critical values: -1.96 and 1.96

We can conclude that the current percentage of all college graduates who will say that college education prepared them for a job is different from 55%.

(Because p value is less than the significance level , so significant results )