According to a Pew Research Center nationwide telephone survey of American adults, 75% of adults said that college education has become too expensive for most people, and they cannot afford it. One wants to find the probability that in a random sample of 1400 adult Americans, more than 76.5% of the adults in this sample will hold this opinion. Find the z score only.

We are given that according to a Pew Research Center nationwide telephone survey of American adults, 75% of adults said that college education has become too expensive for most people, and they cannot afford it.

Also, a random sample of 1400 adult Americans is taken.

Let p = % of adults who said that college education has become too expensive according to a Pew Research Center nationwide telephone survey.

Now, the z score probability distribution for sample proportion is given by;

Z = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = % of the adults in the sample of 1400 adult Americans who hold this opinion = 76.5%

n = sample of Americans = 1400

Now, probability that in a random sample of 1400 adult Americans, more than 76.5% of the adults will hold this opinion is given by = P([tex]\hat p[/tex] > 76.5%)

The z-score is = [tex]\frac{0.765-0.75}{\sqrt{\frac{0.765(1-0.765)}{1400} } }[/tex]

= 1.324

Therefore, the z score is 1.324.