Question

Please answer and show all work, thanks!

In a random sample of 68 people, 51 are classified as "successful." a. Determine the sample proportion, p, of "successful" people b. If the population proportion is 0.70, determine the standard error of the proportion. a. p(Round to two decimal places as needed.) b. ơp-D (Round to four decimal places as needed.) Enter your answer in each of the answer boxes.

Answer 1

a) the sample proportion p of "successful" people

$=\frac{51}{68}$

=0.75

b) Population proportion is p=0.70

Therefore, standard error of proportion

$=\sqrt{\frac{p*(1-p)}{n}}$

$=\sqrt{\frac{0.70*(1-0.70)}{68}}$

=0.0556