Question

7. A survey of 300 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. Suppose that based on the ​survey, 120 of the 300 students responded​ "yes."

​a) What is the value of the sample proportion p (with the ^ on top of p)?

​b) What is the standard error of the sample​ proportion?

c) ​c) Construct an approximate 95​% confidence interval for the true proportion p by taking ±2 SEs from the sample proportion.

Answer 1

7)

Solution :

Given that,

n = 300

x = 120

(a)

= x / n = 120 / 300 = 0.400

1 - = 1 - 0.400 = 0.600

(c) At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 2

(B) Standard error = (((0.400 * 0.600) / 300) = 0.057

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2 * (((0.400 * 0.600) / 300)

= 0.057

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.400 - 0.057 < p < 0.400 + 0.057

0.343 < p < 0.457

(0.343 , 0.457)