Question

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, 135 of the 300 students responded​ "yes."

​a) What is the value of the sample proportion p^?

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

​c) Construct an approximate 95 confidence interval for the true proportion p by takingplus or minus 2 SEs from the sample proportion.

Answer 1

Solution :

Given that,

n = 300

x = 135

(a)

= x / n = 135 / 300 = 0.450

1 - = 1 - 0.450 = 0.550

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 = 1.96

(b)

Standard error = (((0.450 * 0.550) / 300) = 0.02872

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

= 1.96 * (((0.450 * 0.550) / 300)

= 0.056

(c)

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.450 - 0.056 < p < 0.450 + 0.056

0.394 < p < 0.506

(0.394 , 0.506)