Question

A survey of 200 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 200 students responded​ "yes." ​a) What is the value of the sample proportion ModifyingAbove p with caret​? ​b) What is the standard error of the sample​ proportion? ​c) Construct an approximate 95​% confidence interval for the true proportion p by taking plus or minus 2 SEs from the sample proportion.

Answer 1

Solution :

Given that,

(a)

Point estimate = sample proportion = = x / n = 120 /200 = 0.600

Z_/2 = 2

(b)

Standard error = (( * (1 - )) / n) = (((0.600 * 0.400) / 200) = 0.0364

Margin of error = E = Z _{/ 2} * SE

= 2 * 0.03464

= 0.069

(c)

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.600 - 0.069 < p < 0.600 + 0.069

0.531 < p < 0.669

(0.531 , 0.669)