Question

In a recent survey of 850 students, it was found that 476 of them have laptops. a. Find the sample proportion p ̂. b. Find a 95% confidence interval for the actual proportion of students who have laptops. 

Answer 1

We have sample of 850 students.

Out of which 476 have laptops.

a.)Sample proportion $p = \dfrac{476}{850}$

b.) Confidence interval can be given as $p \pm z\sqrt{\dfrac{p(1-p)}{n}}$

Value of z at 95% confidence level is 1.96

Hence, Confidence interval $= \dfrac{476}{850} \pm 1.96 \sqrt{\dfrac{0.56\times 0.44}{850}}$

$= 0.56 \pm 0.03$