Question

In order to estimate the average time high school students are currently spending on social media, data were collected from a sample of 81 students over a one-week period. Based on prior studies, the population standard deviation can be assumed to be 1.2 hours.

With a .95 probability, what is the margin of error (approximately)?

Select one:

a. .26

b. 1.64

c. .21

d. 1.96

Answer 1

Solution :

Given that,

Population standard deviation =     = 1.2

Sample size n =81

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   ( Using z table )

Margin of error = E =   Z / 2     * ( /n)
= 1.96 * (1.2 / 81 )

E= 0.26