Question

A survey sent to 483 university students found 597 of these students admitted to texting during class. Construct a 99% confidence interval for the proportion of all university students who text during class.

Answer 1

Solution :

Given that,

n = 597

x = 483

Point estimate = sample proportion = = x / n = 483 / 597=0.809

1 -   = 1-0.809 =0.191

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z  0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.809*0.191) / 597)

E = 0.041

A 99% confidence interval for proportion p is ,

- E < p < + E

0.809 - 0.041< p < 0.809+0.041

0.768< p < 0.850

(0.768 , 0.850)