Question

From a survey of 282 McMaster students, you find that 52% take the bus to McMaster every day. Construct an approximate 99% confidence interval for the true proportion p and report the lower limit of the interval (assume z0.005=2.58).

Answer 1

Solution :

Given that,

n = 282

Point estimate = sample proportion = = 52%=0.52

1 -   = 1-0.52 =0.48

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 = 2.58 ( Using z table )

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

= 2.58* (((0.52*0.48) / 282)

E = 0.077

A 99% confidence interval is ,

- E < p < + E

0.52 - 0.077 < p < 0.52 + 0.077

0.443< p < 0.597

lower limit = 0.443

upper  limit = 0.597