Question

The head of Health Services at Goodheart University (GU) would like to investigate the proportion of smokers at GU. In order to do so, the head of Health Services chooses a random sample of 500 Goodheart students, and finds that 80 of them are smokers.

Construct a 90% confidence interval for the proportion of smokers at Goodheart University.

Answer 1

Solution :

Given that,

n = 500

x = 80

Point estimate = sample proportion = = x / n = 80/500=0.16

1 -   = 1- 0.16 =0.84

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.16*0.84) / 500)

E = 0.027

A 90% confidence interval for proportion p is ,

- E < p < + E

0.16 -0.027 < p < 0.16+0.027

0.133< p < 0.187

The 90% confidence interval for the proportion p is : (0.133 , 0.187)