Question

Suppose that a consumer advocacy group would like to conduct a survey to find the proportion p of smartphone users that were happy with their smartphone. The advocacy group took a random sample of 1,000 smartphone users and found that 400 were happy with their smartphone. A 90% confidence interval for p is closest to:

Answer 1

Solution :

Given that,

n = 1000

x = 400

Point estimate = sample proportion = = x / n =400/1000=0.4

1 - = 1-0.4=0.6

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 (((0.4*0.6) / 1000)

= 0.096

A 90% confidence interval for p is ,

- E < p < + E

0.4-0.096 < p < 0.4+0.096

0.304< p < 0.496