Question

16 out of 151 randomly selected faculty members who were surveyed at a community college know sign language. What would be a 90% confidence interval for the proportion of all faculty members at the community college who know sign language? (Round to three decimal places and enter the interval in the format (lower bound, upper bound).)

Answer 1

Solution :

Given that,

n = 151

x = 16

Point estimate = sample proportion = = x / n = 16/151=0.106

1 -   = 1-0.106 =0.894

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z  0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.106*0.894) / 151)

E = 0.041

A 90% confidence interval proportion p is ,

- E < p < + E

0.106 - 0.041< p < 0.106 + 0.041

0.065< p < 0.147

lower bound 0.065

upper bound 0.147