Question

       Suppose we wish to make an estimate of the proportion of the population that has a learning disability.  We take a random sample of 100 people.  In this sample, 15% of the respondents have a learning disability.

       1.  Construct a 95% confidence interval for the proportion of people in the population who have a learning disability.

       2.  Interpret, in words, the meaning of this interval.       

Answer 1

Solution :

Given that,

n = 100

Point estimate = sample proportion = = 0.15

1 -   = 1- 0.15 =0.85

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.15*0.85) /100 )

= 0.06999

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.15-0.06999 < p < 0.15+0.06999

0.0800< p < 0.2200

The 95% confidence interval for the population proportion p is : 0.0800, 0.2200

Solution :