Question

# Suppose we wish to make an estimate of the proportion of the population that has a...

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.

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 :

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