Question

A survey of 490 women revealed that 49​% wear flat shoes to work. a. Use this sample information to develop a 95​% confidence interval for the population proportion of women who wear flat shoes to work. b. Suppose that the people who administered this survey now wish to estimate the proportion of women who wear athletic shoes to work with a margin of error of 0.01 with 99​% confidence. Determine the sample size required.

Answer 1

Solution :

Given,

n = 490 ....... Sample size

   = 0.49

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [ 0.49 *(1 - 0.49)/490]

= 0.044

Now the confidence interval is given by

( - E)   ( + E)

( 0.49 - 0.044 )   ( 0.49 + 0.044 )

0.446   0.534  

Required 95% Confidence Interval is ( 0.446 , 0.534 )

b) Solution :

Given,

E = 0.01

c = 99% = 0.99

or p = 0.5

1 - 1- p = 1 - 0.5 = 0.5

Now,

= 1 - c = 1 - 0.99 = 0.01

/2 = 0.005

= 2.576 (using z table)

The sample size for estimating the proportion is given by

n =

= (2.576)² * 0.5 * 0.5 / (0.01²)

= 16589.44

= 16590 ..(round to the next whole number)

Answer : Required Sample size is n = 16590