Question

Out of 400 people sampled, 252 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. As in the reading, in your calculations: --Use z = 1.645 for a 90% confidence interval --Use z = 2 for a 95% confidence interval --Use z = 2.576 for a 99% confidence interval. Give your answers to three decimals Give your answers as decimals, to three decimal places. [,]

Answer 1

Solution :

Given,

n = 400 ....... Sample size

x = 252 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 252/400 = 0.63  

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645 (use z table)

Now , the margin of error is given by

E =   *  

= 1.645 * [ 0.63 *(1 - 0.63)/400]

= 0.0397

Now the confidence interval is given by

( - E)   ( + E)

( 0.63 - 0.0397 )   ( 0.63 + 0.0397 )

0.590   0.670  

Required 90% Confidence Interval is ( 0.590 , 0.670 )