Question

1) In a sample of 160 adults, 147 had children. Construct a 99% confidence interval for the true population proportion of adults with children.

Give your answers as decimals, to three places

< p <

2)

You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p∗=35%p∗=35%. You would like to be 98% confident that your esimate is within 2.5% of the true population proportion. How large of a sample size is required?

n =

Answer 1

a)

Solution :

Given that,

n = 160

x = 147

Point estimate = sample proportion = = x / n = 147/160=0.919

1 -   = 1- 0.919 =0.081

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.919*0.081) / 160)

= 0.056

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.919-0.056 < p < 0.919+0.056

0.863< p < 0.975

The 99% confidence interval for the population proportion p is : 0.863,0.975

b)

Solution :

Given that,

= 35%=0.35

1 - = 1 - 0.35 = 0.65

margin of error = E =2.5 % = 0.025

At 98% confidence level the z is,

= 1 - 98%

= 1 - 0.98 = 0.02

/2 = 0.01

Z_/2 = 2.326 ( Using z table )

Sample size = n = (Z_/2 / E)² * * (1 - )

= (2.326 / 0.025)² * 0.35 * 0.65

=1969.340464

Sample size =1970