Question

Suppose you want to estimate the true proportion of persons who regularly eat oatmeal for breakfast. You get a random sample of n=100 responses. From the sample, you find that 23 people regularly eat oatmeal.

A) Construct a 99% confidence interval for the population proportion.

B) Interpret the meaning of a confidence interval

Answer 1

Solution :

Given that,

n = 100

x = 23

Point estimate = sample proportion = = x / n = 23/100=0.23

1 -   = 1-0.23 =0.77

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.23*0.77) /100 )

E = 0.1084

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.23- 0.1084< p < 0.23+0.1084

0.1216< p < 0.3384

The 99% confidence interval for the population proportion p is : 0.1216,0.3384