Question

4.         Suppose a random sample of 2100 people reveals that 1870 of them have cell phones.

            a.   Construct a 95% confidence interval for the true proportion of people who have cell phones.

            b.   Repeat for confidence level 99%.

Answer 1

Solution :

Given that,

n = 2100

x = 1870

Point estimate = sample proportion = = x / n =1870/2100=0.890

1 -   = 1-0.890 =0.110

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 (((0.890*0.110) / 2100)

E = 0.0133

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.890-0.0133 < p < 0.890+0.0133

0.8767< p < 0.9033

The 95% confidence interval for the population proportion p is : 0.8767, 0.9033

(B)

Solution :

Given that,

n = 2100

x = 1870

Point estimate = sample proportion = = x / n =1870/2100=0.890

1 -   = 1-0.890 =0.110

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 (((0.890*0.110) / 2100)

E = 0.0178

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.890-0.0178 < p < 0.890+0.0178

0.8722 p < 0.9078

The 99% confidence interval for the population proportion p is : 0.8722 ,0.9078