Question

In a random sample of 200 people, 135 said that they watched educational TV. Find the 95% confidence interval of the true proportion of people who watched educational TV.

• A.

  0.325 <  < 0.675

• B.

  –1.96 <  < 1.96

• C.

  0.6101 <  <  0.7399

• D.

  0.6704 < < 0.6796

Answer 1

Solution :

Given that,

n = 200

x = 135

Point estimate = sample proportion = = x / n = 135/200=0.675

1 - = 1-0.675=0.325

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )   

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

= 1.96 (((0.675*0.325) /200 )

= 0.0649

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.675-0.0649 < p < 0.675+0.0649

0.6101 < p<  0.7399