Question

Educational Television In a random sample of 186 people, 138 said that they watched educational television. Find the 94%confidence interval of the true proportion of people who watched educational television. Use a graphing calculator. Round the answers to at least three decimal places.

Answer 1

Given:

x= 138, n = 186

Using TI-84

Hit ---"STAT" --->> by using right arrow ----->> "Tests"

By using down arrow hit ---->> A: 1- PropZInt..

Enter values of x = 138, n = 186 C-Level = 0.94 and hit Calculate

Confidence Interval of the true proportion of people who watched educational television is (0.682 , 8.802)

Regular Method :

Given: x=number of success = 138, n=sample size = 186, 94 % confidence interval.

So Zc = 1.88    using standard normal table

=x/n = 138 / 186 = 0.7419

= 0.7419

= 1-     = 1- 0.7419

= 0.2581

Confidence interval for proportion p is

-E < p < + E

Where E = standard error

E = Zc *

E = 1.88 *

E = 0.0603

Confidence interval for proportion p is

-E < p < + E

0.7419 - 0.0603 < p < 0.7419 + 0.0603

0.682 < p < 0.802

Confidence interval for p is ( 0.682, 0.802)

Estimate of p = 0.7419 , Standard error = 0.0603, 94 % CI 0.682 , 0.802

With 94% confidence the proportion of all of people who watched educational television is (0.682 , 0.802)