Question

ducational Television In a random sample of people, said that they watched educational television. Find the confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.

Answer 1

Complete Question

Educational Television In a random sample of 200 people, 159 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.

Answer:

The 90% confidence interval is
`0.748  <  p <  0.842`

Explanation:

From the question we are told that

The sample size is n = 200

The number of people that watched the educational television is
`k = 159`

Generally the sample proportion is mathematically represented as

`\^ p = (159)/(200)`

=>
`\^ p = 0.795`

From the question we are told the confidence level is 90% , hence the level of significance is

`\alpha = (100 - 90) \%`

=>
`\alpha = 0.10`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.645`

Generally the margin of error is mathematically represented as

`E =  Z_{(\alpha )/(2) } * \sqrt{(\^ p (1- \^ p))/(n) }`

=>
`E =  1.645  * \sqrt{(0.795 (1- 0.795))/(200) }`

=>
`E = 0.04696`

Generally 95% confidence interval is mathematically represented as

`\^ p -E <  p <  \^ p +E`

=>
`0.795  -0.04696  <  p <  0.795 + 0.04696`

=>
`0.748  <  p <  0.842`