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 [tex]0.748 < p < 0.842[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 200
The number of people that watched the educational television is [tex]k = 159[/tex]
Generally the sample proportion is mathematically represented as
[tex]\^ p = \frac{159}{200}[/tex]
=> [tex]\^ p = 0.795[/tex]
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.645[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} } [/tex]
=> [tex]E = 1.645 * \sqrt{\frac{0.795 (1- 0.795)}{200} } [/tex]
=> [tex]E = 0.04696 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\^ p -E < p < \^ p +E[/tex]
=> [tex]0.795 -0.04696 < p < 0.795 + 0.04696[/tex]
=> [tex]0.748 < p < 0.842[/tex]
