Answer:
The 90% confidence interval is [tex]0.199 < p < 0.261 [/tex]
The sample size to develop a 95% confidence interval is [tex]n = 2032 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n =500
The sample proportion is [tex]\^ p = 0.23[/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{0.23 (1- 0.23)}{500} } [/tex]
=> [tex]E = 0.03096 [/tex]
Generally 90% confidence interval is mathematically represented as
[tex]\^ p -E < p < \^ p +E[/tex]
=> [tex]0.23 -0.03096 < p < 0.23 + 0.03096 [/tex]
=> [tex]0.199 < p < 0.261 [/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
The margin of error is given as [tex]E = 0.01[/tex]
Generally the sample size is mathematically represented as
[tex]n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p ) [/tex]
=> [tex]n = [\frac{1.96 }{0.01} ]^2 *0.23 (1 - 0.23 ) [/tex]
=> [tex]n = 2032 [/tex]
