Answer:
The 99% confidence interval is [tex] 0.4023 < p < 0.5157[/tex]
Step-by-step explanation:
From the question we are told that
The sample size is n = 514
The sample proportion is [tex]\^ p = 0.459[/tex]
From the question we are told the confidence level is 99% , hence the level of significance is
[tex]\alpha = (100 - 99 ) \%[/tex]
=> [tex]\alpha = 0.01[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 2.58[/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 =2.58 * \sqrt{\frac{0.459 (1- 0.459)}{514} } [/tex]
=> [tex]E =0.0567 [/tex]
Generally 99% confidence interval is mathematically represented as
[tex] 0.459 -0.0567 < p < 0.459 + 0.0567[/tex]
=> [tex] 0.4023 < p < 0.5157[/tex]
