Answer:
the 90% confidence interval is
[tex]0.5305 < p < 0.5695 [/tex]
Step-by-step explanation:
From the question we are told that
The sample proportion is [tex]\^ p = 0.55[/tex]
The sample size is n = 1754
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{ 0.55 (1- 0.55)}{1754} } [/tex]
=> [tex]E = 0.0195 [/tex]
Generally 90% confidence interval is mathematically represented as
[tex]\^ p -E < p < \^ p +E[/tex]
=> [tex]0.55 -0.0195 < p < 0.55 + 0.0195[/tex]
=> [tex]0.5305 < p < 0.5695 [/tex]
