Answer:
The  95% confidence interval is [tex]0.1193 < p <0.4407[/tex]
Step-by-step explanation:
From the question we are told that
  The sample size is  [tex]m = 30[/tex]
  The sample proportion is  [tex]\r p = 0.28[/tex]
 Â
Given that the confidence interval is  95% then the level of significance is mathematically represented as
      [tex]\alpha = 100 - 95[/tex]
      [tex]\alpha = 5\%[/tex]
     [tex]\alpha = 0.05[/tex]
Next we obtain the critical value of the [tex]\frac{\alpha }{2}[/tex] from the normal distribution table, the value is
    [tex]Z_{\frac{ \alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
    [tex]E = Z_{\frac{ \alpha }{2} } * \sqrt{\frac{ ( \r p (1 - \r p ))}{n} }[/tex]
=> Â Â [tex]E = 1.96 * \sqrt{\frac{ (0.28 (1 - 0.28 ))}{ 30} }[/tex]
=> Â [tex]E = 0.1607[/tex]
The  95% confidence interval is Â
   [tex]\r p - E < p < \r p + E[/tex]
=> [tex]0.28 - 0.1607 < p < 0.28 + 0.1607[/tex]
=> [tex]0.1193 < p <0.4407[/tex]
