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]
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]
=> [tex]E = 1.96 * \sqrt{\frac{ (0.28 (1 - 0.28 ))}{ 30} }[/tex]
=> [tex]E = 0.1607[/tex]
The 95% confidence interval is
[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]
