Respuesta :
Answer:
We can conclude that the true population proportion at 95% of confidence is between (0.221;0.339)
Step-by-step explanation:
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]p[/tex] represent the real population proportion of interest
[tex]\hat p =0.28[/tex] represent the estimated proportion for the sample
n=225 is the sample size required
[tex]z[/tex] represent the critical value for the margin of error
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
The margin of error is given by :
[tex]Me=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]Me=1.96 \sqrt{\frac{0.28(1-0.28)}{225}}=0.0587[/tex]
And replacing into the confidence interval formula we got:
[tex]0.28 - 1.96 \sqrt{\frac{0.28(1-0.28)}{225}}=0.221[/tex]
[tex]0.28 + 1.96 \sqrt{\frac{0.108(1-0.28)}{225}}=0.339[/tex]
And the 95% confidence interval would be given (0.221;0.339).
We can conclude that the true population proportion at 95% of confidence is between (0.221;0.339)
