Respuesta :
Answer:
The estimated proportion on this case is [tex]\hat p =\frac{300}{1200}=0.25[/tex]
The margin of error is given by:
[tex] ME = Â 1.96\sqrt{\frac{0.25(1-0.25)}{1200}}= 0.0245 \approx 0.025[/tex]
So then the claim makes sense for this case since the estimated proportion of 0.25 and the margin of error 0.025 or 2.5%
If we replace the values obtained we got:
[tex]0.25 - 1.96\sqrt{\frac{0.25(1-0.25)}{150}}=0.226[/tex]
[tex]0.25 + 1.96\sqrt{\frac{0.25(1-0.25)}{150}}=0.275[/tex]
The 95% confidence interval would be given by (0.226;0.275)
Explanation:
Previous concepts
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". Â
The margin of error is the range of values below and above the sample statistic in a confidence interval. Â
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". Â
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]
The confidence interval for the mean is given by the following formula: Â
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
The estimated proportion on this case is [tex]\hat p =\frac{300}{1200}=0.25[/tex]
The margin of error is given by:
[tex] ME = Â 1.96\sqrt{\frac{0.25(1-0.25)}{1200}}= 0.0245 \approx 0.025[/tex]
So then the claim makes sense for this case since the estimated proportion of 0.25 and the margin of error 0.025 or 2.5%
If we replace the values obtained we got:
[tex]0.25 - 1.96\sqrt{\frac{0.25(1-0.25)}{150}}=0.226[/tex]
[tex]0.25 + 1.96\sqrt{\frac{0.25(1-0.25)}{150}}=0.275[/tex]
The 95% confidence interval would be given by (0.226;0.275)
