Respuesta :
Answer:
[tex] ME= 1.96 \sqrt{\frac{0.757(1-0.757)}{185} +\frac{0.581 (1-0.581)}{155}}= 0.0993[/tex]
Step-by-step 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".
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_A[/tex] represent the real population proportion for 18-21
[tex]\hat p_A =\frac{140}{185}=0.757[/tex] represent the estimated proportion for 18-21
[tex]n_A=185[/tex] is the sample size required for 18-21
[tex]p_B[/tex] represent the real population proportion for 40-43
[tex]\hat p_B =\frac{90}{155}=0.581[/tex] represent the estimated proportion for 40-43
[tex]n_B=155[/tex] is the sample size required for 40-43
[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 for the difference of two proportions would be given by this formula
[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/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]
And the Margin of error is given by:
[tex] ME= z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]
If we replace we got:
[tex] ME= 1.96 \sqrt{\frac{0.757(1-0.757)}{185} +\frac{0.581 (1-0.581)}{155}}= 0.0993[/tex]
