Respuesta :
Answer:
The 99% confidence interval would be given (0.004;0.436).
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_A[/tex] represent the real population proportion A
[tex]\hat p_A =0.64[/tex] represent the estimated proportion A
[tex]p_B[/tex] represent the real population proportion B
[tex]\hat p_B =0.42[/tex] represent the estimated proportion B
[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 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.576[/tex]
The standard error is given by:
[tex]\sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}=0.084[/tex]
And replacing into the confidence interval formula we got:
[tex](0.64-0.42) - 2.576(0.084)=0.004[/tex]
[tex](0.64-0.42) + 2.576(0.084)=0.436[/tex]
And the 99% confidence interval would be given (0.004;0.436).
