Respuesta :
Answer:
[tex](0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942[/tex]
[tex](0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978[/tex]
And the 90% confidence interval would be given (0.0942;0.09978).
We are confident at 90% that the difference between the two proportions is between [tex]0.0942 \leq p_A -p_B \leq 0.09978[/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 female for Biology
[tex]\hat p_A =\frac{84199}{144796}=0.582[/tex] represent the estimated proportion female for biology
[tex]n_A=144796[/tex] is the sample size for A
[tex]p_B[/tex] represent the real population proportion female for calculus AB
[tex]\hat p_B =\frac{102598}{211693}=0.485[/tex] represent the estimated proportion female for Calculus AB
[tex]n_B=211693[/tex] is the sample size required for 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 90% confidence interval the value of [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2=0.05[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.64[/tex]
And replacing into the confidence interval formula we got:
[tex](0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942[/tex]
[tex](0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978[/tex]
And the 90% confidence interval would be given (0.0942;0.09978).
We are confident at 90% that the difference between the two proportions is between [tex]0.0942 \leq p_A -p_B \leq 0.09978[/tex]
