Respuesta :
Answer:
[tex](0.274-0.1875) - 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.0318[/tex]
[tex](0.274-0.1875) + 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.141[/tex]
And the 95% confidence interval would be given (0.032;0.141).
We are confident at 95% that the difference between the two proportions is between [tex]0.032 \leq p_A -p_B \leq 0.141[/tex]
The upper bound would be 0.141.
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".
Solution to the problem
[tex]p_B[/tex] represent the real population proportion of business school graduates who go on to graduate school within five years after graduation
[tex]\hat p_B =\frac{75}{400}=0.1875[/tex] represent the estimated proportion of business school graduates who go on to graduate school within five years after graduation
[tex]n_B=400[/tex] is the sample size
[tex]p_A[/tex] represent the real population proportion of non-business school graduates who attend graduate school
[tex]\hat p_A =\frac{137}{500}=0.274[/tex] represent the estimated proportion of non-business school graduates who attend graduate school
[tex]n_A=500[/tex] is the sample size required for Brand 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]
Solution to the problem
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 replacing into the confidence interval formula we got:
[tex](0.274-0.1875) - 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.0318[/tex]
[tex](0.274-0.1875) + 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.141[/tex]
And the 95% confidence interval would be given (0.032;0.141).
We are confident at 95% that the difference between the two proportions is between [tex]0.032 \leq p_A -p_B \leq 0.141[/tex]
The upper bound would be 0.141.
