Respuesta :
Answer:
[tex](0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412[/tex]
[tex](0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318[/tex]
And the upper limit would be -0.0318
Step-by-step explanation:
Assuming this complete question :A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates who go on to graduate school within five years after graduation and the proportion of non-business school graduates who attend graduate school. A random sample of 400 business school graduates showed that 75 had gone to graduate school while in a random sample of 500 non-business graduates, 137 had gone on to graduate school. Based on a 95 percent confidence level, what is the upper limit of the confidence interval estimate?
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 graduate school
[tex]\hat p_A =\frac{75}{400}=0.1875[/tex] represent the estimated proportion for graduate school
[tex]n_A=400[/tex] is the sample size required for graduate school
[tex]p_B[/tex] represent the real population proportion for non graduate school
[tex]\hat p_B =\frac{137}{500}=0.274[/tex] represent the estimated proportion for non graduate school
[tex]n_B=500[/tex] is the sample size required for non graduate school
[tex]z[/tex] represent the critical value for the margin of error
Solution to the problem
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 replacing into the confidence interval formula we got:
[tex](0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412[/tex]
[tex](0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318[/tex]
And the upper limit would be -0.0318
