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? Round to three decimal places.

Answer 1

Answer:

$(0.274-0.1875) - 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.0318$  

$(0.274-0.1875) + 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.141$  

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 $0.032 \leq p_A -p_B \leq 0.141$

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

$p_B$ represent the real population proportion of business school graduates who go on to graduate school within five years after graduation

$\hat p_B =\frac{75}{400}=0.1875$ represent the estimated proportion of business school graduates who go on to graduate school within five years after graduation

$n_B=400$ is the sample size

$p_A$ represent the real population proportion of​ non-business school graduates who attend graduate school

$\hat p_A =\frac{137}{500}=0.274$ represent the estimated proportion of​ non-business school graduates who attend graduate school

$n_A=500$ is the sample size required for Brand B

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.274-0.1875) - 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.0318$  

$(0.274-0.1875) + 1.96 \sqrt{\frac{0.274(1-0.274)}{500} +\frac{0.1875(1-0.1875)}{400}}=0.141$  

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 $0.032 \leq p_A -p_B \leq 0.141$

The upper bound would be 0.141.