Question

Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an​ endowment, and the college spends only the interest earned by the fund. A recent survey of 8 private colleges in the United States revealed the following endowments​ (in millions of​ dollars): 60.2,​ 47.0, 235.1,​ 490.0, 122.6,​ 177.5, 95.4, and 220.0. Summary statistics yield Upper X overbarequals180.975 and Sequals143.042. Calculate a​ 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments.

Answer 1

Answer:

$180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370$  

$180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580$  

So on this case the 95% confidence interval would be given by (61.370;300.580)  

Step-by-step explanation:

Previous concepts

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".

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".

Solution to the problem

$\bar X=180.975$ represent the sample mean  

$\mu$ population mean (variable of interest)  

$s=143.042$ represent the sample standard deviation  

n=8 represent the sample size  

The confidence interval on this case is given by:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$   (1)

We can find the degrees of freedom and we got:

$df = n-1= 8-1=7$

The next step would be find the value of $\t_{\alpha/2}$, $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$  

Using the t table with df =7, excel or a calculator we see that:  

$t_{\alpha/2}=2.365$

Since we have all the values we can replace:

$180.975 - 2.365\frac{143.042}{\sqrt{8}}=61.370$  

$180.975 + 2.365\frac{143.042}{\sqrt{8}}=300.580$  

So on this case the 95% confidence interval would be given by (61.370;300.580)