Question

A random survey of enrollment at 35 community colleges across the United States yielded the following figures:

6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000;

9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080;

11,622. Assume the underlying population is normal.

a. i.$\stackrel{-}{x}$ = __________

ii. sx = __________

iii. n = __________

iv. n – 1 = __________

b. Define the random variables X and $\stackrel{-}{X}$in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United

States.

i. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

e. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why?

Short answer

(a) The result of part (a)

i. $\overline{x}=8628.74$

ii.$s_x=6944$

iii.$n=35$

iv.$n-1=34$

(b) The mean enrollment for the sample of $35$community colleges is $X$, and the enrollment for the community college is $\overline{X}$.

(c) With the parameters localid="1650242739329" $tn-1,$the distribution is.

localid="1650242761305" $t35-1=t34$

(d) The final result is

i. localid="1650242764893" $CI=(6243.4,11014)$.

ii. Shown through Diagram

iii. E B M=2385.3 .

(e) Both will shrink in size as a result of this.

Step-by-step solution

Explanation (a) 

i. We need to know the mean and standard deviation in order to calculate them. To enter the stat List editor using the SETUP Editor command, press STAT followed by 1.

$\overline{x}=8628.74$

ii. The standard deviation of community college enrolment,$s_x=6944$.

iii. The number of colleges that were polled,

$n=35$

iv. If there are $35$community colleges in all, the value equals $n-1=34$.

Explanation (b)

The mean enrollment for the sample of 35 community colleges is X, and the enrollment for the community college is X.

Explanation (c)

With the parameters$tn-1,$ the distribution is.

$t35-1=t34$

Explanation (d)

i. The confidence interval should be stated.

The confidence interval's output,

$CI=(6243.4,11014)$

ii. The graph is as follows:

iii. The formula is used to compute the error bound.

$EBM=t_{n-1}\left(\frac{\alpha }{2}\right)\frac{s}{\sqrt{n}}$

$EBM=t_{35-1}\left(\frac{0.05}{2}\right)\frac{6944}{\sqrt{35}}$

$EBM=2385.3$

Explanation (e)

As we all know, when the sample size grows by a significant amount, the variability lowers as well. We have a smaller Confidence interval with fewer error boundaries to be taken the genuine population parameter as the variability diminishes. Both will shrink in size as a result of this.