Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

6,416; 1,550; 2,110; 9,351; 21,830; 4,299; 5,945; 5,722; 2,827; 2,046; 5,481; 5,202; 5,855; 2,749; 10,011;
6,356; 27,000; 9,415; 7,683; 3,202; 17,502; 9,200; 7,380; 18,315; 6,557; 13,714; 17,767; 7,491; 2,769;
2,861; 1,264; 7,284; 28,165; 5,081; 11,624.

Assume the underlying population is normal.

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

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

Sagot :

fichoh

Answer:

(6243.99, 11014.53) ; 2385.27 ; (7998.17, 9260.34) ;

Step-by-step explanation:

Given the data:

6416; 1550; 2110; 9351; 21830; 4299; 5945; 5722; 2827; 2046; 5481; 5202; 5855; 2749; 10011; 6356; 27000; 9415; 7683; 3202; 17502; 9200; 7380; 18315; 6557; 13714; 17767; 7491; 2769; 2861; 1264; 7284; 28165; 5081; 11624

Using calculator :

Sample mean, m= 8629.25714

Sample standard deviation, s = 6943.92362

1.) T test distribution ;

Sample size, n = 35

Confidence interval (C. I) : m ± Zcritical * s/sqrt(n)

n = sample size = 35

Tn-1,0.025 = t34, 0.025 = 2.0322

C.I = 8629.25714 ± 2.0322 * (6943.92362 / sqrt(35))

C.I = 8629.25714 ± 2385.2689

Lower bound = 8629.25714 - 2385.2689 = 6243.98824

Upper bound = 8629.25714 + 2385.2689 = 11014.52604

(6243.99, 11014.53)

Error bound :

E = t34, 0.025 * (s/sqrt(n))

E = 2.0322 * 6943.92362 / sqrt(35)

E = 2.0322 * 1173.7373

E = 2385.27

C.)

If n = 500

C.I = 8629.25714 ± 2.0322 * (6943.92362 / sqrt(500))

C.I = 8629.25714 ± 631.08285

Lower bound = 8629.25714 - 631.08285 = 7998.17429

Upper bound = 8629.25714 + 631.08285 = 9260.33999

(7998.17, 9260.34)

Error bound :

E = t34, 0.025 * (s/sqrt(n))

E = 2.0322 * 6943.92362 / sqrt(500)

E = 2.0322 * 310.54170

E = 631.08

Both the error margin and the confidence interval reduces due to large sample size.

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.