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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 :

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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.

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