Question

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

Answer 1

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.