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.
