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EXERCISE 38 (PAGE 381 )- American Auto Institute --L(11-3 The president of the American Insurance Institute wants to compareBcker 3 Berry 4 Cobb 5 Debuck 6 Dibucci 7 Eckroate 8 German 9 Glasson 10 King 11 Kucic 2090 1683 1402 1830 930 697 1741 1129

EXERCISE 38 (PAGE 381 )- American Auto Institute --L(11-3 The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The data for the sample of families and their two insurance quotes are reported below (SEE DATA SET). . .At the.10 significance level, can we conclude that there is a difference in the amounts quoted?
Bcker 3 Berry 4 Cobb 5 Debuck 6 Dibucci 7 Eckroate 8 German 9 Glasson 10 King 11 Kucic 2090 1683 1402 1830 930 697 1741 1129 1018 1881 12 Meridieth 1571 874 1579 1577 860 1 Family Progressive GEICO 1610 1247 2327 1367 1461 1789 1621 1914 1956 1772 1375 1527 1767 1636 1188 13 Obeid 14 Price 15 Phillips 16 Tresize
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Answer #1
Progressive GEICO
2090 1610
1683 1247
1402 2327
1830 1367
930 1461
697 1789
1741 1621
1129 1914
1018 1956
1881 1772
1571 1375
874 1527
1579 1767
1577 1636
860 1188
n = 15 15
x-bar = 1390.80 1637.13
s = 437.76 299.06

Data:        

n1 = 15       

n2 = 15       

x1-bar = 1390.8       

x2-bar = 1637.13       

s1 = 437.76       

s2 = 299.06       

Hypotheses:        

Ho: μ1 = μ2        

Ha: μ1 ≠ μ2        

Decision Rule:        

α = 0.1       

Degrees of freedom = 15 + 15 - 2 = 28      

Lower Critical t- score = -1.701130908       

Upper Critical t- score = 1.701130908       

Reject Ho if |t| > 1.701130908       

Test Statistic:        

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   √(((15 - 1) * 437.76^2 + (15 - 1) * 299.06^2)/(15 + 15 -2)) =     374.9

SE = s * √{(1 /n1) + (1 /n2)} = 374.88044840989 * √((1/15) + (1/15)) = 136.8869853      

t = (x1-bar -x2-bar)/SE = -1.79951366       

p- value = 0.082723597       

Decision (in terms of the hypotheses):        

Since 1.79951366 > 1.701130908 we reject Ho

There is sufficient evidence of a significant difference in the amounts quoted.

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