f98Key1a

# f98Key1a - Answer Key for Exam 1 ( Version A) 1. Use the...

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Answer Key for Exam 1 ( Version A) 1. Use the pooled variance formula, [S p 2 = (n 1 -1)*S 1 ² + (n 2 -1)*S 2 ²] /[n 1 +n 2 -2]. From the given information, S p 2 = [(14-1)*(1.94)² + (17-1)*(2.25)²] / [14+17-2] = 4.48 Ans. A 2. With a sample size of 14 and 17 we use the t-test statistic. t* = [( 1 - 2 )-Do] / √(S p 2 /n 1 + S p 2 / n 2 ) . From the given information, (16.3-18.3) / √(4.48 / 14 + 4.48 / 17 ) = -2.618 = 2.618 Ans. E 3. Given t* = 1.74 and df=29, we note that in the 29 df row of the tables 1.699 <1.74<2.045. The tail area corresponding to 1.74 is between the tail areas corresponding to 1.699 and 2.045 i.e. . 025 < a < .05. To obtain the p-value for a two tailed test we multiply by 2 .→.05 < p < .1. Ans. A 4. A 90% C.I. is related to a two-tailed test at α = 10%. If the p-value(>.10) is larger than α (=.10), the decision would be to fail to reject Ho and conclude µ 1 = µ 2. This situation would be consistent with the C.I. for the difference containing zero. Ans. E 5. Since we are testing that the average price of a steak lunch in Dallas is more than 10 dollars

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## This note was uploaded on 03/09/2011 for the course DSCI 3710 taught by Professor Dr.jay during the Spring '08 term at North Texas.

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f98Key1a - Answer Key for Exam 1 ( Version A) 1. Use the...

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