{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Sample_Exam2_Key

Sample_Exam2_Key - significance 3 a H p = 0.44 H a p ≠...

This preview shows pages 1–3. Sign up to view the full content.

MGMT 305 Spring 2007 Sample Exam 2 Solution Key Part I: D D B D D D E D D C C B A B A B B Part II: 1. (a) H 0 : μ ≤ 160 H a : μ > 160 4 . 2 16 25 160 175 0 = - = - = n s x t μ (b) 0.01 < p-value < 0.025, since p-value = P(t > 2.4). (c) Reject H 0 since t = 2.4 > t .05 = 1.753. At 5% level of significance, conclude that the average amount of withdrawals per customer transaction has increased recently. (d) Do not reject H 0 since t = 2.4 < t .01 = 2.602. At 1% level of significance, conclude that the average amount of withdrawals per customer transaction did not increase recently.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(e) 95% Confidence Interval for μ: 175 ± (2.131)(25/√16) = 175 ± 13.32 = (161.68, 188.32). 2. a. H 0 : μ 1 - μ 2 = 0 H a : μ 1 - μ 2 0 b. 2 1 x x - = 76 – 71.625 = 4.375 c. 55 . 48 2 8 8 1 . 49 ) 1 8 ( 48 ) 1 8 ( 2 n n s ) 1 n ( s ) 1 n ( s 2 1 2 2 2 2 1 1 2 = - + - + - = - + - + - = 484 . 3 ) 8 1 8 1 ( 55 . 48 ) n 1 n 1 ( s s 2 1 2 x x 2 1 = + = + = - ( 29 1 2 1 2 0 4.375 1.2557 3.484 x x x x t s - - - = = = d. ) 747 . 14 , 997 . 5 ( 372 . 10 375 . 4 484 . 3 977 . 2 375 . 4 s t x x 2 1 x x 2 / 2 1 - = ± = × ± = ± - - α e. Since “0” is within the 99% confidence interval, do not reject H 0 . Conclude that the average scores of male and female members are not different at 1% level of
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: significance. 3. a. H : p = 0.44, H a : p ≠ 0.44 b. p = 205/500 = 0.41 z = (0.41-0.44)/sqrt{p *(1-p )/n} =-1.35 p-value = 2*(0.5-0.4115) = 0.177 Since 0.177 > 0.05, we do not reject the null hypothesis. c. Since 0.177>0.01, we do not reject the null hypothesis. 4. (a) Define μ d as the difference before and after the instruction, (i.e. μ d = μ Posttest-Pretest ) = [2+0+6+(-2)+3]/5 = 1.8 (b) s d = = = 3.0332 ± t α/2,n-1 = 1.8 ± t 0.025,4 = 1.8 ± 2.776 = 1.8 ± 3.7655= (-1.9655, 5.5655) (c) H : μ d ≤ 0 H a : μ d > 0 t = = = 1.327 At α=0.05, critical value is t 0.05,4 = 2.132. Conclusion: Do not reject H because t = 1.327 < 2.132. The instruction is not effective at α=0.05....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

Sample_Exam2_Key - significance 3 a H p = 0.44 H a p ≠...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online