stat homework - 15.96-16=-.04...

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1. One-Sample Z Test of mu = 42 vs > 42 The assumed standard deviation = 2.64 90% Lower N Mean SE Mean Bound Z P 65 42.954 0.327 42.534 2.91 0.002 Alpha P-value .10 .002…………………………………. .reject H0 because p is less than alpha .05 .002…………………………………… reject H0 because p is less than alpha .01 .002…………………………………. .reject because p is less than alpha .001 .002……………………………………Fail to reject because p is greater than alpha Very strong evidence it exceeds 42 2. 16.05-16=.05
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.1/Srt36=.1/6=.0166 .05/.0166=3.012 Z=3 Failed to reject the null H because Z (3) is greater than 2.576
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Unformatted text preview: 15.96-16=-.04 .1/Srt36=.1/6=.0166-.04/.016=-2.5 Reject the null -2.5 is less than 2.576; greater than 16 oz based on the sample size 16.02-16=.02 .1/Srt36=.1/6=.0166 .02/.016=1.25 Reject the because 1.25 is less than 2.576; greater than 16 oz based on sample size 15.94-16=-.06 .1/Srt36=.1/6=.0166-.06/.016=-3.75 Reject because -3.75 is less than 2.576; greater than 16 oz based on sample size 3. A. Ho <u =42 ; Ha u >42 B. 42.95-42=.95 2.6424/8.06= 8.06 .95/8.06=2.899 Reject null H because p of .0025 is less than alpha .01. very strong evidence that they are very satisfied Problem 5...
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This note was uploaded on 04/07/2012 for the course ACC 515 taught by Professor Willaims during the Spring '12 term at Keller Graduate School of Management.

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stat homework - 15.96-16=-.04...

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