z = -1.9 for H 0 : m = 0.55 liter and H a : 0.55
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z = -1.9 for H0: m = 0.55 liter and Ha: μ 0.55 liter
Using P-Values to Reject or Not Reject. In Exercise, use Table A-1 to find the P-value that corresponds to the given standard score, and determine whether to reject the null hypothesis at the 0.05 significance level. Is the alternative hypothesis supported?
Table A-1
Transcribed Image Text:
TABLE A-1 Standard Normal (2) Distribution: Cumulative Area from the LEFT Z .00 .01 .02 .03 .04 .05 -3.50 and lower .0001 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 -3.0 .0013 .0013 .0013 .0012 .0012 .0011 -2.9 .0019 .0018 .0018 .0017 .0016 .0016 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 -2.6 .0047 .0045 .0044 .0043 .0041 .0040 -2.5 .0062 .0060 .0059 .0057 .0055 .0054 -2.4 .0082 .0080 .0078 .0075 .0073 .0071 -2.3 .0107 .0104 .0102 .0099 .0096 .0094 -2.2 .0139 .0136 .0132 .0129 .0125 .0122 -2.1 .0179 .0174 .0170 .0166 .0162 .0158 -2.0 .0228 .0222 .0217 .0212 .0207 .0202 -1.9 .0287 .0281 .0274 .0268 .0262 .0256 -1.8 .0359 .0351 .0344 .0336 .0329 .0322 -1.7 .0446 .0436 .0427 .0418 .0409 .0401 .06 .0003 .0004 .0006 .0008 .0011 .0015 .0021 .0029 .0039 .0052 .0069 .0091 .0119 .0154 .0197 .0250 .0314 .0392 .07 .0003 .0004 .0005 .0008 .0011 .0015 .0021 .0028 .0038 .0051 .0068 0.0089 .0116 .0150 .0192 .0244 .0307 .0384 .08 .0003 .0004 .0005 .0007 .0010 .0014 .0020 .0027 .0037 .0049 .0066 .0087 .0113 .0146 .0188 .0239 .0301 .0375 .09 .0002 .0003 .0005 .0007 .0010 .0014 .0019 .0026 .0036 .0048 .0064 .0084 .0110 .0143 .0183 .0233 .0294 .0367
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Statistical Reasoning For Everyday Life
ISBN: 978-0134494043
5th Edition
Authors: Jeff Bennett, William Briggs, Mario Triola
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