Question

References Mailings Review View Help Search [10] The MINITAB data set in Angel Tests Folder gives the following data values for the actual weights of 120 1-kg sugar packets. Is there any evidence that it is not reasonable to model these weights with a normal distribution with a mean of 1.03 kg and a standard deviation ofơ: 0.01 kg? 4. 0 Weights 1.001 1.004 1.007 1.007 1.007 1.010 1.011 1.012 1.013 1.013 1.013 1.015 1.016 1.016 1.017 1.017 1.018 1.018 1.018 1.019 1.021 1.021 1.022 1.022 1.023 1.024 1.024 1.024 1.025 1.025 1.025 1.025 1.026 1.026 1.026 1.026 1.026 1.026 1.027 1.027 1.027 1.027 1.027 1.028 1.028 1.028 1.028 1.028 1.028 1.028 1.028 1.029 1.030 1.030 1.030 1.030 1.030 1.031 1.031 1.031 1.032 1.032 1.032 1.032 1.032 1.033 1.033 1.033 1.034 1.034 1.034 1.034 1.035 1.035 1.035 1.035 1.035 1.036 1.036 1.036 1.036 1.036 1.036 1.037 1.037 1.037 1.038 1.038 1.038 1.038 1.039 1.040 1.041 1.041 1.041 1.041 1.042 1.042 1.042 1.042 1.043 1.043 1.044 1.045 1.045 1.045 1.046 .046 1.046 1.047 1.047 1.048 1.048 1.050 1.050 1.050 1.050 1.051 1.054 1.057 α = 0.05 Complete the following table. Region Under x-weights Observed Expected (O-E)/E weights from Normal Normal Curve 0.15 x <1.00 2.15% 13.6% 34.1% 34.1% 13.6% 2.158 0.15 Chi-squared- p-value: Conclusion: ㈡i e圏曲욱@.-

Answer 1

here we will use use minitab to varify

to test

H0: distribution is same as expected proportion

vs

H1: they are not same

LCB UCB Freq Expercted proportion (O-E)A2/E 0.00 1.00 0.0015 0.10200 6 1.01 1.00 0.0215 8.56186 1.01 1.02 0.1360 0.16842 1.03 1.02 0.3410 0.20646 1.03 1.04 31 0.3410 0.75640 2 1.04 21 1.05 0.1360 0.81893 1.05 1.06 0.0215 1.61795 0 10.00 1.06 0.0015 0.10200

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: Freq Test Contribution to Chi-Sq Category Observed Proportion Expected 0.0015 0.1605 0.16050 0.0215 2.3005 3.16770 0.02094 0.33325 14 0.1360 14.5520 0.3410 36.4870 31 0.3410 36.4870 0.82515 21 2.85711 0.1360 14.5520 0.0215 2.3005 0.21269 0.16050 0.0015 0.1605 N DF Chi-Sq P-Value 7 8 107 7 7.73784 0.356

chi-sq=7.73784

p-value= 0.356

decision : since p-value > 0.05 we can say that the distribution is identical (accept H0)