A simple random sample of the weights of 19 green M&Ms has a mean of 0.8635 g
Question:
A simple random sample of the weights of 19 green M&Ms has a mean of 0.8635 g (as in Data Set 18 in Appendix B). Assume that ? is known to be 0.0565 g. Use a 0.05 significance level to test the claim that the mean weight of all green M&Ms is equal to 0.8535 g, which is the mean weight required so that M&Ms have the weight printed on the package label. Do green M&Ms appear to have weights consistent with the package label?
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Data Set 18: M&M Plain Candy Weights (grams) STATDISK: Minitab: Excel: Data set name is M&M. Worksheet name is M&M.MTW. Workbook name is M&M.XLS. App name is MM, and the file names are the same as for text files. TI-83/84 Plus: Text file names: Text file names are RED, ORNG, YLLW, BROWN, BLUE, GREEN. Orange Yellow 0.735 0.883 0.895 0.769 0.859 0.784 Red 0.751 0.841 0.856 0.865 0.799 0.864 0.966 0.852 0.859 0.866 0.857 0.859 0.942 0.838 0.873 0.863 0.888 0.809 0.890 0.925 0.878 0.793 0.905 0.977 0.850 0.830 0.856 0.842 0.778 0.786 0.853 0.864 0.873 0.880 0.882 0.931 0.824 0.858 0.848 0.851 Brown 0.696 0.876 0.855 0.806 0.840 0.810 0.868 0.858 Blue 0.881 0.863 0.775 0.854 0.859 0.982 0.818 0.868 0.803 Green 0.925 0.914 0.881 0.865 0.865 1.015 0.932 0.848 0.842 0.940 0.876 0.809 0.865 0.832 0.833 0.807 0.845 0.841 0.852 0.932 0.778 0.833 0.814 0.881 0.818 0.855 0.942 0.825 0.869 0.912 0.887 0.886 0.864 0.881 0.825 0.791 0.810
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Hypothesis H o 08535 All green MMs have a mean of ...View the full answer
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