Question

Listed below are the weights (in grams) of a sample of M&M's Plain candies, classified according to color. Doints Red 0.946 1. 107 0913 0904 0.926 0.926 1. 006 0.914 0.922 1.052 0903 0.895 Orange 0.902 0.943 0.916 0.910 0.903 0.901 0.919 0.901 0.930 0.883 Yellow 0.929 0.960 0.938 0.933 0932 0.999 0.907 0.906 0.930 0.952 0.939 0.940 0 882 0.906 Brown 0.896 99 0.906 0941 183B 0.892 0.005 0.824 0908 0.833 Tan Green 0.845 0935 09090903 0 873 0965 0.902 0822 0.956 0.871 0.959 1905 0.916 0905 08220 852 0.965 098 Click here for the Excel Data File 1. Identify the null hypothesis and the alternate hypothesis Null hypothesis: a. Ho Hi Hat Hati Ho b. Ho Hi= H₂-H-Ho Ho He < Prey 7 of 15 Next > Grow

Answer 1

Answer: ---- 1) The null and alternative hypotheses are: Date: ----29/11/2019 H:4 = 4y = 13 = 44 = Ms = Ho H.: Not all the treatments are same There are three treatments. So k = 3. Degrees of freedom in the numerator is =k-1 = 6-1 Degrees of freedom in the denominator is =n-k = 64-6 = 58 Significance level = 0.05. At (5, 58) df and 0.05 significance level, the critical value is 2.37. So our decision rule is to reject the null hypothesis if the computed F is greater than 2.37. 4) Using excel, we get the following ANOVA table:

ANOVA Source of Variation as of MS P-value F crit Between Groups 0.034785 0.00696 3.864463 0.004325 2.37378 Within Groups 0.10439 58 0.0018 Total 0.13917 63

ANOVA Source of Variation SS di MS F P-value F crit Between Groups 0.035 5 0.007 3.864 0.004 2.374 Within Groups 0.104 58 0.002 Total 0.139

We see that 3.86 > 2.37. So we reject the null hypothesis that the all the treatments are same