Question

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17. (10pts) The table shows the cost per ounce for a sample of rice with short, medium, or long grain type. At a 0.05 can you conclude that at least one mean price per omnce is different than the others? Short 0.35 0.49 Medium 0.60 0.44 Long 0.75 0.65 0.60 0.63 0.68 0.48 0.69 0.70 0.55 0.72 0.69 2 points Extra Credit: Fill out the incomplete ANOVA table. Sum of squares degrees of freedom 0.06-19 mean squares Variation Class Between Error(within Total F F N/A NA Pr>F) p.val N/A N/A .1899 14 N/A (a) Identify the null and alternative hypothesis. (b) Find the test statistic, decide whether to reject or fail to reject the null hypothesis. (e) Interpret the decision in context.

Answer 1

Now first we calculate Sum of squares (Between) SSB {ni (22-2² SSB - 6 (0-538333 - 0.601333) + 5(0-596-0-601333)? 4 (0.7025-0-601333)

SSR 0.0649 Now, we compute Sum of squares (Total) TSS = { } (x - 2)2 TSS = (0:35 -0.601333) + ... +10.42-0-601333) +(0-60-0.601333)? + t(0.55-0.601333)2 +(0.75-0-601333) + + (0.69-0-601333) TSS = o-189973 Now, Sum of squares Costhaus) (within) SSE= TSS - SSB = 0.189973 -0.0649 0.125 Pr (>F) Now, the ANOVA TABLE is :- mean Sum of Variation degree of squares squares freedom K-1=3-1 SSB = = 0.0649 d-f = 2 0-0649 2 =0.03245 IMSB = SSB F=MSB IP value class (Between) 0-0314 MSE = 0.03245 0.01042 = 3-1142 Error SSE=0-125 (within) IN-k=15-3/MSESSE die =12 = 0.125 12 = 0.01042 Total TSS=0-189973/N-1 = 15-1

since, 0.0814 the pvalue is [Rcode for pualue: pvalue ! - pf ( 3.1142,2,12) null Since, the value is greater reater than o-os so do not reject the hypother's heris (Ho) at 2:0.05 level of significance Hence, there is not sufficient evidence to conclude that at least price per sunce is different than the 9 one mean others.