Question

Test al the true average yield for this kind 10.27 A random sample from a certain piece of machinery were 9 days. Use the 0.01 level of s I t contention th ge yield for this kind of alfalfa is less than 2.0 tons pe ple from a company's very extensive files shows that orders for machinery were filled, respectively, in 12, 10, 17, 14, 13, 18, 11. a 0.01 level of significance to test the claim that, on average, su s. Choose the alternative hypothesis in such a way th orders are me 9.5 days implies that it takes longer than th rejection of the null hypothesis i = 9.5 days implies that it takes alue and indicate whether use of the p-value leads you to the sa conclusion regarding significance at the 0.01 level. Aforist is considering the effects of a new fertilizer on the fruit size of ornamer potted orange trees. The use of standard fertilizers produces, on average, fruits diameter 4.4 cm. As an experiment, she tries the fertilizer on 20 potted orange tr Find the p-value and indicate whet

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Answer 1

Null hypothesis: Ho: u=9.5 Alternative hypothesis: Hu> 9.5 Rejection rule: If p-value = a(=0.01), then reject the null hypothesis H. EXCEL (Mega Stat) output: Hypothesis Test: Mean vs. Hypothesized Value 9.500 hypothesized value 13.000 mean Sample 3.207 std. dev. 1.134 std. error 8n 7 df 3.09 t .0088 p-value (one-tailed, upper)

P-value = 0.0088 Conclusion: Use a significance level, a = 0.01. Here, p-value is 0.0088, which is less than the value of a =0.01. That is, p-value (=0.0088) sa(=0.01). Therefore, by the rejection rule, reject the null hypothesis H. Interpretation: There is sufficient evidence at level of significance a = 0.01 Therefore, H: U> 9.5