Question

Important instructions: For all situations requiring a hypothesis test (z test, one-sample t test, one-sample variance test or two-sample t test), you must 1. Choose the appropriate test based on the information you are given 2. State null and alternative hypotheses 3. Choose a one or two tailed test and explain why you chose that test 4. Calculate the appropriate test statistic, showing all work neatly. This includes calculations of means, standard deviations, etc. 5. Draw the appropriate conclusions (i.e., do you reject or fail to reject Ho?) To help you know for sure which questions require this, I have marked the questions requiring these steps with an asterisk (*). You may assume data are normal and have equal variances - that necessary assumptions have been fulfilled. Use your own paper and label each question clearly. You may not receive full credit if the work is disorganized or difficult to follow. 1. *(10 points) You may assume these data are normal and have equal variances. You are comparing acorn production in two different locations. Does the Mesic Soil area lead to a greater number of acorns per five square meter sample than the Sand Bottoms area? Mesic Soil Sand Bottoms 95 93 83

Answer 1

Conclusion: We can conclude that Mesic Soil area leads to a greater number of acorns per five square meter than the Sand Bottoms area at the 0.05 level of significance.

Sample size Sample mean Sample Standard deviation Mesic soil = 1 n = = (x; } 95.00 Band Bottoms n2 =10 22 - 1 Enri = 85.00 = 3.69 Pooled standard deviation , Ap= 87 (,-) + 5 (12-1) = 3.3698 -3.02 n, + n2-2 1 These two data are independent and follow nonmal distribution with equal varianus. Hence, the suitable test statistic is too-sample t-test, 2) Hall hypothesis, Ho: M,=M2 Alternative hypothens, Hi M, >Mo .f one tail test) 3) One tail test because we want to test that the Mesic soil area head to a greater number of acorns per five square miler sample than the Sand Bottoms area. The appropriate test statistic is fa a 42 dni tn 2 6.24 = 95 - 85.60 E 3.36981 tot to p-value = P(&> 6.24) = 0.0000 Decision. The p-vahu is less than Henu, reject the null hypothesis at 20.2 = 18 d. 0.05 level of significance