Question

Joan's Nursery specializes in custom-designed landscaping for residential areas. The estimated labor cost associated with a particular landscaping proposal is based on the number of plantings of trees, shrubs, and so on to be used for the project. For cost estimating purposes, managers use one and half hours of labor time for the planting of a medium-sized tree. Actual times from a sample of 10 plantings during the past month follow (times in hours). 1.4 2.0 1.6 1.6 1.3 2.5 1.6 1.3 2.9 2.2 With a 0.01 level of significance, test to see whether the mean tree-planting time differs from one and half hours. a. State the null and alternative hypotheses. HO: - Select your answer - 9 1.5 H: Select your answer - 1.5 b. Compute the sample mean. (to 1 decimal) c. Compute the sample standard deviation. (to 3 decimals) d. What is the p-value? t-value (to 3 decimals) p-value - Select your answer - e. What is your conclusion? We - Select your answer - reject the null. There - Select your answer - enough evidence to prove that the mean tree-planting time differs from 1.5 hours.

Answer 1

Standard Deviation, s: 0.54405882034942 Count, N: 10 Sum, Ex: 18.4 Mean, X: 1.84 Variance, s2: 0.296 Steps 1 N S = (I; - 7)?, N-1 i=1 s2 = {(xi-x)2 N-1 (1.4 - 1.84)2 + ... + (2.2 - 1.84)2 10-1 2.664 9 = 0.296 S = 70.296 = 0.54405882034942

The provided sample mean is X = 1.84 and the sample standard deviation is s = 0.544, and the sample size is n = 10. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: A = 1.5 Ha: u 1.5 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a two-tailed test is te = 3.25. The rejection region for this two-tailed testis R={t: 1t| > 3.25} (3) Test Statistics The t-statistic is computed as follows: X-MO t= 1.84 – 1.5 0.544/10 1.976 s/n (4) Decision about the null hypothesis Since it is observed that t = 1.976 <te = 3.25, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p= 0.0795, and since p= 0.0795 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean nis different than 1.5, at the 0.01 significance level.

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