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 two 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). 1.6 1.4 2.7 2.3 2.5 2.3 2.6 3.0 1.3 2.3 With a 0.05 level of significance, test to see whether the mean tree-planting time differs from two hours. (a) State the null and alternative hypotheses. H0: ???? = 2 Ha: ???? ≠ 2 H0: ???? < 2 Ha: ???? ≥ 2 H0: ???? ≤ 2 Ha: ???? > 2 H0: ???? ≥ 2 Ha: ???? < 2 H0: ???? > 2 Ha: ???? ≤ 2 (b) Compute the sample mean. (c) Compute the sample standard deviation. (Round your answer to three decimal places.) (d) What is the test statistic? (Round your answer to three decimal places.) What is the p-value? (Round your answer to four decimal places.) p-value = (e) What is your conclusion? Reject H0. We cannot conclude that the mean tree-planting time differs from two hours. There is no reason to change from the two hours for cost estimating purposes.Do not reject H0. We cannot conclude that the mean tree-planting time differs from two hours. There is no reason to change from the two hours for cost estimating purposes. Do not reject H0. We can conclude that the mean tree-planting time differs from two hours. There is a reason to change from the two hours for cost estimating purposes.Reject H0. We can conclude that the mean tree-planting time differs from two hours. There is a reason to change from the two hours for cost estimating purposes.

We cannot reject the null hypothesis and conclude that no significant evidence exists to confirm that the mean tree-planting time differs from two hours.

Given :

X = 23.71,17.79, 29.87,18.78,28.76

b)Sample mean, x = Σx / n = 22/10 = 2.2

c) Standard deviation, s = 0.516

H0 : μ = 2

H1 : μ > 2

Test statistic :

(x - μ) ÷ (s/√(n))

n = 10

d) Test statistic :

=(2.2 - 2) ÷ (0.516/√(10))

=0.2 / 0.1631735

Test statistic = 1.23

Using the P-value from T-score we get the degrees of freedom:

df

= n - 1

= 10 - 1

= 9

P-value (1.23, 9) = 0.1249

e) Since, P-value > α ; We fail to reject the null and conclude that no significant evidence exists to support that mean tree-planting time differs from two.

A hypothesis is thetheory put up to explain a phenomenon. If a hypothesis cannot be tested by the scientific method, it cannot be referred to as a scientific hypothesis.

Scientific theories frequently start with historical observations that can't be adequately explained by the body of knowledge at this time. A hypothesis is a tested assertion regarding the relationship between two or more variables in the scientific area or a theory put out to explain an observed phenomena.

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