Question

1. The Mastic tree (Pistacia lentiscus) is used in reforestation efforts in southeastern Spain. The article "Nutri ent Deprivation Improves Field Performance of Woody Seedlings in a Degraded Semi-arid Shrubland" (R. Trubata, J. Cortina, and A. Vilagrosaa, Ecological Engineering, 2011:1164-1173) presents a study that in vestigated the effect of adding slow-release fertilizer to the usual solution on the growth of trees. Following are the heights, in cm, of 10 trees grown with the usual fertilizer (the control group), and 10 trees grown with the slow-release fertilizer (treatment). These data are consistent with the mean and standard deviation reported in the article. Can you conclude that the mean height of plants grown with slow-release fertilizer is greater than that of plants with the usual fertilizer? Usual 17.3 22.0 19.5 18.7 19.5 18.5 18.6 20.3 20.3 20.3 Slow-release 25.2 23.2 25.2 26.2 25.0 25.5 25.2 24.1 24.8 23.6 (a) Which method should be used for conducting hypothesis tests and generating confidence intervals for (b) What assumptions need to be checked in order to justify the use of the statistical method? Do this and (c) Find a 98% confidence interval for the difference in the mean height of plants grown using the usual this data? Why? include any appropriate plots and output from software. fertilizer and the slow-release fertilizer. Interpret your confidence interval in the context of the study appropriately. (d) Conduct the appropriate test at the 2% significance level. Be sure to show the hypothesis statements (e) Suppose someone has made the claim that the mean height for plants grown with the slow-release fertilizer is greater than the mean height of plants grown with the usual fertilizer. Find a 98% one sided confidence interval, and conduct the one-sided hypothesis test at the 2% significance level. conduct the or Plants growt

Answer 1

a) We will do independent t test and use t critical value for generating confidence interval.

Because population standard deviations are unknown and we are assuming population variances are equal.

b) Assumptions:

1) Normality

2) Homogeneity of Variances .

From above normality plot and boxplot the data are approximately normally distributed.

Since P value=0.084> 0.05 therefore NOT significant. We can conclude that population variances are equal.

c)

FROM ABOVE TABLE 98% CONFIDENCE INTERVAL FOR $\mu1-\mu2$ where $\mu1$ is population mean for Usual fertilizer and $\mu2$ for slow release fertilizer.

98% confidence interval is (-6.58805,-4.01195)

d) Null hypothesis H0: $\mu1-\mu2=0$

ALTERNATIVE HYPOTHESIS Ha: $\mu1-\mu2\ne0$

alpha=0.02

t= -10.502

df= n1+n2-2= 18

P value=0.00

Since P value <0.02 therefore SIGNIFICANT.

Decision: Reject H0.

Conclusion: We have sufficient evidence to conclude that there is difference in means.

NOTE: As per rule I have done the first four question. Please repost question e along with the above data. Thank You.