Question

A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean growth of 1.35 inches/year if a fertilization treatment is applied to the stand. He conducts his experiment, collects data from a sample of 32 plots, and gets a sample mean diameter growth of 1.6 in./year with the standard deviation of 0.46 in./year. Does he have enough evidence to support his claim? (Show WORK: NULL and ALTERNATIVE HYPOTHESIS; summarize data into a single T-STATISTICS and determine how likely the t-statistics would be if NULL hypothesis is TRUE (Use test statistics or p-value approach); and make decision about NULL HYPOTHESIS and CONCLUSION about the original question on the problem.)

Answer 1

Claim : The mean diameter growth will be different from the known mean growth of 1.35 inches/year.

: µ = 1.35  vs : µ  ≠ 1.35

Given : = 1.6 , S = 0.46 , n = 32

Population standard deviation σ is not known therefore we use t statistic.

Test statistic:

​ t =

t =

t = 0.25 / 0.0813

t = 3.07

Critical value:     α = 0.05

As Ha contain ≠  sign , this is two tail test,

df = n - 1 = 32 -1 = 31

For two tail test , critical value t( 0.05,31 ) = 2.034 --- ( from t distribution table )

| t | = 3.07

Decision : As | t | is greater than 2.034 , we reject the null hypothesis H0

Conclusion : There is significant evidence that the mean diameter growth will be different from the known mean growth of 1.35 inches/year.