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. What did the forester discovered?

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ([tex]\alpha=0.01,0.05, 0.1, 0.15[/tex]).

Step-by-step explanation:

Data given and notation

[tex]\bar X=1.6[/tex] represent the sample mean

[tex]s=0.46[/tex] represent the sample deviation

[tex]n=32[/tex] sample size

[tex]\mu_o =1.35[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:

Null hypothesis:[tex]\mu =1.35[/tex]

Alternative hypothesis:[tex]\mu \neq 1.35[/tex]

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07[/tex]

P-value

The degrees of freedom are given by:

[tex] df =n-1= 32-1=31[/tex]

Since is a two-sided test the p value would be:

[tex]p_v =2*P(t_{31}>3.07)=0.0044[/tex]

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example ([tex]\alpha=0.01,0.05, 0.1, 0.15[/tex]).