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. What did the forester discovered?

Answer 1

`t=(1.6-1.35)/((0.46)/(√(32)))=3.07`

The degrees of freedom are given by:

`df =n-1= 32-1=31`

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

`p_v =2*P(t_(31)>3.07)=0.0044`

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 (
`\alpha=0.01,0.05, 0.1, 0.15`).

Explanation:

Data given and notation

`\bar X=1.6` represent the sample mean

`s=0.46` represent the sample deviation

`n=32` sample size

`\mu_o =1.35` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` 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:
`\mu =1.35`

Alternative hypothesis:
`\mu \\eq 1.35`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Calculate the statistic

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

`t=(1.6-1.35)/((0.46)/(√(32)))=3.07`

P-value

The degrees of freedom are given by:

`df =n-1= 32-1=31`

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

`p_v =2*P(t_(31)>3.07)=0.0044`

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 (
`\alpha=0.01,0.05, 0.1, 0.15`).