Question

A field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are the results of his random sampling. Can he conclude, at the 0.10 level of significance, that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree?Pine trees Spruce treesSample size 20 30Mean trunk diameter (cm) 45 39Sample variance 100 150What is the test value for this hypothesis test?Test value:Round your answer to three decimal places.What is the critical value?____________Critical value: ________________________Round your answer to three decimal places.

Answer 1

`t_(crit)= 1.299`

Explanation:

Data given and notation

`\bar X_(P)=45` represent the mean for the sample Pine

`\bar X_(S)=39` represent the mean for the sample Spruce

`s_(P)=√(100)=10` represent the sample standard deviation for the sample Pine

`s_(P)=√(150)=12.247` represent the sample standard deviation for the sample Spruce

`n_(P)=20` sample size selected for Pines

`n_(S)=30` sample size selected for Spruce

`\alpha=0.1` represent the significance level

t would represent the statistic (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree, the system of hypothesis would be:

Null hypothesis:
`\mu_(P)-\mu_(T)\leq 0`

Alternative hypothesis:
`\mu_(P)-\mu_(S)>0`

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

`t=\frac{(\bar X_(P)-\bar X_(S))-12}{\sqrt{(s^2_(P))/(n_(P))+(s^2_(S))/(n_(S))}}` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

`t=\frac{(45-39)}{\sqrt{(100)/(20)+(150)/(30)}}}=1.897`

Critical value

In order to find the critical value we need to calculate first the degrees of freedom given by:

`df = n_P +n_S -2 = 20+30-2=48`

Now we weed to look in the t distribution with 48 degrees of freedom a quantile that accumulates 0.9 of the area on the left and 0.1 of the area on the right. And this value on this case is
`t_(crit)= 1.299`

And we can use the following excel code: "=T.INV(0.9,48)"

Since our calculated value is higher than the critical value we have enough evidence on this case to reject the null hypothesis. And then makes sense the claim the the mean for Pines is greater than the mean for Spruce trees.