Growth of Douglas Fir Seedlings

An experiment was conducted to compare the growth of Douglas fir seedlings under three different levels of vegetation control: 0%, 50%, and 100%. Sixteen seedlings were randomized to each level of control. The resulting sample means for stem volume were 58, 73, and 105 cubic centimeters ([tex]$cm^3$[/tex]), respectively, with [tex]$sp = 17 \, cm^3$[/tex]. The researcher hypothesized that the average growth at 50% control would be less than the average of the 0% and 100% levels.

(a) What are the coefficients for testing this contrast?

(b) Perform the test and report the test statistic, degrees of freedom, and P-value. Do the data provide evidence to support this hypothesis?

(a) The coefficients for testing this contrast are -1, 2, and -1. (b) [tex]n_{1}[/tex]= [tex]n_{2}[/tex] = [tex]n_{3}[/tex] = 16, Degrees of freedom = 45, If the P-value is smaller than the significance level (e.g., α = 0.05), we reject the null hypothesis and conclude that there is evidence to support the hypothesis that the average growth at 50% vegetation control is less than the average growth at 0% and 100% control levels.

(a) To test the contrast hypothesis that the average growth at 50% vegetation control is less than the average growth at 0% and 100% control levels,

we can set up the following contrast coefficients:

Contrast coefficients: c = [-1, 2, -1]

which indicate the weight or contribution of each group mean to the contrast. The first coefficient (-1) represents the weight for the 0% control group, the second coefficient (2) represents the weight for the 50% control group, and the third coefficient (-1) represents the weight for the 100% control group.

(b) To perform the test,

we can use the contrast coefficients to calculate the test statistic and P-value.

Test statistic (t-value):

t = ([tex]c_{1}[/tex] × [tex]X_{1}[/tex] + [tex]c_{2}[/tex] × [tex]X_{2}[/tex] + [tex]c_{3}[/tex] × [tex]X_{3}[/tex]) / √ ([tex]sp^2[/tex] × ([tex]c_{1}^{2} /n_{1}[/tex] + [tex]c_{2} ^{2} /n_{2}[/tex] + [tex]c_{3} ^{2} /n_{3}[/tex]))

where:

[tex]c_{1}[/tex], [tex]c_{2}[/tex], [tex]c_{3}[/tex] are the contrast coefficients

[tex]X_{1}[/tex], [tex]X_{2}[/tex], [tex]X_{3}[/tex] are the sample means for each control level

sp is the pooled standard deviation

[tex]n_{1}[/tex], [tex]n_{2}[/tex], [tex]n_{3}[/tex] are the sample sizes for each control level

Using the given values:

[tex]c_{1}[/tex] = -1,

[tex]c_{2}[/tex] = 2,

[tex]c_{3}[/tex] = -1

[tex]X_{1}[/tex] = 58,

[tex]X_{2}[/tex]= 73,

[tex]X_{3}[/tex] = 105

sp = 17

[tex]n_{1}[/tex] = [tex]n_{2}[/tex] = [tex]n_{3}[/tex] = 16

Calculating the t-value:

t = (-1 × 58 + 2 × 73 - 1 × 105) / √ ([tex]17^2[/tex] × ([tex]-1^2/16[/tex] +[tex]2^2/16[/tex] + [tex]-1^2/16[/tex]))

Degrees of freedom:

df = [tex]n_{1}[/tex] +[tex]n_{2}[/tex] +[tex]n_{3}[/tex] - 3

= 16 + 16 + 16 - 3

= 45

Using the calculated t-value and degrees of freedom,

we can determine the P-value from a t-distribution table or statistical software.

Learn more about hypothesis here:

https://brainly.com/question/32562440

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