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8. In order to estimate the effectiveness of thinning, five plots were established in thinned and unthinned stands (on simila

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In order to estimate the effectiveness of thinning plots were established in thinned and unthinned stands (on similar sites and in similar forest types). On each plot ,the volume increment per year hectare was measured The results follow:

Volume Increment (m3/year/hectare) :

  Thinned 8 6 5 10 11

Unthinned 9 4 4 6 2

a) estimate the real standard deviation in the thinned stand with 90% confidence limits.?

By using excel ,the sample standard deviation for Thinned stand is SThinned = 2.55.

we are given sample size , n = 5 . Therefore ,the number of degrees of freedom is:

k=n-1=5–1=4

The critical values 105 = 9.488 and x 5 = 0.711. we construct the 95% confidence interval by evaluating the following:

(n-1) s2 (n-1) s2

(5 - 1)(2.55)2 (5 - 1) (2.55)2 V <O<V - 9.488 0.711

1.66 <σ< 6.05

b) Using the proper technique, find out if the real variancesfrom both stands are similar or significantly different with 90%confidence.?

   The hypotheses are:

Ho : 01 hinned = Onthinner

Ho : oi hinned to inthinned

By using excel ,the sample standard deviation for Thinned stand SThinned = 2.55. and for unthinned stand is Sunthinned = 2.65

The test statistic is:

Strinn F = Sithin Thinned 2.552 2.652 = 0.93 hinned

dfi = 5–1 = 4

df2 = 5–1 = 4

Critical value for two tailed test at 90% confidence is 9.60 Since test statistic is less then critical value ,fail to reject the null hypotheses .There is insufficient evidence to support the claim that the real variances from both stands are significantly different.

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