In order to estimate the effectiveness of thinning, five plots were established

Question

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

Volume Increment (m3/hectare/year)

Thinned

Unthinned

8

9

6

4

5

4

10

6

11

2

a.Estimate the real standard deviation in the thinned stand with 90% confidence limits.

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

Volume Increment (m3/hectare/year)

Thinned

Unthinned

8

9

6

4

5

4

10

6

11

2

Explanation / Answer

A)

USing technology, the sample standard deviation is

s = 2.549509757

As              
              
df = n - 1 =    4          
alpha = (1 - confidence level)/2 =    0.05          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    9.487729037          
chi^2(alpha/2) =    0.710723021          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    2.740381803          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    36.58246493          
              
Thus, the confidence interval for the variance is              
              
(   2.740381803   ,   36.58246493   )
              
Also, for the standard deviation, getting the square root of the bounds,              
              
(   1.65540986   ,   6.048343982   ) [ANSWER]

***********************

b)

Using technology, the sample standard deviations are

s1 = 2.549509757  
s2 = 2.645751311

Formulating the null and alternative hypotheses,              
              
Ho:   sigma1^2 / sigma2^2   =   1  
Ha:    sigma1^2 / sigma2^2   =/   1  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical F, as alpha =    0.1   ,      
alpha/2 =    0.05          
df1 = n1 - 1 =    4         
df2 = n2 - 1 =    4          

F (crit) =    0.156537812   and   6.388232909  
              
Getting the test statistic, as              
s1 =    2.549509757          
s2 =    2.645751311          
              
Thus, F = s1^2/s2^2 =    0.928571429          
              
As F is between the two critical values, we FAIL TO REJECT THE NULL HYPOTHESIS.              

Hence, there is no significant evidence that the real variances from both standard are different, at 0.10 significance. [CONCLUSION]

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.