A researcher wants to estimate the mean diameter at breast height (dbh) of Dougl

Question

A researcher wants to estimate the mean diameter at breast height (dbh) of Douglas fir trees in the western Washington Cascades as a part of a study in the structural development of an old-growth Douglas fir/western hemlock stand. A pilot sample of 12 Douglas firs yielded a sample mean of 147.3 cm and a sample standard deviation of 18.8 cm. How many more trees need to be sampled in order of to estimate the mean dbh within plus or minus 5.0 cm and 95% confidence? If you can only sample 50 trees and you still need to be 95% confident, what is your new confidence interval width? If you can only sample 50 trees and you still need to be within plus or minus 5.0 cm, what is your new confidence level?

Explanation / Answer

a.

TRADITIONAL METHOD
given that,
sample mean, x =147.3
standard deviation, s =18.8
sample size, n =12
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 18.8/ sqrt ( 12) )
= 5.4271
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 2.201
margin of error = 2.201 * 5.4271
= 11.945
III.
CI = x ± margin of error
confidence interval = [ 147.3 ± 11.945 ]
= [ 135.355 , 159.245 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =147.3
standard deviation, s =18.8
sample size, n =12
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 11 d.f is 2.201
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 147.3 ± t a/2 ( 18.8/ Sqrt ( 12) ]
= [ 147.3-(2.201 * 5.4271) , 147.3+(2.201 * 5.4271) ]
= [ 135.355 , 159.245 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 135.355 , 159.245 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

b.
sample size = 50 then new confidence interval,
TRADITIONAL METHOD
given that,
sample mean, x =147.3
standard deviation, s =18.8
sample size, n =50
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 18.8/ sqrt ( 50) )
= 2.6587
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 49 d.f is 2.01
margin of error = 2.01 * 2.6587
= 5.344
III.
CI = x ± margin of error
confidence interval = [ 147.3 ± 5.344 ]
= [ 141.956 , 152.644 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =147.3
standard deviation, s =18.8
sample size, n =50
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 49 d.f is 2.01
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 147.3 ± t a/2 ( 18.8/ Sqrt ( 50) ]
= [ 147.3-(2.01 * 2.6587) , 147.3+(2.01 * 2.6587) ]
= [ 141.956 , 152.644 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 141.956 , 152.644 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

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