Leaf length is normally distributed for leaves on a large oa

Leaf length is normally distributed for leaves on a large oak tree. A sample of is collected with sizes 3, 3, 5, 6, 6, 6, 7, 8, 8, 9 and 9. Answers should include any written information necessary, and use R or R Studio for confidence interval calculations.

What is the degrees of freedom for this sample?

Write the expression for the confidence interval of mean leaf size using the sample mean, t-statistic, standard deviation, and sample size.

What is the 95% confidence interval for the mean of leaf size?

What is the 95% confidence interval for variance of leaf size?

Solution

a)
d.f = n-1 = 11-1 = 10
b)
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
Mean(x)=6.364
Standard deviation( sd )=2.111
Sample Size(n)=11
c)
Confidence Interval = [ 6.364 ± t a/2 ( 2.111/ Sqrt ( 11) ) ]
= [ 6.364 - 2.228 * (0.636) , 6.364 + 2.228 * (0.636) ]
= [ 4.946,7.782 ]
d)
Confidence Interval
CI = (n-1) S^2 / ^2 right < ^2 < (n-1) S^2 / ^2 left
Where,
S = Standard Deviation
^2 right = (1 - Confidence Level)/2
^2 left = 1 - ^2 right
n = Sample Size

Since aplha =0.05
^2 right = (1 - Confidence Level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 10 df are 20.4832 , 3.247
S.D( S^2 )=2.111
Sample Size(n)=11
Confidence Interval = [ 10 * 4.4563/20.4832 < ^2 < 10 * 4.4563/3.247 ]
= [ 44.5632/20.4832 < ^2 < 44.5632/3.247 ]
[ 2.1756 , 13.7244 ]