A journal article reports that a sample of size 5 was used a

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.566, 233.302). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? [Hint: Use the center of the interval and its width to determine x and s.] (Round your answers to three decimal places.)

=   (         ,        )Hz

Solution

Margin of error = (Upper- Lower)/ 2 = (233.302-229.566)/2 = 1.868
Mean =    231.434
Margin of Error = t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
& we know that ta/2 at 0.05 for n-1 = 4 d.f is 2.776

AT 95% ME
1.868 = 2.776 * (sd/ Sqrt(n))
(sd/ Sqrt(n)) = 1.868/2.776 = 0.6729

AND Now, we calculate margin of error at 99% C.I
ta/2 at 0.01   for n-1 = 4 d.f is 4.604
Margin of Error = 4.604 * 0.6729 = 3.098

Now, the confidence interval at 99% C.I is
[Lower, Upper] = [ 231.434-3.098, 231.434+3.098] = [ 228.336, 234.532]