Answer:
(229.5450, 234. 3230)
Step-by-step explanation:
Given that 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.
95% confidence interval was
![(230.061, 233.807)\\](https://tex.z-dn.net/?f=%28230.061%2C%20233.807%29%5C%5C)
From confidence interval we find mean as the average of lower and upper bound.
Mean = ![\frac{463.868}{2} \\=231.934](https://tex.z-dn.net/?f=%5Cfrac%7B463.868%7D%7B2%7D%20%5C%5C%3D231.934)
Margin of error = upper bound -mean
=![1.873](https://tex.z-dn.net/?f=1.873)
i.e. 1.96 * std error = 1.873
For 99% interval margin of error would be 2.58*std error
so margin of error for 99% = ![2.58*\frac{1.873}{1.96} \\=2.3890](https://tex.z-dn.net/?f=2.58%2A%5Cfrac%7B1.873%7D%7B1.96%7D%20%5C%5C%3D2.3890)
Confidence interval 99% lower bound = Mean - 2.3890 =229.5450
Upper bound = Mean +2.3890 = 234.3230