Respuesta :
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
[tex](230.061, 233.807)\\[/tex]
From confidence interval we find mean as the average of lower and upper bound.
Mean = [tex]\frac{463.868}{2} \\=231.934[/tex]
Margin of error = upper bound -mean
=[tex]1.873[/tex]
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% = [tex]2.58*\frac{1.873}{1.96} \\=2.3890[/tex]
Confidence interval 99% lower bound = Mean - 2.3890 =229.5450
Upper bound = Mean +2.3890 = 234.3230
Using confidence interval concepts, it is found that the limits of the 99% confidence interval are: (229.71, 234.158).
The standard format of a confidence interval is:
[tex]\overline{x} \pm M[/tex]
In which
- [tex]\overline{x}[/tex] is the sample mean.
- M is the margin of error.
The sample mean is the mean of the two bounds, thus:
[tex]\overline{x} = \frac{230.061 + 233.807}{2} = 231.934[/tex]
The margin of error is half of the difference, thus:
[tex]M = \frac{233.807 - 230.061}{2} = 1.873[/tex]
The standard format of a margin of error is:
[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]
In this situation, [tex]\sigma[/tex] and [tex]\sqrt{n}[/tex] remaining constant, just the critical value z changes.
- Looking at the z-table, the critical value for a 95% CI is z = 1.96.
- For a 99% CI, it is z = 2.327.
Thus, the new margin of error is:
[tex]M = \frac{2.327}{1.96} \times 1.873 = 2.224[/tex].
Thus, the new limits are:
[tex]\overline{x} - M = 231.934 - 2.224 = 229.71[/tex]
[tex]\overline{x} + M = 231.934 + 2.224 = 234.158[/tex]
The new limits are (229.71, 234.158).
A similar problem is given at https://brainly.com/question/23215076
