Question

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 (229.266, 233.002). 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.)

Answer 1

Answer:

$Lower = 231.134- 3.098=228.036$

$Upper = 231.134+ 3.098=234.232$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

And for this case we know that the 95% confidence interval is given by:

$\bar X=\frac{233.002 +229.266}{2}= 231.134$

And the margin of error is given by:

$ME = \frac{233.002 -229.266}{2}= 1.868$

And the margin of error is given by:

$ME= t_{\alpha/2} \frac{s}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1 = 5-1=4$

And the critical value for 95% of confidence is $t_{\alpha/2}= 2.776$

So then we can find the deviation like this:

$s = \frac{ME \sqrt{n}}{t_{\alpha/2}}$

$s = \frac{1.868* \sqrt{5}}{2.776}= 1.506$

And for the 99% confidence the critical value is: $t_{\alpha/2}= 4.604$

And the margin of error would be:

$ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098$

And the interval is given by:

$Lower = 231.134- 3.098=228.036$

$Upper = 231.134+ 3.098=234.232$