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 (229.764, 233.504). 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 \(\overline x \) and s.)

Short answer

The interval is \((228.533,234.733)\).

Step-by-step solution

Step 1:Confidence bound.

Upper confidence bound for \(\mu \) and lower confidence bound for \(\mu \) are given by,

\(\overline x + {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-upper bound,

\(\overline x - {t_{\alpha /2,n - 1}} \cdot \frac{s}{{\sqrt n }}\)-lower bound.

With confidence level of \(100(1 - \alpha )\% \). Then distribution the random sample is taken from is normal.

Step 2:Sample standard deviation.

In order to obtain confidence interval with \(\mu \) confidence level there are a couple of values that are missing , \(\overline x ,s,{t_{\alpha /2,n - 1}}\). From the resulting interval \((229.764,233.504)\). The sample mean can be obtained as the centre of the interval,

\(\begin{array}{l}\overline x = \frac{{229.764 + 233.504}}{2}\\\overline x = 231.633\end{array}\)

The sample standard deviation can be obtained using the width of the interval,

\(w = 2 \cdot {t_{\alpha /2}} \cdot \frac{s}{{\sqrt n }}\)

Where in case of \(95\% \) confidence level, \(\alpha \) is \(0.05\), which means that \(\alpha /2 = 0.025\) and the sample size is \(5\). From the appendix table, the \(t\) value can be obtained and it is ,

\({t_{0.025,4}} = 2.776\)

The width \(w\)is the difference between upper and lower limit

\(\begin{array}{l}w = 233.504 - 229.764\\w = 3.738\end{array}\)

Now the sample standard deviation can be obtained from

\(w = 2 \cdot {t_{\alpha /2}} \cdot \frac{s}{{\sqrt n }}\)

As follows,

\(\begin{array}{l}s = \frac{{w\sqrt n }}{{2{t_{\alpha /2}}}}\\s = \frac{{3.738\sqrt 5 }}{{2(2.776)}}\\s = 1.5055\end{array}\)

Step 3:Confidence interval.

To obtain the confidence interval for \(\mu \) with \(\mu \) confidence level, the only missing value is \({t_{\alpha /2,n - 1}}\) which for

\(\begin{array}{l}\alpha = 0.01\\(\alpha /2) = 0.005\end{array}\)

And

\(\begin{array}{l}n - 1 = 5 - 1\\ = 4\end{array}\)

From the table in the appendix of the book. Therefore the interval becomes,

\(\begin{array}{l}\left( {231.633 - 4.604\frac{{1.5055}}{{\sqrt 5 }},231.633 + 4.604\frac{{1.5055}}{{\sqrt 5 }}} \right)\\ = (228.533,234.733)\end{array}\)

Hence, The interval is \((228.533,234.733)\).