Question

An article in the Journal of Composite Materials (Vol 23, 1989, p. 1200) describes the effect of delamination on the natural frequency of beams made from composite laminates. Five such delaminated beams were subjected to loads, and the resulting frequencies were as follows (in Hz):

230.66, 233.05, 232.58, 229.48, 232.58

(a) Find a 90% two-sided CI on mean natural frequency. Round your answers to 2 decimal places..
(b) Do the results of your calculations support the claim that mean natural frequency is 235 Hz?

Answer 1

Answer:

a) The 90% confidence interval would be given by (230.212;233.128)

b) For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.

Step-by-step explanation:

Previous concepts  and data given

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".

Data: 230.66, 233.05, 232.58, 229.48, 232.58

We can calculate the mean and the deviation from these data with the following formulas:

$\bar X= \frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=231.67$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=1.531 represent the sample standard deviation

n=5 represent the sample size  

Part a) Confidence interval

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

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

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$231.67-2.13\frac{1.531}{\sqrt{5}}=230.212$    

$231.67+2.13\frac{1.531}{\sqrt{5}}=233.128$

So on this case the 90% confidence interval would be given by (230.212;233.128)

Part b)  Do the results of your calculations support the claim that mean natural frequency is 235 Hz?  

For this case if we analyze the confidence interval we see that not contains the value of 235. So we can't support the claim that the true mean is higher than 235 Hz at 10% of significance.