Question

The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel (in ksi in. , given in increasing order)]: 65.1 71.9 72.7 73.1 73.4 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.3 79.6 79.7 79.9 80.1 82.2 83.6 93.8 Calculate a 99% CI for the standard deviation of the fracture toughness distribution. (Round your answers to one decimal place.) , ksi in Is this interval valid whatever the nature of the distribution? Explain. The interval is always valid. The distribution needs to be skewed to the right. The distribution needs to be approximately uniform. The distribution needs to be approximately normal.

Answer 1

Answer:

The 99% confidence interval is given by:  

3.9<σ<8.8  

The distribution needs to be approximately normal.

Step-by-step explanation:

1) Data given and notation  

Data: 65.1 71.9 72.7 73.1 73.4 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.3 79.6 79.7 79.9 80.1 82.2 83.6 93.8

We can calculate the sample standard deviation with this formula:

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

s=5.438 represent the sample standard deviation  

$\bar x$ represent the sample mean  

n=22 the sample size  

Confidence=99% or 0.99  

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 mean or variance 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.  

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .  

2) Calculating the confidence interval  

The confidence interval for the population variance is given by the following formula:  

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$  

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:  

$df=n-1=22-1=21$  

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.005,21)" "=CHISQ.INV(0.995,21)". so for this case the critical values are:  

$\chi^2_{\alpha/2}=41.401$  

$\chi^2_{1- \alpha/2}=8.034$  

And replacing into the formula for the interval we got:  

$\frac{(21)(5.438)^2}{41.401} \leq \sigma \frac{(21)(5.438)^2}{8.034}$  

$15.0 \leq \sigma^2 \leq 77.3$  

Now we just take square root on both sides of the interval and we got:  

$3.9 \leq \sigma \leq 8.8$  

So the 99% confidence interval is given by:  

3.9<σ<8.8  

For the conditions in order to calculate this interval we assume this:

The distribution needs to be approximately normal.