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)]: 65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8 Calculate a 99% CI for the standard deviation of the fracture toughness distribution. (Round your answers to one decimal place.)

Answer 1

`((21)(5.437)^2)/(41.402) \leq \sigma^2 \leq ((21)(5.437)^2)/(8.034)`

`14.996 \leq \sigma^2 \leq 77.278`

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8

Explanation:

Data given:

65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8

The sample mean would be given by:

`\bar X = (\sum_(i=1)^n X_i)/(n)`

We can calculate the sample deviation with this formula:

`s = (\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)`

And we got:

s=5.437 represent the sample standard deviation

`\bar x` represent the sample mean

n=22 the sample size

Confidence=99% or 0.99

The confidence interval for the population variance is given by:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

The degrees of freedom given by:

`df=n-1=22-1=21`

The Confidence is 0.99 or 99%, the value of significance is
`\alpha=0.01` and
`\alpha/2 =0.005`, and the critical values are:

`\chi^2_(\alpha/2)=41.402`

`\chi^2_(1- \alpha/2)=8.034`

And the confidence interval would be:

`((21)(5.437)^2)/(41.402) \leq \sigma^2 \leq ((21)(5.437)^2)/(8.034)`

`14.996 \leq \sigma^2 \leq 77.278`

And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:

3.9<σ<8.8