Question

In a​ study, researchers compared the variability in product of a particular piece of equipment to a known analytic capability to decide whether product consistency could be improved. The following data were obtained for

10

batches of product. Preliminary test results suggest the data are normally distributed and that

sequals=0.546.

Construct a

90​%

confidence interval for the process variation of the piece of equipment under consideration.

29.2

29.7

30.3

29.1

29.7

29.5

28.8

29.5

30.4

30.3

The

90​%

confidence interval is from

nothing

to

nothing.

​(Round to three decimal places as​ needed.)

Answer 1

Solution:

The 90% confidence interval for the process variation of the piece of equipment under consideration is:

$\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\frac{0.1}{2}}}}<\sigma<\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\frac{0.1}{2}}}}$

Where:

$\chi^{2}_{\frac{0.1}{2}}=16.919$

$\chi^{2}_{1-\frac{0.1}{2}}=3.325$

$\therefore \sqrt{\frac{(10-1)0.546^{2}}{16.919}}<\sigma<\sqrt{\frac{(10-1)0.546^{2}}{3.325}}$

  $\boldsymbol{0.398<\sigma<0.898}$

Therefore, the 90% confidence interval is $\boldsymbol{0.398<\sigma<0.898}$

Note: If we are required to the confidence interval for the variance, then it is given below:

$\frac{(10-1)0.546^{2}}{16.919}<\sigma^{2}<\frac{(10-1)0.546^{2}}{3.325}$

$\boldsymbol{0.159<\sigma^{2}<0.807}$