Question

You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 15 containers. Estimate the population standard deviation at a 98% level of confidence.
12.05 12 11.96 12.14 12.12
11.97 11.96 12.11 11.94 11.82
12 12.18 11.91 12.06 11.99

(Data checksum: 180.21)

Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer.

a) Find the sample standard deviation:

b) Find the lower and upper χ2χ2 critical values at 98% confidence:
Lower: Upper:

c) Report your confidence interval for σσ: ( , )

Answer 1

Answer:

Step-by-step explanation:

weight (x)                         $( x - \overline x)^2$

12.05                               0.00123

12                                     0.00019

11.96                                0.00292

12.14                                0.0159

12.12                                0.00112

11.97                                0.0019

11.96                               0.0029

12.11                                 0.0092

11.94                                 0.0055

11.82                                0.038

12                                     0.00019

12.18                                0.028

11.91                                  0.0108

12.06                                0.0021

11.99                                  0.0029

$\sum x = 180.21$                    $\sum (x - \overline x)^2 = 0.12285$

Sample size = 15

$\overline x = \dfrac{180.21}{15} = 12.014$

Sample standard deviation

$x = \sqrt{\dfrac{0.12285}{15-1}}$

$x = 0.0937$

$\text{degree of freedom = n - 1}$

$degree of freedom = 15- 1=14$

$\text{confidence interval} \alpha = 1- 0.98 = 0.02 \\ \\ \alpha/2 = 0.02/2 = 0.01$

$From \ X^2 \ table}$$\text{, the critical value} X^2 \text{=0.99 for degree of freedom =14 is given by : 4.660}$

$\text{, the critical value} X^2 \text{=0.01 for degree of freedom =14 is given by : 29.141}$

$\text{Thus, the lower limit = 4.660, the upper limit = 29.141}$

$\text{the confidence interval of the standard deviation} \ \sigma \ is:$

$\sqrt{\dfrac{(n -1)s^2}{X^2_{\dfrac{\sigma}{2}}} } \le \sigma \le \sqrt{\dfrac{(n -1)s^2}{X^2_{1-\dfrac{\sigma}{2}}} }$

replacing our values:

$\sqrt{\dfrac{(15 -1)0.0937^2}{29.141} } < \sigma$

$= 0.06495 < \sigma < 0.1624$

$\mathbf{Thus; \ \sigma (0.06495 , 0.1624)}$