Question

Plastic fill process. University of Louisville operators examined the process of filling plastic pouches of dry blended biscuit mix (Quality Engineering, Vol. 91, 1996). The current fill mean of the process is set at $\mathbf{\mu }$= 406 grams, and the process fills standard deviation is $\mathbf{\sigma }$= 10.1 grams. (According to the operators, “The high level of variation is since the product has poor flow properties and is, therefore, difficult to fill consistently from pouch to pouch.”) Operators monitor the process by randomly sampling 36 pouches each day and measuring the amount of biscuit mix in each. Consider$\overline{\mathbf{x}}$ the mean fill amount of the sample of 36 products. Suppose that on one particular day, the operators observe $\overline{\mathbf{x}}$= 400.8. One of the operators believes that this indicates that the true process fill mean for that day is less than 406 grams. Another operator argues that$\mathbf{\mu }$ = 406, and the small observed value is due to random variation in the fill process. Which operator do you agree with? Why?

Short answer

The operator, who got the mean value of 406 grams, can be considered.

Step-by-step solution

Given information

There is the process of filling plastic pouches with dry-blended biscuit mix. The mean of the filling is 406 grama, and the standard deviation is 10.1 grams.

Two operators took 36 pouches as the sample, and one of them observed that the mean of the filling is 400.8 grams, and another one observed that the mean is 406 grams.

Null hypothesis: Ho

$\mathrm{\mu }=406$

Alternate hypothesis: Ha

$\mathrm{\mu }<406$

Significance level:$\alpha =0.05$

Statistic test:

$\begin{array}{l}\text{Z=}\frac{-\mathrm{\mu }}{\mathrm{\sigma }/\sqrt{\mathrm{n}}}\\ \mathrm{Z}=\frac{400.8-406}{10.1/\sqrt{36}}\\ \mathrm{Z}=-3.08910891\end{array}$

P value:

Z = - 3.089 and Z test p-value will be 0.001

Conclusion

Simply by the outputs of those two operators, there can be easily concluded that the second operators get the absolute mean compared to the original population of pouches. He gets the mean value of 406 grams, the same as the population. Other while the first operator gets the mean value of 400.8 grams.So, there can be agreed with the operator who got the mean value of 407 grams.