Answer to Question #319788 in Statistics and Probability for gel

Question

You are the operations manager for XYZ Cereal Company and is responsible for monitoring the amount of cereal box filled. You select and weigh a random sample of 50 boxes in order to calculate a sample mean and investigate how close the fill weights are to the company’s specifications of a mean of 368 grams. Suppose the sample of 50 boxes indicates a sample mean of 372.5 grams and the population standard deviation is assumed to be 15 grams. This time you must make a decision and conclude whether(or not) the fill mean weight in the entire process is equal to 368 grams in order to know whether the fill process needs adjustment. How could you rationally make this decision? 

Answer 1

$H_0:\mu =368\\H_1:\mu \ne 368\\Z=\sqrt{n}\frac{\bar{x}-\mu}{\sigma}=\sqrt{50}\frac{372.5-368}{15}=2.12132\sim N\left( 0,1 \right) \\P-value:\\P\left( \left| Z \right|\geqslant 2.12132 \right) =2\varPhi \left( -2.12132 \right) =2\cdot 0.01695=0.0339\\Since\,\,the\,\,P-value\,\,is\,\,small\,\,\left( less\,\,than\,\,0.05 \right) \,\,the\,\,null\,\,hypothesis\,\,is\,\,declined. \mu \ne 368 (depends\ on\ level\ of\ significance\, it\ may\ not\ be\ declined\ on\ 0.01\ level)$