A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting. It is believed that the machine is underfilling the bags. A 19 bag sample had a mean of 443 grams with a standard deviation of 21. A level of significance of 0.025 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer:A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 447 gram setting. [tex]H_0:\mu = 447\\H_a:\mu\neq 447[/tex]Mean = [tex]\mu = 447\\s = 21\\x = 443[/tex]n = 19Since n < 30 , so we will use t test[tex]t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]Substitute the values :[tex]t=\frac{443-447}{\frac{21}{\sqrt{19}}}[/tex][tex]t=−0.8302[/tex]t calculated = -0.830degree of freedom = n-1 = 19-1 = 18A level of significance=α=0.025 [tex]t_{(df,\frac{\alpha}{2})}=2.093[/tex]t critical = 2.093t calculated < t critical So, We failed to reject null hypothesis Decision rule  -0.830< 2.093 So, We failed to reject null hypothesis