A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420.0420.0 gram setting. It is believed that the machine is underfilling the bags. A 3333 bag sample had a mean of 417.0417.0 grams. A level of significance of 0.010.01 will be used. State the hypotheses. Assume the variance is known to be 576.00576.00. Enter the hypotheses:

Answer:[tex]H_{0}[/tex] : μ=420 gram[tex]H_{a}[/tex] : μ≠420 gram we fail to reject the null hypothesis at 0.01 significanceStep-by-step explanation:Let μ be the mean value for the bag filling machine. Then [tex]H_{0}[/tex] : μ=420 gram[tex]H_{a}[/tex] : μ≠420 gramz-score of the sample mean can be calculated as:z=[tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] where X is the sample mean  (417)M is the mean assumed in null hypothesis (420)s is the standard deviation ([tex]\sqrt{576}[/tex] = 24) N is the sample size (33)then z=[tex]\frac{417-420}{\frac{24}{\sqrt{33} } }[/tex] =-0.718since p(z)=  0.858 > 0.01 we fail to reject the null hypothesis.