Question

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 1

Answer:

$H_{0}$ : μ=420 gram

$H_{a}$ : μ≠420 gram

we fail to reject the null hypothesis at 0.01 significance

Step-by-step explanation:

Let μ be the mean value for the bag filling machine. Then

$H_{0}$ : μ=420 gram

$H_{a}$ : μ≠420 gram

z-score of the sample mean can be calculated as:

z=$\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

• X is the sample mean  (417)
• M is the mean assumed in null hypothesis (420)
• s is the standard deviation ($\sqrt{576}$ = 24)
• N is the sample size (33)

then z=$\frac{417-420}{\frac{24}{\sqrt{33} } }$ =-0.718

since p(z)=  0.858 > 0.01 we fail to reject the null hypothesis.