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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:

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Answer:


H_(0) : μ=420 gram


H_(a) : μ≠420 gram

we fail to reject the null hypothesis at 0.01 significance

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=
(X-M)/((s)/(√(N) ) ) where

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

then z=
(417-420)/((24)/(√(33) ) ) =-0.718

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

User Tony Kiernan
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