Question

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 439.0 gram setting. It is believed that the machine is underfilling the bags. A 47 bag sample had a mean of 433.0 grams. A level of significance of 0.05 will be used. Determine the decision rule. Assume the standard deviation is known to be 21.0. Enter the decision rule.

Answer 1

Answer with explanation:

Let $\mu$ be the average weight of chocolate chips in each bag.

As per given , we have

$H_0: \mu =439.0\\\\ H_a: \mu<439.0$

Since $H_a$ is left-tailed , so we perform a left-tailed test.

Also, the population standard deviation is known $\sigm=21.0$.

So we use z-test.

Test statistic : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqt{n}}}$

Substitute given value ,$\overline{x}=433.0\ , n=47,\ \sigma=21 \ , \mu=439$

$z=\dfrac{433-439}{\dfrac{21}{\sqrt{47}}}\approx-1.96$

Significance level = 0.05

Decision rule : Reject $H_0$ when p-value < 0.05.

By z-table , P- value for left-tailed test  = P(z<-1.96)= 1- P(z<1.96)

[∵ P(Z<-z)=1-P(Z<z)]

= 1- 0.9750

=0.025

Since , P-value(0.025) < 0.05 , so we reject the null hypothesis.

We support the claim that the machine is underfilling the bags at 0.05 significance level .