MATH SOLVE

2 months ago

Q:
# The Campbell’s Soup Company has just purchased an automated filling machine to fill bottles with its V8 tomato juice. The machine will fill bottles according to a normally distributed process, with a mean of 47 fluid ounces and a standard deviation of 0.4 fluid ounces. (a) Upon installation of the machine, a technician decides to test whether the machine is working properly. He takes a sample of 50 bottles and determines the fill level of the bottles. In looking at the data, he decides that the filling process variability can be represented as normally distributed with a standard deviation of 0.4 fluid ounces. He finds that the sample mean is 46.9 fluid ounces. What should he deduce concerning whether the mean is indeed 47 fluid ounces, given a two-sided hypothesis test and a significance level of .05?

Accepted Solution

A:

Answer:We fail to reject the null hypothesis because -1.7678 does not fall inside the rejection region {z | z < -1.96 or z >1.96}Step-by-step explanation:We have that the machine will fill bottles according to a normally distributed process, let's say [tex]X[/tex] is a random variable that represents this process. We know that [tex]X[/tex] has a mean of 47 fluid ounces and a standard deviation of 0.4 fluid ounces. We have a sample of n = 50 bottles and the observed value [tex]\bar{x} = 46.9[/tex].
We want to test
[tex]H_{0}: \mu = 47[/tex] vs [tex]H_{1}: \mu \neq 47[/tex] (two-tailed alternative)
We have n = 50 large enough, and the point estimator of the mean [tex]\bar{X}[/tex], which is normally distributed with the same mean than [tex]X[/tex], i.e., [tex]\mu[/tex], and with standard deviation given by [tex]\sigma/\sqrt{n} = 0.4/\sqrt{50}[/tex]. Therefore, our test statistic is
[tex]Z = \frac{\bar{X}-47}{\sigma/\sqrt{n}}[/tex] and
[tex]z = \frac{46.9 - 47}{0.4/\sqrt{50}} = -1.7678[/tex].
As we want a significance level of 0.05, we should find the values [tex]z_{0.025}[/tex] and [tex]z_{0.975}[/tex], i.e., the 2.5th quantile and the 97.5th quantile for the standard normal distribution. These values are -1.96 and 1.96 respectively. The rejection region is given by {z | z < -1.96 or z >1.96}. We fail to reject the null hypothesis because -1.7678 does not fall inside the rejection region.