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.
