Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: volume of root beer in a Windsor Bottling Company can.
A sample of n=24 cans was taken and their contents measured, resulting:
X[bar]= 11.4 oz
S= 0.62 oz
Assuming that the variable has a normal distribution X~N(μ;σ²), the parameter of interest is the average contents of the root beer cans of the Windsor Bottling Company (μ)
The claim is that the population mean content of the cans is different from 12 oz, symbolically: μ ≠ 12
The statistical hypothesis (Null and alternative) have to be complementary, exhaustive and mutually exclusive. The null hypothesis is the "no change" hypothesis and always carries the "=" sign.
If the claim is μ ≠ 12, its complement is μ = 12, the expression carrying the "=" sign will be the null hypothesis and its complement will be the alternative hypothesis:
H₀: μ = 12
H₁: μ ≠ 12
α: 0.05
To test the population mean of this normal population, you have to apply a one sample t-test, with statistic:
![t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BS%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~t_%7Bn-1%7D)
This test is two-tailed, using the critical value approach, you have to determine two rejection regions. Meaning, you'll reject the null hypothesis to small values of the statistic or to high values of the statistic.
The decision rule is:
If 
≤ -2.069 or if ≥ 2.069, then you reject the null hypothesis.
If -2.069 < < 2.069, then you do not reject the null hypothesis.
The value is less than the left critical value, the decision is to reject the null hypothesis.
Then you can say that with a 5% significance level, there is significant evidence to reject the null hypothesis, then the average amount of root beer of the Windsor Bottling Company is different from 12 oz, this means that the claim about the amount of root beer in the cans is correct.
I hope it helps!
