Question

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. Following are the amounts measured in a simple random sample of eight cans. 11.98 12.08 11.87 11.95 11.89 12.01 11.97 12.10 Assume that you want to test to determine whether the mean volume is not 12 ounces. Use LaTeX: \alpha=0.05α = 0.05 level of significance level. a) Find the test statistics.

Answer 1

Answer:

The mean volume is 12 ounces.

Step-by-step explanation:

In this case we need to test whether the mean volume the machine fills the beverage cans with is 12 ounces or not.

The hypothesis for this test can be defined as:

H₀: The mean volume is 12 ounces. i.e. μ = 12 ounces.

Hₐ: The mean volume is not 12 ounces. i.e. μ ≠ 12 ounces.

As the population standard deviation is not known, we will use a t-test for single mean.

The sample mean and sample standard deviation is:

$\bar x=\frac{1}{n}\sum X\\=\frac{1}{8}\times [11.98+12.08+11.87+11.95+11.89+12.01+11.97+12.10]\\=11.98$

$\text{s} = \sqrt{\dfrac{\sum_{i=1}^{n}(x_i - \overline{x})^{2}}{n - 1}}\\=\sqrt{\dfrac{\sum_{i=1}^{n}(11.98 - 11.98)^{2}}{8 - 1}}+\sqrt{\dfrac{\sum_{i=1}^{n}(12.08- 11.98)^{2}}{8 - 1}}+...+\sqrt{\dfrac{\sum_{i=1}^{n}(12.10- 11.98)^{2}}{8 - 1}}\\=0.0815$Compute the test statistic as follows:

$t=\frac{\bar x-\mu}{\s/\sqrt{n}}=\frac{11.98-12}{0.0815/\sqrt{8}}=-0.69$

The test statistic value is -0.69.

Decision rule:

If the p-value of the test is less than the significance level, then the null hypothesis will be rejected.

Compute the p-value of the test as follows:

$p-value=2P(t_{7}$

*Use a t-table.

The significance level of the test is, α = 0.05.

p-value = 0.512 > α = 0.05

The null hypothesis was failed to be rejected.

Hence, it can be concluded that the mean volume is 12 ounces.