Question

Beer shelf life is a problem for brewers and distributors because when beer is stored at room temperature, its flavor deteriorates. When the average furfuryl ether content reaches 6 μg per liter, a typical consumer begins to taste an unpleasant chemical flavor. At α = 0.05, would the following sample of 12 randomly chosen bottles stored for a month convince you that the mean furfuryl ether content exceeds the taste threshhold?

6.94 7.50 5.87 7.92 8.66 7.78 7.95 6.35 6.16 6.28 5.68 7.02
For the following hypothesis:
a) H0: μ ≤ 6 versus H1: μ > 6, what is the value of the test statistic?
b) What is the p-value?
c) At α = .05, we would
i. reject the null hypothesis.
ii.fail to reject the null hypothesis.

Answer 1

a) 3.52

b) 0.002399

c) At α = .05 we reject the null hypothesis.

Explanation:

We are given the following information:

Population mean, μ = 6 μg per liter

Sample size, n = 12

Alpha, α = 0.05

6.94, 7.50, 5.87, 7.92, 8.66, 7.78, 7.95, 6.35, 6.16, 6.28, 5.68, 7.02

`Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}`

`Mean =\displaystyle(84.11)/(12) = 7.009`

`\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}`

where
`x_i` are data points,
`\bar{x}` is the mean and n is the number of observations.

`SS.D = \sqrt{(10.03)/(11)} = 0.95`

First, we design the null and the alternate hypothesis

`H_(0): \mu \leq 6\\H_A: \mu > 6`

We use One-tailed t test to perform this hypothesis.

a) Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n-1)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(7.009 - 6)/((0.95)/(√(11)) ) = 3.52`

b) P-value for one tailed t-test at 11 degree of freedom and α = 0.05 is 0.002399

c) Since P-value < 0.05

The result is insignificant.

Thus, at α = .05 we reject the null hypothesis.