In: Statistics and Probability
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 = .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?
6.72 | 5.94 | 8.83 | 8.86 | 6.38 | 5.36 | 7.79 | 7.21 | 8.40 | 8.09 | 5.52 | 5.42 | |
(a) | For the following hypothesis |
H0: μ ≤ 6 versus H1: μ > 6, what is the value of the test statistic? (Round your answer to 3 decimal places.) |
Test statistic |
(b) | What is the p-value? (Round your answer to 4 decimal places.) |
p-value |
(c) | At α = .05, we would | ||||
|
Values ( X ) | ||
6.72 | 0.1045 | |
5.94 | 1.2173 | |
8.83 | 3.1923 | |
8.86 | 3.3004 | |
6.38 | 0.44 | |
5.36 | 2.8335 | |
7.79 | 0.5576 | |
7.21 | 0.0278 | |
8.4 | 1.8406 | |
8.09 | 1.0956 | |
5.52 | 2.3204 | |
5.4 | 2.6351 | |
Total | 84.52 | 19.5651 |
Test Statistic :-
t = 2.7098
Test Criteria :-
Reject null hypothesis if
Result :- Reject null hypothesis
P value
Looking for the value t = 2.7098 in t table across 11 degree of freedom
t = 2.7098 lies between the values 2.201 and 2.718 respective P vlaues are 0.025 and 0.010.
Using excel to calculate P value = 0.0101
Decision based on P value
Reject null hypothesis if P value < level of significance
0.0101 < 0.05, reject null hypothesis
We would reject null hypothesis.