In cases where the population standard deviation is not known, t-value is used
as the test statistic for one-tailed tests.
(Refer section t-test for details on the test and the t-value).
Example: To assess whether the proportions of surfactant in
the detergent packs are correct, a study was conducted by technicians from a contracted lab.
The study involved examining samples from 80 packs. According to the specifications, the packs
should contain an average of 15% surfactant.
The findings from the sample indicate that the average surfactant composition in
the sampled packs is 14.7 grams per 100 grams of detergent, with a standard deviation of 1.2
grams. Based on these findings, can we infer that the packs contain less than the specified
level of surfactant?
H0: μ ≥ 15 gm per 100 gm of detergent
HA: μ < 15 gm per 100 gm of detergent
α = 5%
$$ t = \frac{\bar x-μ_0}{σ/\sqrt n} = \frac{14.7-15}{1.2/8.94} = -2.24 $$
Degrees of freedom = 79.
p-value = 0.014.
The probability of obtaining a t-value of −2.24 or lower, when sampling 80 packs
from the population, has been determined to be 0.014 or 1.4%. This probability is lower than the
significance level α of 5%. In other words, if the average surfactant concentration is assumed
to be 15%, the probability that our sample would have an average of 14.7% or less is only
0.014 (1.4%). Based on this analysis, the null hypothesis is rejected, indicating that the
data suggests that the manufacturer is not meeting the specified concentration of 15%.
Note: A p-value from t-value calculator is provided on this
webpage.
