Since the population standard deviation is not known, t-value is used as test statistic. (Refer section t-test for details on the test and the t-value).

*Example:* For its detergents house brand, a retailer procures packs from a
manufacturer. According to specifications, these packs contain on average 15% of a surfactant. To
verify that the proportions are correct, technicians at a contracted lab take samples from 80 packs
and examine their ingredient composition. They find that for the sampled packs, the average
surfactant composition is 14.7 gm per 100 gm of detergent, and the standard deviation is 1.2 gm.
Based on these findings, can we infer that the packs contain less than the specified level of surfactant?

H_{0}: μ ≥ 15 gm per 100 gm of detergent

H_{A}: 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 a population is low (0.014 <α = 0.05). More specifically, if the surfactant concentration was actually 15%, the chance that our sample would average ≤14.7% is only 0.014 (1.4%). The null hypothesis is rejected, the data suggests that manufacturer is not meeting the specified concentration of 15%.

*Note: A p-value from t-value calculator is provided on this
web page.*

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