Sample from Normally Distributed Population, s2 Unknown

Sample from Normally Distributed Population, s2 Unknown, n £ 30:
l Example: Dissolving (in sec.) of a drug in agitated gastric juice
(42.7, 43.4, 44.6, 45.1, 45.6, 45.9, 46.8, 47.6). Do these data
provide sufficient evidence to indicate that the populations mean for
dissolving a drug greater than 45 sec.? Let ? = 0.05.
l H0: m = 45 , H1:m > 45
l From Percentiles of t Distribution Table
l t (1 - a) , v = t(1 - 0.05), 7 = 1.8946
l The critical value of test statistics is 1.8946
l Statistical decision: Accept H0, since t < 1.8946.
l Clinical decision: Conclude that, on the bases of these data, there is
an indication that the mean for dissolving a drug less than 45 sec.
Paired t- test:
In previous discussion involving the difference between two
population means, it was assumed the samples were independent.
A method employed for assessing the effectiveness of treatment or
experimental procedure is one that makes use of related
observations resulting from nonindependent samples. A hypothesis
test based on this type of data is known as paired comparisons test.