Small-Sample Hypothesis Tests for µ and 𝝁𝟏 − 𝝁𝟐
Example 1: Directional hypothesis test for a population mean.
Suppose that the lengths (in millimeters) of metal fibers produced by a certain process have
a normal distribution for which both the mean µ and the variance 𝜎 2 are unknown. We want to test
whether the mean length of such fibers exceeds 5.2 mm. We want to test this claim, using a
significance level of 0.05, using a sample of 15 fibers randomly selected from the population of
fibers produced by this process.
Step 1: H0:  ≤ 5.2 mm.
versus
Ha:  > 5.2 mm.
Step 2: n = 15,  = 0.05
X  5.2mm
Step 3: The test statistic is T 
, which under H0 has a t distribution with d.f. = 14.
 S 


 15 
Step 4: The rejection region is a right-hand tail of the t-distribution. The area of this left-hand tail
is 0.05, our chosen significance level. The boundary point of the rejection region is 1.7613.
Step 5: Now, we select the random sample from the population, collect the data, and do the
calculations. We find that x  5.4 mm, s = 0.4226 mm., t = 1.8329, and p-value = 0.0441. The pvalue is less than our chosen significance level.
Step 6: We reject H0 at the 0.05 level of significance. We have sufficient evidence to conclude that
the mean length of such fibers exceeds 5.2 mm.
Example 2: Comparing two independent population means.
Two methods, A and B, were used to determine of the latent heat of fusion of ice (the latent
heat of fusion is the amount of heat a substance absorbs in changing phase from solid to liquid). The
investigators wished to find out the amount by which the methods differed. In such a situation,
where several measurements are made of the same quantity by a measuring instrument, it is
reasonable to assume that the population distribution of such measurements is normal. The
following table gives the change in total heat from ice at -0.72°C to water at 0°C in calories per
gram of mass.
Method A
Method B
79.98
79.97
80.02
79.97
80.04
80.05
79.94
80.02
80.03
79.98
80.04
80.02
79.97
80.03
80.00
79.97
80.03
80.02
80.03
80.04
79.95
We want to test whether Method B tends to yield lower measurements, on average, than Method A.
Step 1: H0: 𝜇𝐴 − 𝜇𝐵 ≤ 0
Step 2: n = 13, m = 8,  = 0.05
Step 3: The test statistic is
versus
𝑇=
Ha: 𝜇𝐴 − 𝜇𝐵 > 0
(𝑌̅𝐴 − 𝑌̅𝐵 ) − 0
,
1
1
√𝑆𝑃2 ( + )
13 8
which under H0 has a t distribution with 19 d.f.
Step 4: The rejection region is a right-hand tail of the t-distribution, with boundary point 𝑡19,0.05 =
1.7291.:
Step 5: From the data, we obtain 𝑦̅𝐴 = 80.02076923 cal/gm, 𝑠𝐴 = 0.0239657876 cal/gm, 𝑦̅𝐵 =
79.97875 cal/gm,and 𝑠𝐴 = 0.0313676357 cal/gm. Then the pooled variance estimate is
(12)(0.0239657876)2 + (7)(0.0313676357)2
𝑆𝑃2 =
= 0.0003625.
19
The value of the test statistic is then
80.02076923 − 79.97875
𝑇=
= 3.4722.44847.
1
1
√(0.0003625) ( + )
13 8
and p-value = 0.001276. The p-value is smaller than our chosen significance level.
Step 6: We reject H0 at the 0.05 level of significance. We have sufficient evidence to conclude that
the mean heat of fusion of ice as measured by Method B is lower than the mean heat of fusion of ice
as measured by Method A.