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The t statistic is constructed assuming the null hypothesis to be true (i.e., η = η 0 ):

y –
t 0 = y – η 0 ----------------
---------------- =
η 0
s y s/ n
On the assumption of random sampling from a normal distribution, t 0 will have a t-distribution with
y
ν = n − 1 degrees of freedom. Notice that t 0 may be positive or negative, depending upon whether is
greater or less than η 0 .
For a one-sided test that η > η 0 (or η < η 0 ), the null hypothesis is rejected if the absolute value of
the calculated t 0 is greater than t ν,α where α is the selected probability point of the t distribution with ν =
n − 1 degrees of freedom.
For a two-sided test (η > η 0 or η < η 0 ), the null hypothesis is rejected if the absolute value of the
calculated t 0 is greater than t ν,α/2 , where α/z is the selected probability point of the t distribution with ν =
n − 1 degrees of freedom. Notice that the one-sided test uses t α and the two-sided test uses t α /2 , where
the probability α is divided equally between the two tails of the t distribution.

Constructing the Confidence Interval
The (1 − α)100% confidence interval for the difference y η is constructed using t distribution as follows:
–

–
– t ν,a /2 s y < y η < +t ν,a/2 s y
If this confidence interval does not include (y η 0 ) , the difference between the known and measured
–
values is so large that it is unlikely to arise from chance. It is concluded that there is a difference between
the estimated mean and the known value η 0 .
A similar confidence interval can be defined for the true population mean:

y – t ν,a /2 s y < η < y + t ν,a /2 s y

If the standard η 0 falls outside this interval, it is declared to be different from the true population mean
η, as estimated by , which is declared to be different from η 0 .
y

Case Study Solution
The concentration of the standard specimens that were analyzed by the participating laboratories was
1.2 mg/L. This value was known with such accuracy that it was considered to be the standard: η 0 =
1.2 mg/L. The average of the 14 measured DO concentrations is = 1.4 mg/L, the standard deviation is
y
= 0.083 mg/L. The difference between the known and
s = 0.31 mg/L, and the standard error is s y
measured average concentrations is 1.4 − 1.2 = 0.2 mg/L. A t-test can be used to assess whether 0.2
mg/L is so large as to be unlikely to occur through chance. This must be judged relative to the variation
in the measured values.
The test t statistic is t 0 = (1.4 − 1.2)/0.083 = 2.35. This is compared with the t distribution with ν =
13 degrees of freedom, which is shown in Figure 16.1a. The values t = −2.16 and t = +2.16 that cut off
5% of the area under the curve are shaded in Figure 16.1. Notice that the α = 5% is split between 2.5%
on the upper tail plus 2.5% on the lower tail of the distribution. The test value of t 0 = 2.35, located by
the arrow, falls outside this range and therefore is considered to be exceptionally large. We conclude
that it is highly unlikely (less than 5% chance) that such a difference would occur by chance. The
estimate of the true mean concentration, = 1.4, is larger than the standard value, η 0 = 1.2, by an amounty
that cannot be attributed to random experimental error. There must be bias error to explain such a large
difference.

© 2002 By CRC Press LLC