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4.2 Test Errors and Test Power   115


              p  = P ( ≤  ( − µ B  / ) σ X  ) = P ( ≤  ( 1260 − 1300  / )  77 . 94 ) =  . 0  304 .
                        x
                                       Z
                    Z

              If  we are basing  our conclusions  on a 5% level of significance,  and since
           p > 0.05, we then have no evidence to reject the null hypothesis.
              Note that until now  we have assumed that we knew the true value of the
           standard deviation. This, however, is seldom the case. As already discussed in the
           previous chapter, when using the sample  standard  deviation – maintaining the
           assumption  of normality of the random variable  −  one  must  use  the  Student’s     t
           distribution. This is the usual procedure, also followed by statistical software
           products, where these parametric tests of means are called t tests.



           4.2  Test Errors and Test Power

           As described in the previous section, any decision derived from hypothesis testing
           has, in general, a certain degree of uncertainty. For instance, in the drill example
           there is always a chance that the null hypothesis is incorrectly rejected. Suppose
           that a sample from the good quality of drills has x =1190 hours. Then, as can be
           seen  in Figure  4.1, we would  incorrectly reject the  null hypothesis  at a 10%
           significance level. However, we would not reject the null hypothesis at a 5% level,
           as shown in Figure 4.3. In general, by lowering the chosen level of significance,
           typically 0.1, 0.05 or 0.01, we decrease the Type I Error:

              Type I Error: α = P(H 0 is true and, based on the test, we reject H 0).

              The price to be paid for the decrease of the Type I Error is the increase of the
           Type II Error, defined as:

              Type II Error: β = P(H 0 is false and, based on the test, we accept H 0).

              For instance, when in Figures 4.1 and 4.3 we decreased α from 0.10 to 0.05, the
           value of β increased from 0.10 to:

              β = P ( ≥  (x α − µ A  / ) σ X  ) = P ( ≥  ( 1172 8 . − 1100  / )  77 . 94 ) =  . 0  177 .
                    Z
                                        Z

              Note that a high value of β  indicates that when the observed statistic does not
           fall in the critical region there is a good chance that  this is due not to the
           verification  of the  null  hypothesis itself  but, instead, to the verification  of a
           sufficiently close alternative hypothesis. Figure 4.4 shows that, for the same level
           of significance, α, as the alternative hypothesis approaches the null hypothesis, the
           value of  β increases, reflecting a decreased  protection against an alternative
           hypothesis.
              The degree of protection against alternative hypotheses is usually measured by
           the so-called power of the test, 1–β, which measures the probability of rejecting the
           null hypothesis when it is false (and thus should be rejected). The values of the
           power for several alternative values of  µ A, using the computed values of  β as
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