TESTS ABOUT ONE MEAN AND ONE VARIANCE In the previous paragraphs it was assumed that we were sampling from a normal distribution and the variance was known. The null hypothesis was generally of the form H 0 : μ= μ 0 . There are essentially tree possibilities for the alternative hypothesis, namely that μ has increased, H 1 : μ> μ 0 ; μ has decreased, H 1 : μ< μ 0 ; μ has changed, but it is not known if it has increased or decreased, which leads to a two-sided alternative hypothesis H 1 ;μ≠ μ 0 . To test H 0 ;μ= μ 0 against one of these tree alternative hypotheses, a random sample is taken from the distribution, and an observed sample mean, x ¯ , that is close to μ 0 supports H 0 . The closeness of x ¯ to μ 0 is measured in term of standard deviations of X ¯ , σ/ n which is sometimes called the standard error of the mean. Thus the statistic could be defined by Z= X ¯ − μ 0 σ2 /n = X ¯ − μ 0 σ/ n , and the critical regions, at a significance level α , for the tree respective alternative hypotheses would be: z≥ z α z≤ z α | z |= z α/2 In terms of x ¯ these tree critical regions become x ¯ ≥ μ 0 + z α σ/ n , x ¯ ≤ μ 0 − z α σ/ n , | x ¯ − μ 0 |≥ z α σ/ n

These tests and critical regions are summarized in TABLE 1 . The underlying assumption is that the distribution is N( μ, σ 2 ) and σ 2 is known. Thus far we have assumed that the variance σ 2 was known. We now take a more realistic position and assume that the variance is unknown. Suppose our null hypothesis is H 0 ;μ= μ 0 and the two-sided alternative hypothesis is H 1 ;μ≠ μ 0 . If a random sample X 1 , X 2 ,..., X n is taken from a normal distribution N( μ, σ 2 ) ,let recall that a confidence interval for μ was based on T= X ¯ −μ S 2 /n = X ¯ −μ S/ n .

TABLE 1 H 0 H 1 Critical Region μ= μ 0 μ> μ 0 z≥ z α or x ¯ ≥ μ 0 + z α σ/ n μ= μ 0 μ< μ 0 z≤− z α or x ¯ ≤ μ 0 − z α σ/ n μ= μ 0 μ≠ μ 0 | z |≥ z α/2 or | x ¯ − μ 0 |≥ z α/2 σ/ n

This suggests that T might be a good statistic to use for the test H 0 ;μ= μ 0 with μ replaced by μ 0 . In addition, it is the natural statistic to use if we replace σ 2 /n by its unbiased estimator S 2 /n in ( X ¯ − μ 0 )/ σ 2 /n in a proper equation. If μ= μ 0 we know that T has a t distribution with n-1 degrees of freedom. Thus, with μ= μ 0 , P[ | T |≥ t α/2 ( n−1 ) ]=P[ | X ¯ − μ 0 | S/ n ≥ t α/2 ( n−1 ) ]=α. Accordingly, if x ¯ and s are the sample mean and the sample standard deviation, the rule that rejects H 0 ;μ= μ 0 if and only if | t |= | x ¯ − μ 0 | s/ n ≥ t α/2 ( n−1 ). Provides the test of the hypothesis with significance level α . It should be noted that this rule is equivalent to rejecting H 0 ;μ= μ 0 if μ 0 is not in the open 100( 1−α )% confidence interval ( x ¯ − t α/2 ( n−1 )s/ n , x ¯ + t α/2 ( n−1 )s/ n ). Table 2 summarizes tests of hypotheses for a single mean, along with the three possible alternative hypotheses, when the underlying distribution is N( μ, σ 2 ) , σ 2 is unknown, t=( x ¯ − μ 0 )/( s/ n ) and n≤31 . If n>31, use table 1 for approximate tests with σ replaced by s. TABLE 2 H 0 H 1 Critical Region μ= μ 0 μ> μ 0 t≥ t α ( n−1 ) or x ¯ ≥ μ 0 + t α ( n−1 )s/ n μ= μ 0 μ< μ 0 t≤− t α ( n−1 ) or x ¯ ≤ μ 0 − t α ( n−1 )s/ n μ= μ 0 μ≠ μ 0 | t |≥ t α/2 ( n−1 ) or | x ¯ − μ 0 |≥ t α/2 ( n−1 )s/ n

Let X (in millimeters) equal the growth in 15 days of a tumor induced in a mouse. Assume that the distribution of X is N( μ, σ 2 ) . We shall test the null hypothesis H 0 :μ= μ 0 =4.0 millimeters against the two-sided alternative hypothesis is H 1 :μ≠4.0 . If we use n=9 observations and a significance level of α =0.10, the critical region is | t |= | x ¯ −4.0 | s/ 9 ≥ t α/2 ( 8 )= t 0.05 ( 8 )=1.860. If we are given that n=9, x ¯ =4.3, and s=1.2, we see that t= 4.3−4.0 1.2/ 9 = 0.3 0.4 =0.75. Thus | t |=| 0.75 |<1.860 and we accept (do not reject) H 0 :μ=4.0 at the α =10% significance level. See Figure 1.

Rejection region at the α=10% significance level. In discussing the test of a statistical hypothesis, the word accept might better be replaced by do not reject. That is, in Example 1, x ¯ is close enough to 4.0 so that we accept μ =4.0, we do not want that acceptance to imply that μ is actually equal to 4.0. We want to say that the data do not deviate enough from μ =4.0 for us to reject that hypothesis; that is, we do not reject μ =4.0 with these observed data, With this understanding, one sometimes uses accept and sometimes fail to reject or do not reject, the null hypothesis.

In this example the use of the t-statistic with a one-sided alternative hypothesis will be illustrated.

In attempting to control the strength of the wastes discharged into a nearby river, a paper firm has taken a number of measures. Members of the firm believe that they have reduced the oxygen-consuming power of their wastes from a previous mean μ of 500. They plan to test H 0 :μ=500 against H 1 :μ<500 , using readings taken on n=25 consecutive days. If these 25 values can be treated as a random sample, then the critical region, for a significance level of α =0.01, is t= x ¯ −500 s/ 25 ≤− t 0.01 ( 24 )=−2.492. The observed values of the sample mean and sample standard deviation were x ¯ =308.8 and s=115.15. Since t= 308.8−500 115.15/ 25 =−8.30<−2.492, we clearly reject the null hypothesis and accept H 1 :μ<500 . It should be noted, however, that although an improvement has been made, there still might exist the question of whether the improvement is adequate. The 95% confidence interval 308.8±2.064( 115.15/5 ) or [ 261.27, 356.33 ] for μ might the company answer that question.