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TESTS ABOUT ONE MEAN AND ONE VARIANCE

Module by: Ewa Paszek

Summary: This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by Ewa Paszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

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 H 0 : μ=  μ 0 .

There are essentially tree possibilities for the alternative hypothesis, namely that μ μ has increased,

  1. H 1 : μ>  μ 0 H 1 : μ>  μ 0 ; μ μ has decreased,
  2. H 1 : μ<  μ 0 H 1 : μ<  μ 0 ; μ μ has changed, but it is not known if it has increased or decreased, which leads to a two-sided alternative hypothesis
  3. H 1 ;μ≠ μ 0 H 1 ;μ≠ μ 0 .

To test H 0 ;μ= μ 0 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 ¯ x ¯ , that is close to μ 0 μ 0 supports H 0 H 0 . The closeness of x ¯ x ¯ to μ 0 μ 0 is measured in term of standard deviations of X ¯ X ¯ , σ/ n 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 , Z= X ¯ − μ 0 σ2 /n = X ¯ − μ 0 σ/ n , and the critical regions, at a significance level α α , for the tree respective alternative hypotheses would be:

  1. z≥ z α z≥ z α
  2. z≤ z α z≤ z α
  3. | z |= z α/2 | z |= z α/2

In terms of x ¯ x ¯ these tree critical regions become

  1. x ¯ ≥ μ 0 + z α σ/ n , x ¯ ≥ μ 0 + z α σ/ n ,
  2. x ¯ ≤ μ 0 − z α σ/ n , x ¯ ≤ μ 0 − z α σ/ n ,
  3. | 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 ) N( μ, σ 2 ) and σ 2 σ 2 is known. Thus far we have assumed that the variance σ 2 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 H 0 ;μ= μ 0 and the two-sided alternative hypothesis is H 1 ;μ≠ μ 0 H 1 ;μ≠ μ 0 . If a random sample X 1 , X 2 ,..., X n X 1 , X 2 ,..., X n is taken from a normal distribution N( μ, σ 2 ) N( μ, σ 2 ) ,let recall that a confidence interval for μ μ was based on T= X ¯ −μ S 2 /n = X ¯ −μ S/ n . T= X ¯ −μ S 2 /n = X ¯ −μ S/ n .

TABLE 1
H 0 H 0 H 1 H 1 Critical Region
μ= μ 0 μ= μ 0 μ> μ 0 μ> μ 0 z≥ z α z≥ z α or x ¯ ≥ μ 0 + z α σ/ n x ¯ ≥ μ 0 + z α σ/ n
μ= μ 0 μ= μ 0 μ< μ 0 μ< μ 0 z≤− z α z≤− z α or x ¯ ≤ μ 0 − z α σ/ n x ¯ ≤ μ 0 − z α σ/ n
μ= μ 0 μ= μ 0 μ≠ μ 0 μ≠ μ 0 | z |≥ z α/2 | z |≥ z α/2 or | x ¯ − μ 0 |≥ z α/2 σ/ n | 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 H 0 ;μ= μ 0 with μ μ replaced by μ 0 μ 0 . In addition, it is the natural statistic to use if we replace σ 2 /n σ 2 /n by its unbiased estimator S 2 /n S 2 /n in ( X ¯ − μ 0 )/ σ 2 /n ( X ¯ − μ 0 )/ σ 2 /n in a proper equation. If μ= μ 0 μ= μ 0 we know that T has a t distribution with n-1 degrees of freedom. Thus, with μ= μ 0 μ= μ 0 ,

P[ | T |≥ t α/2 ( n−1 ) ]=P[ | X ¯ − μ 0 | S/ n ≥ t α/2 ( n−1 ) ]=α. P[ | T |≥ t α/2 ( n−1 ) ]=P[ | X ¯ − μ 0 | S/ n ≥ t α/2 ( n−1 ) ]=α.

Accordingly, if x ¯ x ¯ and s are the sample mean and the sample standard deviation, the rule that rejects H 0 ;μ= μ 0 H 0 ;μ= μ 0 if and only if | t |= | x ¯ − μ 0 | s/ n ≥ t α/2 ( n−1 ). | t |= | x ¯ − μ 0 | s/ n ≥ t α/2 ( n−1 ).

Provides the test of the hypothesis with significance level α α . It should be noted that this rule is equivalent to rejecting H 0 ;μ= μ 0 H 0 ;μ= μ 0 if μ 0 μ 0 is not in the open 100( 1−α )% 100( 1−α )% confidence interval ( x ¯ − t α/2 ( n−1 )s/ n , x ¯ + t α/2 ( n−1 )s/ n ). ( 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 ) N( μ, σ 2 ) , σ 2 σ 2 is unknown, t=( x ¯ − μ 0 )/( s/ n ) t=( x ¯ − μ 0 )/( s/ n ) and n≤31 n≤31 . If n>31, use table 1 for approximate tests with σ σ replaced by s.

TABLE 2
H 0 H 0 H 1 H 1 Critical Region
μ= μ 0 μ= μ 0 μ> μ 0 μ> μ 0 t≥ t α ( n−1 ) t≥ t α ( n−1 ) or x ¯ ≥ μ 0 + t α ( n−1 )s/ n x ¯ ≥ μ 0 + t α ( n−1 )s/ n
μ= μ 0 μ= μ 0 μ< μ 0 μ< μ 0 t≤− t α ( n−1 ) t≤− t α ( n−1 ) or x ¯ ≤ μ 0 − t α ( n−1 )s/ n x ¯ ≤ μ 0 − t α ( n−1 )s/ n
μ= μ 0 μ= μ 0 μ≠ μ 0 μ≠ μ 0 | t |≥ t α/2 ( n−1 ) | t |≥ t α/2 ( n−1 ) or | x ¯ − μ 0 |≥ t α/2 ( n−1 )s/ n | x ¯ − μ 0 |≥ t α/2 ( n−1 )s/ n

Example 1

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 ) N( μ, σ 2 ) . We shall test the null hypothesis H 0 :μ= μ 0 =4.0 H 0 :μ= μ 0 =4.0 millimeters against the two-sided alternative hypothesis is H 1 :μ≠4.0 H 1 :μ≠4.0 . If we use n=9 observations and a significance level of α α =0.10, the critical region is | t |= | x ¯ −4.0 | s/ 9 ≥ t α/2 ( 8 )= t 0.05 ( 8 )=1.860. | t |= | x ¯ −4.0 | s/ 9 ≥ t α/2 ( 8 )= t 0.05 ( 8 )=1.860.

If we are given that n=9, x ¯ x ¯ =4.3, and s=1.2, we see that t= 4.3−4.0 1.2/ 9 = 0.3 0.4 =0.75. t= 4.3−4.0 1.2/ 9 = 0.3 0.4 =0.75.

Thus | t |=| 0.75 |<1.860 | t |=| 0.75 |<1.860 and we accept (do not reject) H 0 :μ=4.0 H 0 :μ=4.0 at the α α =10% significance level. See Figure 1.

Figure 1: Rejection region at the α=10% α=10% significance level.
(tails.gif)

REMARK:

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 ¯ x ¯ is close enough to 4.0 so that we accept μ μ =4.0, we do not want that acceptance to imply that μ μ is actually equal to 4.0. We want to say that the data do not deviate enough from μ μ =4.0 for us to reject that hypothesis; that is, we do not reject μ 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.

Example 2

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 μ of 500. They plan to test H 0 :μ=500 H 0 :μ=500 against H 1 :μ<500 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 α =0.01, is t= x ¯ −500 s/ 25 ≤− t 0.01 ( 24 )=−2.492. t= x ¯ −500 s/ 25 ≤− t 0.01 ( 24 )=−2.492.

The observed values of the sample mean and sample standard deviation were x ¯ x ¯ =308.8 and s=115.15. Since t= 308.8−500 115.15/ 25 =−8.30<−2.492, t= 308.8−500 115.15/ 25 =−8.30<−2.492, we clearly reject the null hypothesis and accept H 1 :μ<500 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 ) 308.8±2.064( 115.15/5 ) or [ 261.27, 356.33 ] [ 261.27, 356.33 ] for μ μ might the company answer that question.

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