A Normal Population with Known s

Case I: A Normal Population with Known s

  Although the assumption that the value of s is known is rarely met in practice, this case provides a good starting point because of the ease with which general proce- dures and their properties can be developed. The null hypothesis in all three cases will state that m has a particular numerical value, the null value, which we will

  denote by m 0 . Let 1 , c, X n represent a random sample of size n from the normal

  population. Then the sample mean has a normal distribution with expected value X

  m X 5m and standard deviation s X 5 s 1n . When H 0 is true, m X 5m 0 . Consider now the statistic Z obtained by standardizing under the assumption that H X 0 is true:

  Substitution of the computed sample mean gives z, the distance between and m x 0 expressed in “standard deviation units.” For example, if the null hypothesis is

  H 0 : m 5 100, s X 5 s 1n 5 10 125 5 2.0 , and x 5 103 , then the test statistic

  value is z 5 (103 2 100)2.0 5 1.5 . That is, the observed value of is 1.5 standard x

  deviations (of ) larger than what we expect it to be when H 0 is true. The statistic Z

  is a natural measure of the distance between , the estimator of m, and its expected X

  value when H 0 is true. If this distance is too great in a direction consistent with H a ,

  the null hypothesis should be rejected.

  Suppose first that the alternative hypothesis has the form H a :m.m 0 . Then an value less than m 0 certainly does not provide support for H a . Such an corresponds x

  8.2 Tests About a Population Mean

  to a negative value of z (since x2m 0 is negative and the divisor s 1n is positive). Similarly, an value that exceeds m 0 by only a small amount (corresponding to z,

  which is positive but small) does not suggest that H 0 should be rejected in favor of

  H a . The rejection of H 0 is appropriate only when considerably exceeds m x 0 —that is,

  when the z value is positive and large. In summary, the appropriate rejection region,

  based on the test statistic Z rather than , has the form X zc .

  As discussed in Section 8.1, the cutoff value c should be chosen to control the probability of a type I error at the desired level a. This is easily accomplished

  because the distribution of the test statistic Z when H 0 is true is the standard normal distribution (that’s why m 0 was subtracted in standardizing). The required cutoff c is

  the z critical value that captures upper-tail area a under the z curve. As an example, let c 5 1.645 , the value that captures tail area .05 (z .05 5 1.645) . Then,

  a 5 P(type I error) 5 P(H 0 is rejected when H 0 is true)

  5 P(Z 1.645 when Z , N(0,1)) 5 1 2 (1.645) 5 .05 More generally, the rejection region zz a has type I error probability a. The test

  procedure is upper-tailed because the rejection region consists only of large values of the test statistic.

  Analogous reasoning for the alternative hypothesis H a :m,m 0 suggests a

  rejection region of the form zc , where c is a suitably chosen negative number ( x

  is far below m 0 if and only if z is quite negative). Because Z has a standard normal distribution when H 0 is true, taking c 5 2z a yields P(type I error) 5 a . This is a

  lower-tailed test. For example, z .10 5 1.28 implies that the rejection region z 21.28 specifies a test with significance level .10.

  Finally, when the alternative hypothesis is H a :m2m 0 ,H 0 should be rejected if is too far to either side of m x 0. This is equivalent to rejecting H 0 either if zc or

  if z 2c . Suppose we desire a 5 .05 . Then,

  .05 5 P(Z c or Z 2c when Z has a standard normal distribution)

  5 (2c) 1 1 2 (c) 5 2[1 2 (c)]

  Thus c is such that 12 (c) , the area under the z curve to the right of c, is .025 (and

  not .05!). From Section 4.3 or Appendix Table A.3, c 5 1.96 , and the rejection region is z 1.96 or z 21.96 . For any a, the two-tailed rejection region zz a2

  or z 2z a2 has type I error probability a (since area a2 is captured under each of

  the two tails of the z curve). Again, the key reason for using the standardized test sta-

  tistic Z is that because Z has a known distribution when H 0 is true (standard normal),

  a rejection region with desired type I error probability is easily obtained by using an appropriate critical value.

  The test procedure for case I is summarized in the accompanying box, and the corresponding rejection regions are illustrated in Figure 8.2.

  Null hypothesis: H 0 :m5m 0

  x2m

  Test statistic value: z5

  s 1n

  Alternative Hypothesis

  Rejection Region for Level a Test

  H a :m.m 0 zz a (upper-tailed test)

  H a :m,m 0 z 2z a (lower-tailed test)

  H a :m2m 0 either or (two-tailed zz a2 z 2z a2 test)

  CHAPTER 8 Tests of Hypotheses Based on a Single Sample

  z curve (probability distribution of test statistic Z when H 0 is true)

  Total shaded area P(type I error)

  Shaded area

  P(type I error)

  Shaded area

  Shaded 2 area

  Rejection region: z

  z Rejection region: either

  Rejection region: z

  Figure 8.2 Rejection regions for z tests: (a) upper-tailed test; (b) lower-tailed test; (c) two-tailed test

  Use of the following sequence of steps is recommended when testing hypotheses about a parameter.

  1. Identify the parameter of interest and describe it in the context of the problem sit- uation.

  2. Determine the null value and state the null hypothesis.

  3. State the appropriate alternative hypothesis.

  4. Give the formula for the computed value of the test statistic (substituting the null value and the known values of any other parameters, but not those of any sample- based quantities).

  5. State the rejection region for the selected significance level a.

  6. Compute any necessary sample quantities, substitute into the formula for the test statistic value, and compute that value.

  7. Decide whether H 0 should be rejected, and state this conclusion in the problem

  context. The formulation of hypotheses (Steps 2 and 3) should be done before examining the data.

  Example 8.6

  A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°. A sample of n59 sys- tems, when tested, yields a sample average activation temperature of 131.08°F. If the distribution of activation times is normal with standard deviation 1.5°F, does the data contradict the manufacturer’s claim at significance level a 5 .01 ?

  1. Parameter of interest: m 5 true average activation temperature.

  2. Null hypothesis: H 0 : m 5 130 (null value 5 m 0 5 130) .

  3. Alternative hypothesis: H a : m 2 130 (a departure from the claimed value in

  either direction is of concern).

  4. Test statistic value:

  x2m

  0 5 x 2 130

  z5

  s 1n

  1.5 1n

  8.2 Tests About a Population Mean

  5. Rejection region: The form of H a implies use of a two-tailed test with rejection

  region either zz .005 or z 2z .005 . From Section 4.3 or Appendix Table A.3,

  z .005 5 2.58 , so we reject H 0 if either z 2.58 or z 22.58 .

  6. Substituting and n59 x 5 131.08 ,

  That is, the observed sample mean is a bit more than 2 standard deviations

  above what would have been expected were H 0 true.

  7. The computed value z 5 2.16 does not fall in the rejection region (22.58 ,

  2.16 , 2.58) , so H 0 cannot be rejected at significance level .01. The data does not

  give strong support to the claim that the true average differs from the design value of 130.

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  B and Sample Size Determination The z tests for case I are among the few in statistics for which there are simple formulas available for b, the probability of

  a type II error. Consider first the upper-tailed test with rejection region

  zz . This is equivalent to xm 1z a 0 a s 1n , so H will not be rejected if x,m 1z 0 0 a s 1n . Now let mr denote a particular value of m that exceeds the null value m 0 . Then,

  b(mr) 5 P(H 0 is not rejected when m 5 mr)

  5 P(X , m 1z 0 a s 1n when m 5 mr)

  ,z 1 0 a 2 mr when m 5 mr

  s 1n As m increases, m 0 m becomes more negative, so b(m ) will be small when m

  greatly exceeds m 0 (because the value at which is evaluated will then be quite neg-

  ative). Error probabilities for the lower-tailed and two-tailed tests are derived in an analogous manner.

  If s is large, the probability of a type II error can be large at an alternative value m that is of particular concern to an investigator. Suppose we fix a and also specify b for such an alternative value. In the sprinkler example, company officials

  might view mr 5 132 as a very substantial departure from H 0 : m 5 130 and there-

  fore wish b(132) 5 .10 in addition to a 5 .01 . More generally, consider the two restrictions P(type I error) 5 a and b(mr) 5 b for specified a, mr,

  and b. Then for an upper-tailed test, the sample size n should be chosen to satisfy

  m 0 2 mr az

  a 1 s 1n b5b

  This implies that

  z critical value that

  captures lower-tail area b

  s 1n

  It is easy to solve this equation for the desired n. A parallel argument yields the nec- essary sample size for lower- and two-tailed tests as summarized in the next box.

  CHAPTER 8 Tests of Hypotheses Based on a Single Sample

  Alternative Hypothesis Type II Error Probability b (mr) for a Level a Test m 2 mr

  where (z) 5 the standard normal cdf.

  The sample size n for which a level a test also has b(mr) 5 b at the alternative value m is

  s(z a 1z b ) 2 c for a one-tailed

  m 0 2 mr d (upper or lower) test

  n5 μ s(z a2 1z b ) 2 c for a two-tailed test d

  m 0 2 mr

  (an approximate solution)

  Example 8.7 Let m denote the true average tread life of a certain type of tire. Consider testing

  H 0 : m 5 30,000 versus H a : m . 30,000 based on a sample of size n 5 16 from a

  normal population distribution with s 5 1500 . A test with a 5 .01 requires

  z a 5z .01 5 2.33 . The probability of making a type II error when m 5 31,000 is

  b(31,000) 5 a2.33 1

  Since z .1 5 1.28 , the requirement that the level .01 test also have b(31,000) 5 .1 necessitates

  30,000 2 31,000 d

  n5 c 5 (25.42) 2 5 29.32

  The sample size must be an integer, so n 5 30 tires should be used.

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