15 Propellant Shear Strength Sign Test

EXAMPLE 9-15 Propellant Shear Strength Sign Test

  Montgomery, Peck, and Vining (2006) report on a study in

  1. Parameter of Interest: The parameter of interest is

  which a rocket motor is formed by binding an igniter propel-

  the median of the distribution of propellant shear

  lant and a sustainer propellant together inside a metal housing.

  strength.

  The shear strength of the bond between the two propellant

  2. Null hypothesis: H 0 : ␮苲 ⫽ 2000 psi

  types is an important characteristic. The results of testing 20

  3. Alternative hypothesis: H 1 : ␮苲 ⫽ 2000 psi

  randomly selected motors are shown in Table 9-5. We would

  4. Test statistic: The test statistic is the observed num-

  like to test the hypothesis that the median shear strength is

  ber of plus differences in Table 9-5, or r ⫹ ⫽ 14.

  2000 psi, using ␣ ⫽ 0.05.

  5. Reject H 0 if: We will reject H 0 if the P-value corre-

  sponding to r testing procedure: ⫹ ⫽ 14 is less than or equal to

  This problem can be solved using the eight-step hypothesis-

  ␣ ⫽ 0.05.

  Table 9-5 Propellant Shear Strength Data

  Observation

  Shear Strength

  x i ⫺ 2000

  ⫺321.85

  ⫺291.70

  ⫺

  ⫺215.30

  ⫺220.20

  ⫺234.70

  ⫺246.30

  ⫺

  JWCL232_c09_283-350.qxd 11410 3:08 PM Page 340

  CHAPTER 9 TESTS OF HYPOTHESES FOR A SINGLE SAMPLE

  6. Computations: Since r 14 is greater than

  7. Conclusions: Since P 0.1153 is not less than 0.05,

  n 兾2 20兾2 10, we calculate the P-value from

  we cannot reject the null hypothesis that the median shear strength is 2000 psi. Another way to say this is that the ob-

  served number of plus signs r 2P aR 14 when p

  14 was not large or small

  P

  2 b enough to indicate that median shear strength is different

  20 20 from 2000 psi at the

  0.05 level of significance.

  2 a

  b 10.52 r

  a 10.52 20 r

  r 14 r 0.1153

  It is also possible to construct a table of critical values for the sign test. This table is shown as Appendix Table VIII. The use of this table for the two-sided alternative hypothesis in Equation

  9-54 is simple. As before, let R denote the number of the differences ( X i

  0 ) that are positive

  and let R denote the number of these differences that are negative. Let R min

  (R ,R ). Appendix Table VIII presents critical values r for the sign test that ensure that P (type

  I error) P (reject H 0 when H 0 is true)

  for 0.01, 0.05 and 0.10. If the

  observed value of the test statistic r r , the null hypothesis H 0 :

  0 should be rejected.

  To illustrate how this table is used, refer to the data in Table 9-5 that were used in Example

  9-15. Now r 14 and r 6; therefore, r min (14, 6) 6. From Appendix Table VIII

  with n

  20 and 0.05, we find that r 0.05 5. Since r 6 is not less than or equal to the critical value r 0.05 5, we cannot reject the null hypothesis that the median shear strength is

  2000 psi.

  We can also use Appendix Table VIII for the sign test when a one-sided alternative

  hypothesis is appropriate. If the alternative is H 1 :

  0 , reject H 0 :

  0 if r r ;

  if the alternative is H 1 :

  0 , reject H 0 :

  0 if r r . The level of significance of a

  one-sided test is one-half the value for a two-sided test. Appendix Table VIII shows the one- sided significance levels in the column headings immediately below the two-sided levels.

  Finally, note that when a test statistic has a discrete distribution such as R does in the sign

  test, it may be impossible to choose a critical value r that has a level of significance exactly equal to . The approach used in Appendix Table VIII is to choose r to yield an that is as

  close to the advertised significance level as possible.