Lesson 15 - 3

1. Lesson 15 - 3 Inferences about Measures of Central Tendency

2. Objectives • Conduct a one-sample sign test

3. Vocabulary • One-sample sign test -- requires data converted to plus and minus signs to test a claim regarding the median • Change all data to + (above H0 value) or – (below H0 value) • Any values = to H0 value change to 0

4. Sign Test • Like the runs test, the test statistic used depends on the sample size • In the small sample case, where the number of observations n is 25 or less, we use the number of +’s and the number of –’s directly • In the large sample case, where the number of observations n is more than 25, we use a normal approximation • Conditions: An independent, random sample

5. (k + ½ ) – ½ n z0= --------------------- ½ √n Signs Test for Central Tendencies Test Statistic Small-Sample Case: If n≤ 25, the test statistic in the signs test is k, defined as below. Large-Sample Case: If n > 25 the test statistic is where k = is defined from above and n = number of + and – signs (zeros excluded) Critical Values for a Runs Test for Randomness Small-Sample Case: Use Table VII to find critical value for a one-sample sign test Large-Sample Case: Use Table IV, standard normal table (one-tailed -zα; two-tailed -zα/2).

6. K K Ksm Ksm Rejection Region For a left-tailed test: the greater the number of values above the tested median, the more evidence we have to reject it K = number of + signs (above) For a right-tailed test: the greater the number of values below the tested median, the more evidence we have to reject it K = number of - signs (below) For a two-tailed test: the greater the difference between numbers above and below, the more evidence we have to reject it K = smaller of number of + signs or - signs

7. Hypothesis Tests for Central Tendency Using Signs Test (k + ½ ) – ½ n z0= --------------------- ½ √n Step 0: Convert all data to +, - or 0 (based on H0) Step 1 Hypotheses:Left-tailedTwo-TailedRight-Tailed H0: Median = M0 H0: Median = M0 H0: Median = M0 H1: Median < M0 H1: Median ≠ M0 H1: Median > M0 Step 2 Level of Significance: (level of significance determines critical value) Determine a level of significance, based on the seriousness of making a Type I error Small-sample case: Use Table X. Large-sample case: Use Table IV, standard normal (one-tailed -zα; two-tailed -zα/2). Step 3 Compute Test Statistic: Step 4 Critical Value Comparison: Reject H0 if Small-Sample Case: k ≤ critical value Large-Sample Case: z0 < -zα/2 (two tailed) or z0 < -zα (one-tailed) Step 5 Conclusion: Reject or Fail to Reject Small-Sample: k Large-Sample:

8. Small Number Example A recent article in the school newspaper reported that the typical credit-card debt of a student is $500. Professor McCraith claims that the median credit-card debt of students at Joliet Junior College is different from $500. To test this claim, he obtains a random sample of 20 students enrolled at the college and asks them to disclose their credit-card debt. $6000 $0 $200 $0 $400 $1060 $0 $1200 $200 $250 $250 $580 $1000 $0 $0 $200 $400 $800 $700 $1000 $6000 $0 $200 $0 $400 $1060 $0 $1200 $200 $250 $250 $580 $1000 $0 $0 $200 $400 $800 $700 $1000 + = 8 - = 12 k = 8 n = 20 CV = 5 (from table X) Two-Tailed Test: (Med ≠ 500) so k = number of smaller of the signs We reject H0 if k ≤ critical value (out in the tail). Since 8 > 5, we do not reject H0 and therefore not enough evidence to support median credit card debt different from $500

9. HyCCI for Example 1 • Hyp: H0: Median credit card debt = $500 Ha: Median credit card debt ≠ $500 • Conditions: Assume an independent random sample • Calculations: Two-Tailed Test: (Med ≠ 500) so k = number of smaller of the signs. K = 8 and CV = 5 (from table) • Interpretation: We reject H0 if k ≤ critical value (out in the tail). Since 8 > 5, we do not reject H0 and therefore not enough evidence to support median credit card debt different from $500

10. (k + ½ ) – ½ n z0= --------------------- ½ √n Large Number Example A sports reporter claims that the median weight of offensive linemen in the NFL is greater than 300 pounds. He obtains a random sample of 30 offensive linemen and obtains the data shown in Table 4. Test the reporter’s claim at the α = 0.1 level of significance. Right-Tailed Test: H0: Med = 300 Ha: Med > 300 so k = number of - signs 285 310 300 300 320 308 310 293 329 293 326 310 297 301 315 332 305 340 242 310 312 329 320 300 311 286 309 292 287 305 285 310 300 300 320 308 310 293 329 293 326 310 297 301 315 332 305 340 242 310 312 329 320 300 311 286 309 292 287 305 + = 19 - = 8 0 = 3 n = 30-3 = 27 k = 8 z0 = -10/27 = -1.92 P(z0 < -1.92) = 0.027 Since 0.027 < 0.1 (p-value < α), we would reject the H0 (median = 300) in favor of the alternative, median > 300

11. (k + ½ ) – ½ n z0= --------------------- ½ √n HyCCI for Example 2 • Hyp: H0: Median weight of offensive lineman = 300 lbs Ha: Median weight of offensive lineman > 300 lbs • Conditions: Assume an independent random sample • Calculations: One-Tailed Test: (Med > 300) so k = number of negative signs. • Interpretation: Since 0.027 < 0.1 (p-value < α), we would reject the H0 (median = 300) in favor of the alternative, median weight > 300 lbs for an offensive lineman + = 19 - = 8 0 = 3 n = 30-3 = 27 k = 8 z0 = -10/27 = -1.92 P(z0 < -1.92) = 0.027 (ncdf)

12. Summary and Homework • Summary • The sign test is a nonparametric test for the median, a measure of central tendency • This test counts the number of observations higher and lower than the assumed value of the median • The critical values for small samples are given in tables • The critical values for large samples can be approximated by a calculation with the normal distribution • Homework • problems 5, 6, 10, 12 from the CD

13. Homework Help • 5 M = 8 / M < 8 n = 21 k = 8 (+) CV = 5 • 6 M = 8 / M > 50 n = 19 k = 7 (-) CV = 5 • 10 M = 15 / M < 15 n = 15 k = 7 (+) CV = 3 • 12 M = 26.7 / M < 26.7 n = 20 k = 8 (+) CV = 5