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TOPIC:- Lilliefors test for normality

About Hubert Whitman Lilliefors:

Hubert Whitman Lilliefors (born on February 23-1928, and died in 2008, at Bethesda, Maryland). He was an American statistician, noted for his introduction of the Lilliefors test. He was a professor of statistics at George Washington University for 39 years, and also obtained his PhD there in 1964 under the supervision of Solomon Kullback.

Introduction: Lilliefors test is derived in 1967 by Hubert

Lilliefors. It is used to test the null hypothesis that data

comes from normal distribution. It is based on the Kolmogorov-Smirnov test of

normality. Assumption: The sample is a random sample.

Procedure Of Testing Normality The null and alternative hypotheses are: Null hypothesis:

Data is normally distributed Alternative hypothesis:

Data is not normally distributed Level of significance: =0.05 etc. Test Statistics: d1 = |F(z) – S(z)| and d2 = |F(z) – S(zi-1)|. We then let d = larger of d1 and d2.

Procedure for the test statistic Sort the data from lowest to highest. Compute the sample mean and the sample

standard deviation and for each value of X, compute Z .

For each value, compute F(z) which is the left-tail probability of Z.

For each value, compute S(z) and S(zi-1).

For each value, we need the difference between F(z) and both S(z) and S(zi-1). Let d1 = |F(z) – S(z)| and d2 = |F(z) – S(zi-1)|. We then let d = larger of d1 and d2.

The test statistic is the maximum d. We reject the null hypothesis if the test statistics is greater than the critical value.

If N≤50 then for finding critical values we use lilliefors table. For N>50 the critical value can be found by using fN=0.83+N/√N

Critical region:Dcal≥ Dtab, then reject H0.

Conclusion:

Example: A store wanted to determine if the average sale is significantly more than $250. A

random sample had the following results: Test at a 5% level of significance.

Solution: NULL AD ALTERNATIVE hypothesis: H0: Data is normally distributed. H1: Data is not normally distributed.

2. Significance level: α=0.05

3. Test Statistics: D=max of d1 AND d2 where d1 = |F(z) – S(z)| and d2 = |F(z) – S(zi-1)|

4. Calculations: Arrange the values in the ascending order.

find Z= x-u/σ i-e x-315.5/118.31 find F(z) i-e F(-1.19)=P(z<-1.19)=0.112 find S(z) i-e (commulative frequency)/n find S(zi-1) i-e preceding of S(z).

X F C.F

z P(z)

F(z) S(zi) S(zi-1)

D1=│F(z)-S(z)│

D2=│F(z)-S(zi-1)│

175 1 1 -1.19

0.3880 0.112 1/12=0.083 0 0.112-0.083=0.029 0.122-0=0.122

189 1 2 -1.07

0.3577 0.1423 2/12=0.166 1/12=0.083 0.1423-0.016=0.1257

0.1423-0.083=0.0653

200 1 3 -0.98

0.3365 0.1635 3/12=0.25 2/12=0.166 0.1635-0.25=0.086 0.0025

212 1 4 -0.87

0.3078 0.1922 4/12=0.333 3/12=0.25 0.1922-0.333=0.140 0.578

262 1 5 -0.45

0.1736 0.3264 5/12=0.416 4/12=0.333 0.3264-0.416=0.089 0.0066

298 1 6 -0.15

0.0596 0.4404 6/12=0.5 5/12=0.416 0.4404-0.5=0.0596 0.0244

325 1 7 0.08 0.0319 0.5319 7/12=0.583 6/12=0.5 0.0511 0.0319

355 1 8 0.33 0.1293 0.3707 8/12=0.667 7/12=0.583 0.2963 0.2123

360 1 9 0.38 0.1480 0.352 9/12=0.751 8/12=0.667 0.399 0.315

410 1 10 0.79 0.2852 0.2148 10/12=0.83 9/12=0.751 0.6125 0.5362

428 1 11 0.95 0.3289 0.1711 11/12=0.91 10/12=0.83 0.7389 0.6589

572 1 12 2.17 0.4850 0.015 12/12= 1 11/12=0.91 0.985 0.895

5. Critical value:

If Dcal ≥ Table values ;

then we reject our null hypothesis. DCAL =0.985 , Table value = 0.2426

6. Conclusion: Since the calculated value of D is greater than the table value therefore we reject our null hypothesis and conclude that data is not normally distributed.

Example# 2 An Independent random samples from 6 assistant professors. They asked about

their time used outside the class in the last week.  Data is shown below (in hours) 7, 12, 11, 15, 9, 14.

Solution:1. Hypothesis would be: H0: the data are normally distributed H1: the data are not normally distributed

2. Significance level: α=0.05

3. Test Statistics: D=max of d1 AND d2 where d1 = |F(z) – S(z)| and d2 = |F(z) – S(zi-1)|

4. Calculations: Arrange the values in the ascending order. find Z= x-u/σ i-e x-11.3333/3.0111 find F(z) i-e F(-1.44)=P(z<-1.44)=0.0764 find S(z) i-e (commulative frequency)/n find S(zi-1) i-e preceding of S(z).

X Z P(z) F(z) S(zi) S(zi-1)

D1=│F(z)-S(z)│

D2=│F(z)-S(zi-1)│

7 -1.44 0.4251 0.0764 0.1666 0 0.0764-0.1666= 0.0902

0.0764-0=0.0764

9 -0.77 0.2794 0.2206 0.3333 0.1666 0.1127 0.2206-0.1666= 0.054

11 -0.11 0.0438 0.4562 0.5000 0.3333 0.0438 0.1229

12 0.22 0.0871 0.5871 0.6667 0.5000 0.0796 0.0871

14 0.89 0.3133 0.8133 0.8333 0.6667 0.0200 0.146615 1.22 0.3888 0.8888 1 0.8333 0.1112 0.0555

5. Critical value:

If Dcal ≥Table values ;

then we reject our null hypothesis. DCAL =0.1127 , Table value = 0.2426

6. Conclusion: Since the calculated value of D is less than the table value therefore we do not reject our null hypothesis and conclude that data is normally distributed.

Critical values :

N α = .20 α = .15 α = .10 α = .05 α = .01 4 .3027 .3216 .3456 .3754 .4129 5 .2893 .3027 .3188 .3427 .3959 6 .2694 .2816 .2982 .3245 .3728 7 .2521 .2641 .2802 .3041 .3504 8 .2387 .2502 .2649 .2875 .3331 9 .2273 .2382 .2522 .2744 .3162 10 .2171 .2273 .2410 .2616 .3037 11 .2080 .2179 .2306 .2506 .2905 12 .2004 .2101 .2228 .2426 .2812 13 .1932 .2025 .2147 .2337 .2714 14 .1869 .1959 .2077 .2257 .2627 15 .1811 .1899 .2016 .2196 .2545 16 .1758 .1843 .1956 .2128 .2477 17 .1711 .1794 .1902 .2071 .2408 18 .1666 .1747 .1852 .2018 .2345 19 .1624 .1700 .1803 .1965 .2285 20 .1589 .1666 .1764 .1920 .2226

N α = .20 α = .15 α = .10 α = .05 α = .01 21 .1553 .1629 .1726 .1881 .2190 22 .1517 .1592 .1690 .1840 .2141 23 .1484 .1555 .1650 .1798 .2090 24 .1458 .1527 .1619 .1766 .2053 25 .1429 .1498 .1589 .1726 .2010 26 .1406 .1472 .1562 .1699 .1985 27 .1381 .1448 .1533 .1665 .1941 28 .1358 .1423 .1509 .1641 .1911

29 .1334 .1398 .1483 .1614 .188630 .1315 .1378 .1460 .1590 .184831 .1291 .1353 .1432 .1559 .182032 .1274 .1336 .1415 .1542 .179833 .1254 .1314 .1392 .1518 .177034 .1236 .1295 .1373 .1497 .174735 .1220 .1278 .1356 .1478 .172036 .1203 .1260 .1336 .1454 .169537 .1188 .1245 .1320 .1436 .167738 .1174 .1230 .1303 .1421 .165339 .1159 .1214 .1288 .1402 .163440 .1147 .1204 .1275 .1386 .1616

N α = .20 α = .15 α = .10 α = .05 α = .0141 .1131 .1186 .1258 .1373 .159942 .1119 .1172 .1244 .1353 .157343 .1106 .1159 .1228 .1339 .155644 .1095 .1148 .1216 .1322 .154245 .1083 .1134 .1204 .1309 .152546 .1071 .1123 .1189 .1293 .151247 .1062 .1113 .1180 .1282 .149948 .1047 .1098 .1165 .1269 .147649 .1040 .1089 .1153 .1256 .146350 .1030 .1079 .1142 .1246 .1457 50 0.741/fN 0.775/fN 0.819/fN 0.895/fN 1.035/fN

Where f(n)=

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