Question

13. To study the relationship between nutrition and children's intellectual development, a research firm questioned 1436 children, have the following data. We want to know whether children's intellectual development and nutrition related or not? The significance level a 0.05 智 商 Intelligence quotient Nutrition0 80~90 90-99>100 营养良好 367 Pi>-a 行和 342 266 8.99 8.95 8.00 8.95 a 004 0 103 0211 13 329 8.10 0.016 045 271 0.000 3 34 54 1083 604 GOOD 020 040 4.61 13 82 115 185 0352 a 297 429 0 584 2366 25 712 3357 7.7 营养不良 56 40 16 27 11.34 20 16 0711 1064 949 u47132 8.47 BAD 1 $10 4 351 P 2 10 65 12 49 S 1412246 6.35 0554 0 752 1.07 33 15.09 2052 2 204 535 0 874 1134 1635 列和 1.23 1.564 2167 2833 14 07 12 114 24 37 1551 20 09 26 13 14 68 1692 19 21 47 27 1202 1.646 2032 088 2532 a 558 3059 340 4 65 934 2731 3.490 734 13 36 1325 4.168 is 99 18 31 2116 23 21 (6 marks) You want to coe if throo difforon aolf elube viold difforont dictancoe Vou 14 14. You want to see if three different golf clubs yield different distances. You randomly select five measurements from trials on an automated driving machine for each club. At the 0.05 significance level, is there a difference in mean distance? (Reference Fo.os (2,12)= 3.89 ) Club 2 Club 3 Club 1 200 234 254 222 218 263 197 241 235 206 227 237 204 251 216 6. (6marks)

Answer 1

13. Null Hypothesis: Nutrition and IQ are independent

Alternative Hypothesis: Nutrition and IQ are not independent

Test is chi square test of independence based on the statistic T=
o - E)2/E
. Under the null T~
X
and we reject the null at 5% if observed T>
X3.05 (3,.05

We calculate the values in the following table

   <80 80-90 90-99 >=100 Total

Good 367 342 266 329 1304
384.12 346.89 259.71 313.29
0.7628 0.0688 0.1523 0.7881

Bad 56 40 20 16 132
38.88 35.11 26.29 31.71
7.5352 0.6798 1.5048 7.7855

Total 423 382 286 345 1436

In each cell, the fist entry indicates observed count (O), the second entry indicates expected count (E) and the third entry indicates (O-E)²/E .

Then we get Pearson Chi-Square = 19.277, DF = 3 and =7.81 and thus reject the null.

X3.05 (3,.05

Hence Nutrition and IQ are significantly associated.

14.  Null hypothesis : All golf clubs yield same mean distance
Alternative hypothesis At least one golf clubs yield different mean distance
Then the appropriate method is one way ANOVA. We assume equal variance for the analysis.

Then our test statistic is F=[SS(Between)/df1]/[SSE/df2], where df1 and df2 are the respective degrees of freedom. Now df1=3-1=2, df2=12 and under the null F~F(2,12). Thus we reject the null if observed F>F_.05,2,12

We calculate the sum squares in the following table.

Analysis of Variance Table

Source DF    SS   MS F-Value  
Between 2 4716 2358.20 25.28   
Error 12 1120 93.30
Total 14 5836
Now F_.05,2,12= 3.89 and hence we reject the null.

Thus at least one golf clubs yielded significantly  different mean distance.