Question

This is a 5 number summary for the number of children born.

min	1st Qart	Median	Mean	3rd Quar	Max
0.00	1.00	3.00	3.261	4.00	16.00

Explain why it would be inappropriate to conduct a chi square test for contraceptive method and number of children ever born. Hint look at the output for the following R command (table(dat3$contrMethod, dat3$noKids))

	0	1	2	3	4	5	6	7	8	9	10	11	12	13	16
1	95	143	114	70	57	44	35	18	29	5	9	6	4	0	0
2	0	46	56	70	62	36	27	19	9	3	2	2	0	1	0
3	2	87	106	119	78	55	30	12	9	8	0	3	0	1	1

Answer 1

• For use chi-square test we need sample data are displayed in a contingency table, the expected frequency count for each cell of the table is at least 5.
• In given data 7 cells with expected counts less than 1. Chi-Square approximation probably invalid. 16 cells with expected counts less than 5.

• observed counts 95 143 114 70 57 44 35 18 29 5 Expected counts 41.42 117.86 117.86 110.6 84.12 57.65 39.29 20.92 20.07 6.83 observed counts 0 46 56 70 62 36 27 19 9 3 Expected counts 21.93 62.4 62.4 58.55 44.54 30.52 20.8 11.08 10.63 3.62 observed counts 2 87 106 119 78 55 30 12 9 8 Expected counts 33.65 95.75 95.75 89.85 68.34 46.83 31.92 17 16.3 5.55

• 
  This condition is not satisfied by given data therefore it would be inappropriate to conduct a chi square test for contraceptive method and number of children ever born.

Rcode

(table2)=c("0","1","2","3","4","5","6","7","8","9","10","11","12","13","16")
> rownames(table2)=c("0","1","2")
> table2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 16
0 95 143 114 70 57 44 35 18 29 5 9 6 4 0 0
1 0 46 56 70 62 36 27 19 9 3 2 2 0 1 0
2 2 87 106 119 78 55 30 12 9 8 0 3 0 1 1
> chisq.test(table2)

Pearson's Chi-squared test

data: table2
X-squared = 219.09, df = 28, p-value < 2.2e-16

Warning message:
In chisq.test(table2) : Chi-squared approximation may be incorrect