2. Non parametric test - Chi square test for the following data
29,24,22,19,21,18,19,20,23,18,20,23, Significance Level `alpha=0.05` and One-tailed test

Solution:
Step-1:State the hypothesis
`H_0`: observed data and expected data are independent.

`H_1`: observed data and expected data are not independent.


Step-2:Observed Frequencies `29,24,22,19,21,18,19,20,23,18,20,23`

sum`=256` and `n=12`

Step-3: Calculate Expected Frequency value
`256/12=21.3333`

Step-4: Chi-square

Id	Observed	Expected	`O_i-E_i`	`(O_i-E_i)^2`	`(O_i-E_i)^2/E_i`
1	29	21.3333	7.6667	58.7778	2.7552
2	24	21.3333	2.6667	7.1111	0.3333
3	22	21.3333	0.6667	0.4444	0.0208
4	19	21.3333	-2.3333	5.4444	0.2552
5	21	21.3333	-0.3333	0.1111	0.0052
6	18	21.3333	-3.3333	11.1111	0.5208
7	19	21.3333	-2.3333	5.4444	0.2552
8	20	21.3333	-1.3333	1.7778	0.0833
9	23	21.3333	1.6667	2.7778	0.1302
10	18	21.3333	-3.3333	11.1111	0.5208
11	20	21.3333	-1.3333	1.7778	0.0833
12	23	21.3333	1.6667	2.7778	0.1302
	256				5.0938

`chi^2=sum (O_i-E_i)^2/(E_i)=5.0938`

Step-5: Compute the degrees of freedom (df).
`df=12-1=11`

Step-6: for 11 df, `p(chi^2>=5.0938)=0.9265`

Since the p-value(0.9265) > `alpha(0.05)` (one-tailed test), we can't reject the null hypothesis `H_0`.

---