Question

+36-3310 Detailed Sample Table Total Rain 35 LA SF SD SJ 4 6 22 3 5.2 19.15.2 0.462 0.137 0.45 0.904 21 17 20 No Rain 10.4 17.0 65.0 17.6 0.134 0.04 0.13 0.262 Total 25 85 23 121 156 OvdExpected Contribution to 4. Perform a chi-square test of association to determine if the proportion of rainy days is different among these four cities a. Write the hypotheses.

Answer 1

(a)

H0: Null Hypothesis: The proportion of rainy days is not different among these four cities

HA:Alternative Hypothesis: The proportion of rainy days is different among these four cities (Claim)

Obsrved Frequencies:

	LA	SF	SD	SJ	Total
Rain	4	6	22	3	35
No rain	21	17	63	20	121
Total	25	23	85	23	156

Expected Frequencies:

	LA	SF	SD	SJ	Total
Rain	25X35/156=5.609	23X35/156=5.16	85X35/156=19.071	23X35/156=5.16	35
No rain	25X121/156=19.391	23X121/156=17.84	85X121/156=65.929	23X121/156=17.84	121
Total	25	23	85	23	156

Test Statistic ($\chi ^{2}$) is calculated as follows:

Observed (O)	Expected (E)	(O - E)2/E
4	5.609	0.462
6	5.16	0.137
22	19.071	0.45
3	5.16	0.904
21	19.391	0.134
17	17.84	0.04
63	65.929	0.13
20	17.84	0.262
Total = $\chi ^{2}$ =		2.517

df = (2 - 1)X (4 - 1) = 3

From Table, critical value of $\chi ^{2}$ = 7.815

Since calculated value of $\chi ^{2}$ = 2.517 is less than critical value of $\chi ^{2}$ = 7.815, the difference is not significant. Fail to reject null hypothesis.

Conclusion:
The data do not support the claim that the proportion of rainy days is different among these four cities.