Question

A study looked at number of cavity children per 100 in 18 North American cites before and after public water fluoridation projects. The following table lists the data.

City	Before	After
1	49.2	18.2
2	30	21.9
3	16	5.2
4	47.8	20.4
5	3.4	2.8
6	16.8	21
7	10.7	11.3
8	5.7	6.1
9	23	25
10	17	13
11	79	76
12	66	59
13	46.8	25.6
14	84.9	50.4
15	65.2	41.2
16	52	21
17	50	32
18	12	20

a. Please test whether there is a significant change in number of cavity children per 100 after and before public water fluoridation. Make sure you include all the five steps in the hypothesis testing questions.

b. What is the 95% confidence interval for the change?

Answer 1

Number	Before	After	Difference	( di - d̅ )2
	49.2	18.2	31	383.724568
	30	21.9	8.1	10.9634568
	16	5.2	10.8	0.37345679
	47.8	20.4	27.4	255.644568
	3.4	2.8	0.6	116.880123
	16.8	21	-4.2	243.70679
	10.7	11.3	-0.6	144.26679
	5.7	6.1	-0.4	139.502346
	23	25	-2	179.857901
	17	13	4	54.9245679
	79	76	3	70.7467901
	66	59	7	19.4579012
	46.8	25.6	21.2	95.8223457
	84.9	50.4	34.5	533.09679
	65.2	41.2	24	158.480123
	52	21	31	383.724568
	50	32	18	43.4134568
	12	20	-8	376.791235
Total	675.5	470.1	205.4	3211.37778

Part a)

To Test ;-

H0 :-

H1 :-

Test Statistic :-
S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √( 3211.37778 / 17) = 13.7443
d̅ = Σdi/n = 205.4 / 18 = 11.41111
t = d̅ / ( S(d) / √(n) )
t = 11.4111 / ( 13.7443 / √(18) )
t = 3.5224

Test Criteria :-
Reject null hypothesis if | t | > t(α/2)
Critical value t(α/2) = t( 0.05 /2 ) = 2.10982
| t | > t(α/2) = 3.52243 > 2.10982
Result :- Reject null hypothesis

There is a significant change in number of cavity children per 100 after and before public water fluoridation.

Part b)

Confidence Interval :-
d̅ ± t(α/2, n-1) Sd / √(n)
t(α/2) = t(0.05 /2) = 2.10982
11.4111 ± t(0.05/2) * 13.7443/√(18)
Lower Limit = 11.4111 - t(0.05/2) 13.7443/√(18)
Lower Limit = 4.57632
Upper Limit = 11.4111 + t(0.05/2}) 13.7443/√(18)
Upper Limit = 18.24591
95% Confidence interval is ( 4.5763 , 18.24591 )

SInce the value 0 does not lie in the interval ( 4.5763 , 18.24591 ), hence we reject null hypothesis.