Question

5) Polychlorinated biphenyls (PCBs) are organic chlorine compounds that were widely used as dielectrics and coolants in electrical systems in the past. They were found to be a major environmental contaminant in the 1960s. In a study, the mean PCB content at each of thirteen sites was reported for the years 1982 and 1996 (from “The ratio of DDE to PCB concentrations in Great Lakes herring gull eggs and its use in interpreting contaminants data”, Journal of Great Lakes Research 24 (1): 12-31, 1998). The data are below.

Site:	1	2	3	4	5	6	7	8	9	10	11	12	13
1982	61.48	64.47	45.50	59.70	58.81	75.96	71.57	38.06	30.51	39.70	29.78	66.89	63.93
1996	13.99	18.26	11.28	10.02	21.00	17.36	28.20	7.30	12.80	9.41	12.63	16.83	22.74

a) Which test would be more appropriate in this case: a t-test for the difference between two population means, or a paired t-test Why
b) Do the data provide sufficient evidence to support the claim that the mean PCB level has decreased in the region Be sure to check all assumptions, write the null and alternative hypotheses, calculate the appropriate test statistic, calculate the p-value, and state your conclusion.
c) Construct a 95% confidence interval for the mean decrease in PCB level.

6) Salt sensitivities for ten patients before and after antihypertensive treatment are below.

Patient	1	2	3	4	5	6	7	8	9	10
Before	6.42	-6.71	1.86	11.40	9.39	1.44	9.97	22.86	7.74	15.49
After	10.70	11.40	2.09	10.19	6.99	-0.77	3.29	6.11	-4.02	8.04

a) Are the conditions met for running a paired t-test Explain, addressing each condition. If you need to check for normality, use R to perform the usual diagnostic checks.
b) Regardless of your answer to part (a), run a paired t-test with R. Be sure to state the null and alternative hypotheses, the test statistic, the p-value, and your conclusion.
c) What is a good estimate for the mean sensitivity change

Answer 1

5)

a)

a paired t-test is appropriate because sample are dependent.

b)

Ho :   µd=   0      
Ha :   µd >   0      
              

SAMPLE 1	SAMPLE 2	difference , Di =sample1-sample2	(Di - Dbar)²
61.48	13.99	47.490	75.329
64.47	18.26	46.210	54.749
45.5	11.28	34.220	21.075
59.7	10.02	49.680	118.140
58.81	21	37.810	1.002
75.96	17.36	58.600	391.614
71.57	28.2	43.370	20.787
38.06	7.3	30.760	64.815
30.51	12.8	17.710	445.242
39.7	9.41	30.290	72.604
29.78	12.63	17.150	469.189
66.89	16.83	50.060	126.545
63.93	22.74	41.190	5.661

	sample 1	sample 2	Di	(Di - Dbar)²
sum =	706.36	201.82	504.540	1866.750

mean of difference ,    D̅ =ΣDi / n =   38.811                  
                          
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    12.4725                  
                          
std error , SE = Sd / √n =    12.4725   / √   13   =   3.4592      
                          
t-statistic = (D̅ - µd)/SE = (   38.81076923   -   0   ) /    3.4592   =   11.219
                          
Degree of freedom, DF=   n - 1 =    12                  
  
p-value =        0.000000   [excel function: =t.dist.rt(t-stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                      
data provide sufficient evidence to support the claim that the mean PCB level has decreased in the region

c)

sample size ,    n =    13          
Degree of freedom, DF=   n - 1 =    12   and α =    0.05  
t-critical value =    t α/2,df =    2.1788   [excel function: =t.inv.2t(α/2,df) ]      
                  
std dev of difference , Sd =    √ [ (Di-Dbar)²/(n-1) =    12.4725          
                  
std error , SE = Sd / √n =    12.4725   / √   13   =   3.4592
margin of error, E = t*SE =    2.1788   *   3.4592   =   7.5370
                  
mean of difference ,    D̅ =   38.811          
confidence interval is                   
Interval Lower Limit= D̅ - E =   38.811   -   7.5370   =   31.274
Interval Upper Limit= D̅ + E =   38.811   +   7.5370   =   46.348
                  
so, confidence interval is (   31.2737   < µd <   46.3478   )