This question has multiple parts. Work all the parts to get the most points. The
ID: 3306551 • Letter: T
Question
This question has multiple parts. Work all the parts to get the most points. The homogeneity of the chloride level in a water sample from a lake was tested by analyzing portions drawn from the top and from near the bottom of the lake, with the following results in ppm Cl Top Bottom 26.30 26.22 26.43 26.32 26.28 26.20 26.19 26.11 26.49 26.42 Apply the r test at the 95% confidence level to determine if the chlonde level from the top of the lake is different from that at the bottom. This problem requires values in your textbooks specifie appendices, which you can access through the OWL 2 MindTap Reader. You should not use the OWLv2 References Tables to answer this question as the values will not mateh.) At 95% confidence: and Sincer T tar , we conclude that significant difference exists at the 95% confidence level. Now use the pared t test and determine whether there is a signi cant difference between the top and bottom values at the 95% confidence level. At 95% confidence: and Since t | T tera, we conclude that significant difference exists at the 95% confidence level. . Why is a different conclusion drawn from using the paired z test than from just pooling the data and using the nomal z test for differences in means? O The large sample to sample variability causes sTop and Bottom to be large and masks the differences between the samples taken from the top and the bottom Because there are reduced degrees of freedom with the paired test, the significant difference detected is artificial There are multiple outliers in the Top data that artificially skew the mean, and therefore the teit value.Explanation / Answer
a)
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (1 2)
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (5-1)*0.12^2+(5-1)*0.12^2/4+4
= 0.1152/8
=0.0144
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(26.338-26.254)-0/0.0144(1/5+1/5)
=0.084/0.07589
=1.106
tCRIT is +/- 2.306 and hence do not reject the null hypothesis because tSTAT<tCRIT. No significant difference exists.
b)
0.00092
D=ED/n=0.084
S.D=(E(Di-D)2/n-1)
= 0.00092/4
= 0.015
tSTAT=D- /S.D/(n)
Where tSTAT has n - 1 d.f
tSTAT=0.084-0/0.015/sqrt(5)
=12.52
t critical value=+/- 2.776. Hence reject the null hypothesis.
c)
The large sample to sample variability causes sTop and sBetween to be large and masks the differences between the samples taken from the top and the bottom
Top Bottom D (Di-D) (Di-D)^2 26.3 26.22 0.08 -0.004 1.6E-05 26.43 26.32 0.11 0.026 0.000676 26.28 26.2 0.08 -0.004 1.6E-05 26.19 26.11 0.08 -0.004 1.6E-05 26.49 26.42 0.07 -0.014 0.000196 D 0.084 E(Di-D)^20.00092
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