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32 Solutions Manual for Analytical Chemistry 2.1
 he test statistic for the comparison of the means is texp, for which
.
. .
.
t s
X X
n n
n n
1 63
22 86 22 51
7 7
7 7 0 40
exp
pool
std new
std new
std new
#
#
#
#
=
-
+
=
-
+
=
 with 12 degrees of freedom. he critical value for t(0.05,12) is 2.179. 
Because texp is less than t(0.05,9), we retain the null hypothesis, ind-
ing no evidence, at a = 0.05, that there is a signiicant diference 
between new method and the standard method.
25. his problem is a comparison between two sets of paired data,.he 
diferences, which we deine as (measured – accepted), are
 0.0001 0.0013 –0.0003 0.0015 –0.0006
 he mean and the standard deviation for the diferences are 0.00040 
and 0.00095, respectively. he null hypothesis and the alternative 
hypothesis are
: :H Hd d0 00 A !=
 he test statistic is texp, for which
.
.
.t s
nd
0 00095
0 00040 5
0 942exp= = =
 he critical value for t(0.05,4) is 2.776. Because texp is less than 
t(0.05,4), we retain the null hypothesis, inding no evidence, at 
a = 0.05, that the spectrometer is inaccurate. 
26. his problem is a comparison between two sets of paired data. he 
diferences, which we deine as (ascorbic acid – sodium bisulfate), are
 15 –31 1 20 4 –52 –22 –62 –50
 he mean and the standard deviation for the diferences are –19.7 and 
30.9, respectively. he null hypothesis and the alternative hypothesis 
are
: :H Hd d0 00 A !=
 he test statistic is texp, for which
.
.
.t s
d n
30 9
19 7 9
1 91
–
exp= = =
 he critical value for t(0.10,8) is 1.860. Because texp is greater than 
t(0.10,8), we reject the null hypothesis and accept the alternative 
hypothesis, inding evidence, at a = 0.10, that the two preservatives 
do not have equivalent holding times. 
27. his problem is a comparison between two sets of paired data. he 
diferences, which we deine as (actual – found), are
 –1.8 –1.7 0.2 –0.5 –3.6 –1.7 1.1 –1.7 0.3