HW 2 answers

HW 2 answers - HW
#2
 Due
Friday
9/2/11


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Unformatted text preview: HW
#2
 Due
Friday
9/2/11
 (1)
The
carbohydrate
content
of
a
glycoprotein
is
determined
to
be
12.6,
11.9,
 13.0,
 12.7,
 and
 12.5
 g
 of
 carbohydrate
 per
 100
 g
 of
 protein
 in
 replicate
 analyses.
Find
the
50%
and
90%
confidence
intervals
(µ)
for
the
carbohydrate
 content.
 Write
 out
 in
 words
 what
 these
 calculaJons
 mean
 about
 where
 the
 true
mean
lies.
 Mean = 12.5 s = 0.1 At 50% CL m = 12.5 ± 0.1 wt % At 90% CL m = 12.5 ± 0.3 wt % If we continued to do repeated sets of five measurements many times, half of the sets will have a 50% CL that will include the true mean and 9/10ths of will have a 90% CLs that will include the true mean. That is, consider another 5 measurement set. If you determine its m and think about its ± value as an error bar, you would have a 1 in 2 chance that the error bar would include the true value. If you continue to do sets of 5 measurements, ½ of those will have error bars that include the true mean. HW
#2
 (2)
 Lord
 Rayleigh
 received
 the
 1904
 Nobel
 Prize
 for
 discovering
 argon.
 This
 discovery
 occurred
 when
 he
 noJced
 a
 small
 discrepancy
 between
 two
 sets
 of
 measurements
 of
 density
of
nitrogen
gas.
 In
Rayleigh’s
Jme,
it
was
thought
that
dry
air
was
composed
of
about
one‐fiVh
oxygen
and
 four‐fiVhs
 nitrogen.
 Rayleigh
 removed
 all
 O2
 from
 air
 by
 passing
 the
 air
 through
 red‐hot
 copper
 to
 make
 copper
 oxide.
 He
 then
 measured
 the
 density
 of
 the
 remaining
 gas
 by
 collecJng
it
in
a
fixed
volume
at
constant
temperature
and
pressure.
He
also
prepared
the
 same
 volume
 of
 pure
 nitrogen
 by
 chemical
 decomposiJon
 of
 N2O,
 NO,
 or
 ammonium
 nitrite.

 The
average
mass
of
gas
from
air
was
2.31011
g
with
an
s
of
0.000143
for
7
measurements.
 The
 mass
 of
 gas
 from
 the
 chemical
 sources
 was
 2.29947
 g
 with
 an
 s
 of
 0.00138
 for
 8
 measurements.

 Was
Lord
Rayleigh’s
gas
from
air
different
from
the
N2
produced
chemically?

 spooled
=
0.00102
 Tcalc
=
20.2
 DOF
=
13
 At
95%
CL
tcalc>aable
so
the
difference
is
significant
 HW
#2
 (3)
 Consider
 the
 cholesterol
 content
 of
 six
 sets
 of
 human
 blood
 plasma
 measured
 by
 two
 different
techniques.
Each
sample
is
recorded
as
having
a
different
cholesterol
content.
But
 are
the
two
techniques
yielding
different
answers
at
the
95%
CL?

 Sample
 1
 2
 3
 4
 5
 6
 Method
A
 Method
B
 1.46
 2.22
 2.84
 1.97
 1.13
 2.35
 davg
=
0.06
 Sd
=
0.12
 tcalc
=
1.21
 ttable
>
tcalc
so
methods
give
the
same
answer
 1.42
 2.38
 2.67
 1.80
 1.09
 2.25
 HW
#2
 (4)
Calculate
the
average,
 stdev,
and
coefficient
of
variance
of
the
following
numbers:
 32.76,
30.25,
34.40,
25.39,
40.15.
 Mean
=
32;
Stdev
=
5;
CV
=
15%
 (5)
Can
any
of
the
numbers
in
problem
(4)
be
discarded
from
the
data
set?
 Largest
gap
is
for
the
point
40.15.
Qcalc
for
that
point
is
0.3896.
Qtable
at
the
90
CL
is
0.642
so
 the
point
cannot
be
rejected.
 (6)
Find
the
answer,
absolute
and
percent
relaJve
uncertainty
for
each
of
the
following
and
 express
each
answer
with
the
correct
number
of
significant
figures.

 a. 
15.8(±0.3)
+
207.3(±0.4)
=
223.1
±
0.5
(0.2%)
 b.  6.35(±0.04)
x
10‐3
÷
3.256(±0.002)
x
106
=
1.95x10‐9
±
0.01x109
(0.6)
%
 c.  4.2817(±0.0003)
x
10‐3
÷
{12.62(±0.02)
–
5.312(±0.003)}
=
5.86x10‐4
±
0.02x10‐4
±
0.3
%
 ...
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