A solution

2. H.W#1 Name: Rabil Iglance
Sikder
ID:
& This study is an
experiment because
the factor of interest (diets) is
being
manipulated by the researcher and the
response
variable (milk production) is
being measured
in order to assess the effect of the diets
on milk
production. This manipulation of the
factor
distinguishes this study from an
observational study, where the variables
are
simply observed and recorded without
manipulation.
** The
response variable for this study
is milk
production, which is being measured
daily over a
30-day period for each
cow.
& The factor under
study is the diet,
with five levels:A, B, C, D and E
#
The observational units in each
design
are the cows.

3. *
@ The
experimental units in each
design
are go
Design 1: The farm, as each farm receives
a
unique
diet and thedaily milk production
of all a cows on the farm is being measured.
Design 2: The individual cow
, as each
cow receives a
unique
diet and its daily
milk
production is being measured.
&
o Blocking is used for Design 1, as
the farms are
being used as blocks
to control for sources of variability
within
groups (farms)
⑰& Replication is accomplished for both
designs, although the level of replication
is different for each
design:
*
Design 1: There is a low level of
replication, as each diet is only represented
once across the a farms
*
Design
2: There is a
higher level
of
replication, as each diet is represented
8 times across the cows within each
farm. This allows for a more accurate
estimate of the diet's effect on milk
production-

4. &
This study is an
experiment. It is an
experiment because the investigator
randomly assigned the 10 bottles to
Machines and another 10 bottles to
Machine 2. This manipulation of the
:edependent variable (Maching makes
if an
experiment.
28 Side by side plot for each
machine is constructed and the
file is attacked.
&
We can calculate the standard
reviations
the
informula:
ch - 1
For machine 1, the standard deviation
is
91 = 0.0302769 ounce
For machine 2, the standard deviation
is
& 2
=
0.0254981 ounce

5. The
poor
mandard
dation
is
the
Nf
+
(3x0.025)
=0.0280 ounce
* T
o determine if there is significant
my difference in fill volume, we can
Near
perform
means on
hypothesis
testto compare
the
entire
Data: Machine A Machine I
5 =
16.015 52=16.005
& =
0.0302765
52 =
0.0254981
n
=
10
u2 =
10

6. -test:
We know,
to
-5th-o
Here,
Si =
G+(v
-
1),
-oof
+
(3x0.02,ha
i
sp =
0.0280
O =
to
oooo
= 0.8788

7. Given, x = 0.0 C
degrees offreedom:
n, the
-
2
= 18
Now, we would
reject Itoim, =
M2
if the numerical value of the test
statistic to >
to.025,18
=
2.101 or if
to <-to.025, 18=-2.101.
We find that to =
0.8783<t= 2.101
0.025,18
So, we cannot
reject Ho irl, =
M2
i =0.2G
·
:P-value
will be = 0.28x2
- 0.8
:We compare p-value with <=0.03
We get, p-value? X.

8. So, we fail to
reject the will
hypothesis.
So, Ho: r, =
ut is
supported.
So, we can conclude that the mean
difference in fill volume between the
two
types of machine are
equal.
* We use the formula,
5.-Ja-tak,mitmn2*Na Iee,-en
=j.-.+tax,n+in
=> -0.072h1M,
-
M, = 0.0922
So, 35
percent confidence interval
estimate on the difference in means
extends from -0.0722 to 0.0922
ounce

9. Since, el.-1
=
0
is included
in this interval, so the data
support the
hypothesis that u, =M2
③
we can calculate the standard
reviations
the
reformula:
For 95
the standard deviation
is,
31=2.0995643 KA
For 100° the standard deviation
is,
9 h
=
1.6404266 KA

10. We will do the
hypothesis test to
see whether the
higher backing temperature
results in water with a
lower
mean
photoresist thickness.
Data:
95°C 100°C
5
=
3.367 52 =
6.847
5, =
2.099 S =
1.640
2
n
=
8
nz =
8
-test:
We know,
to
-5th-o
Here,
Si =
G(x
-
1s
n, +
42
-
2
:
+
(x1.s0
8 + 8 -
2

11. v
=
3.5477
=)
Sp
.
9p
= 1.8832KA
O =
to
iGO
T
-
2.6789
Given, x = 0.0 C
degrees offreedom = 8+8-2
= 19
Now, we would
reject Itoim, =
M2
if the numerical value of the test
statistic to >
to.025,16
=
2.145 or if
to(- to.025,1n =
-
2.145
*
We find that to =
2.67893 t =
2.145
0.025, 14
So, we
would reject Ho :r, =
M2
so, the mean is different
I
H. ir
, Elle

12. D I
0.00S
·
O
LO
*
E
-
2.67532.6793
The P-value
will be
between
0.01
and 240,005= 0.01
↓
P-value will be = 0.002X2
-
0
.
01
:We compare p-value with <=0.03
We get, p-value (x &
So, we reject the well hypothesis,
e. =
Mr

13. So, we can
say
there is
enough
support to claim that the
higher
baking temperature results in waters
with a lower mean
photoresist
thicknell.
③
we use the formula,
5.-Ja-tak,mitmn2*Na Iee,-en
=j.-.+tax,n+in
=> -
2.8825x15(u,- M-= 5.0400
So, 35
percent confidence interval
estimate on the difference in means
to
extends from 12.885x105 5,0400KA

14. We are 35% confident that the
mean
difference is between
- 2.8825x10" and 2,0400 KA