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Text Exercise 5'20

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... milk production for man-made shade exceeds the mean milk ... Harvey ... conclude that mean length of fish is larger for North Lake than for Harvey Lake. ... – PowerPoint PPT presentation

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Title: Text Exercise 5'20


1
Homework 17 Score____________
/ 10 Name ______________
Check some answers brefore submitting Homework
17.
Text Exercise 5.20 (a) (b)
You should realize that there is only one
qualitative independent variable to identify.
type of shade (man-made, tree, none)
Do this by first defining the appropriate dummy
variable(s), and then writing a regression model.
1 for man-made shade X1 0 otherwise
1 for tree shade X2 0 otherwise
Y ?0 ?1X1 ?2X2 ? or E(Y) ?0
?1X1 ?2X2
2
(c)
Do this by writing a statement interpreting each
parameter in the model of part (b).
the mean of Y for no shade
?0 ?1 ?2
the amount that the mean milk production for
man-made shade exceeds the mean milk production
for no shade
the amount that the mean milk production for tree
shade exceeds the mean milk production for no
shade
3
Additional HW Exercise 5.2
(a) (b)
The mean length of fish is being studied for
North Lake, Blue Lake, and Harvey Lake. A 0.05
significance level is chosen for a hypothesis
test to see if there is any evidence that mean
length of fish is not the same for the three
lakes. Fish are randomly selected from each
lake, and the lengths in inches are recorded as
follows North 13 17 15 18 17 Blue 15
12 16 11 16 Harvey 14 10 12 13 11
After defining the appropriate dummy variable(s),
write a regression model for the prediction of Y
length of fish from lake.
1 for North Lake X1 0 otherwise
1 for Blue Lake X2 0 otherwise
Y ?0 ?1X1 ?2X2 ? or E(Y) ?0
?1X1 ?2X2
With the model defined in part (a), write the
formula for the population mean length of fish
for each lake.
mean for North
?0 ?1
mean for Blue
?0 ?2
mean for Harvey
?0
4
Do each of the following calculations yN? yB?
yH? y?? SSR SST SSE
(c)
(13 17 15 18 17) / 5
16
(15 12 16 11 16) / 5
14
(14 10 12 13 11) / 5
12
(13 17 13 11) / 15
14
(5)(16 14)2 (5)(14 14)2 (5)(12 14)2
40
(13 14)2 (17 14)2 ,,, (13 14)2 (11
14)2
88
88 40
48
5
1.-continued (d) (e)
Complete the one-way ANOVA table displayed.
Lake
2
40
20
5.00
0.025 lt P lt 0.05
12
48
4
14
88
Summarize the results (Step 4) of the f test to
see if there is sufficient evidence that mean
length of fish is not the same for the for the
three lakes at the 0.05 level.
Since f2,12
have sufficient evidence to reject H0
. We conclude that
5.00 and f2,120.05
3.89, we
mean length of fish is not the same for the for
the three lakes
(0.025 lt P lt 0.05).
6
(f)
Based on the results in part (e), decide whether
or not a multiple comparison method is necessary.
If no, then explain why not if yes, then use
Tukeys HSD multiple comparison method.
Since we have rejected H0 , we need a multiple
comparison method to identify for which lakes
mean length of fish is significantly different.
q?(k,v)
?4
s
2
q0.05(3,12)
3.77
k
v
3
12
s ?2
2 ?2
1 1 5 5
1 1 nN nB
?NB
q?(k,v)
3.4
3.77


s ?2
2 ?2
1 1 5 5
1 1 nN nH
?NH
q?(k,v)
3.77

3.4

s ?2
2 ?2
1 1 5 5
1 1 nB nH
?BH
q?(k,v)
3.77

3.4

yN?
16
yH?
12
yB?
14
2
yN?
yB?
4
yN?
yH?
With ? 0.05, we conclude that mean length of
fish is larger for North Lake than for Harvey
Lake.
yB?
yH?
2
7
Additional HW Exercise 5.3 (a)
(b)
A study is being conducted to see if the drying
time for a brand of outdoor paint can be
predicted from temperature in OF and wind
velocity in miles per hour. Data from random
observation has been stored in the SPSS data file
drypaint.
Write a first-order model for the prediction of
drying time from temperature and wind velocity.
drtm ?0 ?1(tmp) ?2(wnd) ?
Use the Analyze gt Regression gt Linear options in
SPSS to obtain SPSS output displaying the ANOVA
table and coefficients in the least squares
prediction equation for the first-order model in
part (a). To have the mean and standard
deviation displayed for the dependent and
independent variables, click on the Statistics
button, and select the Descriptives option.
Title the output to identify the homework
exercise (Additional HW Exercise 5.3 - part (b)),
your name, todays date, and the course number
(Math 214). Use the File gt Print Preview options
to see if any editing is needed before printing
the output. Attach the printed copy to this
assignment before submission.
8
(c)
Summarize the results (Step 4) of the f test to
see if there is sufficient evidence that the
prediction of drying time from temperature and
wind velocity is significant at the 0.05 level.
Since f2,19 417.907 and f2,190.05 3.52, we
have sufficient evidence to reject H0 . We
conclude that the prediction of drying time from
temperature and wind velocity is significant (P lt
0.01).
OR (P lt 0.001)
9
Additional HW Exercise 5.3 - continued
(d) (e)
In order to see if we can improve prediction, the
complete second order model is now considered.
Write a complete second-order model for the
prediction of drying time using the complete
second order model with independent variables
temperature and wind velocity.
drtm ?0 ?1(tmp) ?2(wnd) ?12(tmp)(wnd)
?11(tmp)2 ?22(wnd)2 ?
Use SPSS to create three new variables, one named
tmp_wnd equal to the product of the variables
temperature and wind velocity, one named tmp2
equal to the square of temperature, and one named
wnd2 equal to the square of wind velocity. Use
the Analyze gt Regression gt Linear options in SPSS
to obtain SPSS output displaying the ANOVA table
and coefficients in the least squares prediction
equation for the complete second-order model in
part (d). Since you only need the ANOVA table
and the coefficients in the least squares
prediction equation, you may choose to delete all
other sections of the SPSS output. Title the
output to identify the homework exercise
(Additional HW Exercise 5.3- part (e)), your
name, todays date, and the course number (Math
214).
10
Use the File gt Print Preview options to see if
any editing is needed before printing the output.
Attach the printed copy to this assignment
before submission.
(f)
Summarize the results (Step 4) of the f test to
see if there is sufficient evidence that the
prediction of drying time using the complete
second order model with independent variables
temperature and wind velocity is significant at
the 0.05 level.
Since f5,16 838.869 and f5,160.05 2.85, we
have sufficient evidence to reject H0 . We
conclude that the prediction of drying time using
the complete second order model with independent
variables temperature and wind velocity is
significant (P lt 0.01).
OR (P lt 0.001)
11
Additional HW Exercise 5.3 - continued (g)
Calculate the partial f statistic for the
hypothesis test to see if there is sufficient
evidence that adding temperature times wind
velocity, temperature squared, and wind velocity
squared after temperature and wind velocity, is
significant at the 0.05 level.
SSR(tmp, wnd, tmp_wnd, tmp2, wnd2) SSR(tmp,
wnd) SSR(tmp_wnd, tmp2, wnd2 tmp, wnd)
MSR(tmp_wnd, rain2, temp2 tmp, wnd)
MSE(tmp, wnd, tmp_wnd, tmp2, wnd2) Numerator
degrees of freedom for f statistic Denominator
degrees of freedom for f statistic
29729.365
29179.456
549.909
29729.365 29179.456
549.909 / (5 2)
183.303
7.088
3
16
183.303 7.088
f3,16
25.861
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