MATH312 Fall 2022 Practice Final Exam
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MATH312
Fall 2022
Practice Final Exam
Note :
1. MAKE SURE TO WRITE DOWN CORRESPONDING NULL AND ALTERNATIVE HYPOTHESES WHENEVER A HYPOTHESIS TESTING TASK IS INVOLVED .
2. Note that the formula for Scheff´e’s procedure on pp.291 is incorrect. The correct formula should be
((a - 1)Fα,a -1,a(n -1) ╱ 、:
1. (65 pts) An auto marketing organization studied the effect of age of automobile owner on the amount of cash offer for a used car by utilizing 12 persons in each of three age groups (young, middle, elderly) who acted as the owner of a used car. A medium price, six-year-old car was selected for the experiment, and the “owners”solicited cash offers for this car from 36 dealers selected at random in the targeted region. Randomization was used in assigning the dealers to the “owners” . The offers (in hundred dollars) follow. The data is as follows:
Age group |
Raw cash offer (in 100s) |
12 z yij j=1 |
12 z yij(2) j=1 |
Young (i=1) Middle (i=2) Elderly (i=3) |
23 25 21 22 21 22 20 23 19 22 19 21 28 27 27 29 26 29 27 30 28 27 26 29 23 20 25 21 22 23 21 20 19 20 22 21 |
258 333 257 |
5580 9259 5535 |
(a) (7 points) Write down the one-way factorial Fixed Effect Model with cash offer being the response
and owner’s age group as factor. Explicitly state the symbols, including the ranges of subscripts,
constraints (of parameters, if applicable) and assumptions (if any).
3
yij = i + i + ij ; 1≤i≤3 , 1≤j≤12 ; i = 0; N(0; 2 ):
i=1
(b) (8 points) Construct the ANOVA table with details for the model in (a).
=
=
=
3 12
yij(2)
i=1 j=1
5580 + 9259 + 5535 = 20374
3 ╱ z yij、2
n
i=1
=
CF =
+ + = 20291:8333
╱ zi(3)=1 z yij、2
an
(258 + 333 + 257)2
3(12)
SSa = A - CF = 20291:8333 - 19975:1111 = 316:7222 SSe = T - A = 20374 - 20291:8333 = 82:1667
SST = T - CF = 20374 - 19975:1111 = 398:8889
Source DF SS MS F
Treatment 2 316.7222 158.3611 63.6014
Error 33 82.1667 2.4899
Total 35 398.889
(c) (5 points) Testing the Age effect at 5% level of significance.
H0 : ) 1 = )2 = )3 = 0< H1 : )1(2) + )2(2) + )3(2) 0 Since F = 63 √6014 3 3 √316 = F0 .05 ,2 ,30 3 F0 .05 ,2 ,33 . Reject H0 , there is a significant difference of mean cash offers among age groups.
(d) (5 points) At 5% level, testing the equality of variances of the three age groups.
sY(2) = = 3< sM(2) = = 1 √659< sE(2) = = 2 √811
3 4 √85 + 4 √ 16
H0 fails to be rejected. There is no significant evidence against homoscedasticity, the variances might be the same.
(e) (15 points) At 5% level, perform the pairwise comparsion by using Fisher’s LSD, Duncan’s, Stu-
dent–Newman–Keuls’, Tukey’s and Scheff´e’s procedure. Report the significant pairs by each of the procedures.
The sample group means in order are zE = = 21 √4167 ← zY = = 21 √ 5 ← zM = = 27 √ 75.
Young Middle
21.5 27.75
Elderly 21.4167 0.0833 6.3333
Young 21.5 6.25
(i) Fisher’s LSD:
t0 .025 ,33 ← = 2 √042 ← = 1 √3131<
(Middle,Elderly) and (Middle,Young) are significantly different.
(ii) Duncan’s:
d0 .05 ,2 ,33 ← = 2 √888 ← = 1 √3155(or use 1.3131 from Fisher’s directly) d0 .05 ,3 ,33 ← = 3 √035 ← = 1 √3825
(Middle,Elderly) and (Middle,Young) are significantly different.
(iii) SNK:
a0 .05 ,2 ,33 ← = 2 √888 ← = 1 √3155(or use 1.3131 from Fisher’s) a0 .05 ,3 ,33 ← = 3 √486 ← = 1 √ 5879
(Middle,Elderly) and (Middle,Young) are significantly different.
(iv) Tukey’s:
a0 .05 ,3 ,33 ← ╱= 3 √486 ← \ = 1 √ 5879<
(Middle,Elderly) and (Middle,Young) are significantly different.
(v) Scheff´e’s
← (3 - 1)F0 .05 , (3 -1) ,33 ← = ←2(3.316) = 1.659
(Middle,Elderly) and (Middle,Young) are significantly different.
(f) (10 points) Construct the most sensible contrast for testing whether the cash offers for middle-aged
owner is higher than the average of the ones for younger and elderly owners. Test the significance at 5% level.
Let u = (uY,uM,uE )t . To test 想 uM = uY - 2uM + uE 想 0, define L = (1,-2,1)t . H0 : Lt u > 0,H1 : Lt u 想 0
t* = = -11.2776 想 -1.697.
Reject H0 , the mean cash offers for middle-aged owner is higher than the average of the ones for younger and elderly owners.
(g) (5 pts) When converted into ranks, the data in ordinal scale is as follows
|
Ranking of cash offer |
12 z Tij j=1 |
Young (i=1) Middle (i=2) Elderly (i=3) |
20.5 23.5 10.5 ~ 16 2 10.5 31.5 28.5 28.5 ~ 28.5 25.5 34 20.5 5.5 23.5 ~ 5.5 16 10.5 |
153.5 366 146.5 |
Use the nonparametric test (Kruskal-Wallis) to test whether any of the age groups outrank the rest in cash offer at 5% level of significance.
H0 : E(i ) = = 18.5, H0 : E(i ) 18.5.
12( - 18.5)2 + 12( - 18.5)2 + 12( - 18.5)2
Reject H0 , there is significant evidence to believe that some age group outranks the others.
(h) (5 points) Repeat (f) but using data in ranks.
From (f) L = (1,-2,1)t . For two sided test on this contrast H0 : Lt u = 0,H1 : Lt u 0
[ -2( ) - ]2
+ +
36(37)/12
H0 will be rejected. Since - 2( ) - = -36, the one-sided test will be significant as well (cutoff is lower).
Another way is to use 义-test
H0 : Lt u > 0,H1 : Lt u 想 0
- 2( ) -
← ( + + )
H0 is rejected again. The mean cash offers for middle-aged owner is higher than the average of the ones for younger and elderly owners.
(i) (5 points) Given the overall median is at 22.5 (in hundreds), for a 3 x 2 contingency table and perform three-sample median test on the equality of medians at 5% level of significance.
Let T = P(Cash Offer > 22.5), H0 : TY = TM = TE H1 : at least one equality fails
oij |
22 5 22 5 Subtotal |
||
Young Middle Elderly |
3 12 3 |
9 0 9 |
12 12 12 |
Subtotal |
18 18 36 |
For all the (i< j)< Eij = = 6.
√ * = + + + + + = 18 3 5 √991 = √0(2) .05 ,2
Reject H0 , some of the age groups have significant different median cash offer.
2. (35 points) The following shows the result of regressing z on y in a simple linear regression model containing intercept term for a set of bivariate data, {yi < zi } , .
Parameter Estimate s.e. t-test statistics p-value
Intercept y
0.203311
0.656040
0.0976
0.1961
2.08
3.35
0.0526
← 0.0038
n = 19 R2 = 0 √397 = 0 √0566 ssR = 0 √0358 ssRes = 0 √0544
(a) (6 pts) Compute sy(2) (the sample variance of z) and ry,x (the sample Pearson correlation coefficient).
sy(2) = = = 0 √005
ry,x = ^R2 = ^0√397 = 0 √63
(b) (5 pts) Find the value of F test statistic and test H0 : é 1 =0 versus H1 : é 1 0 at at the 5% level of
significance.
F* = (t* )2 = (3 √35)2 = 11 √ 223 3 4 √451. Reject H0 , é 1 is significcantly different from 0.
(c) (2 pts) What is the predicted value for z at y = 0 √45?
zˆ = 0 √ 2033 + 0 √656(0 √45) = 0 √4985.
(d) (5 pts) Suppose = 0 √ 5 and sx(2) = 0 √005, construct the 95% prediction interval for z at y = 0 √45.
sx(2) = 0 √005< ÷ sxx = 0 √005(18) = 0 √09.
0 √4985 士 2√ 11(0 √0566) ← 1 + + = (0 √374< 0 √623)
(e) (5 pts) Test the hypothesis: H0 : é 1 <1 versus H1 : é 1 31 at the 5% level of significance.
H0 : é 1 <1< H1 : é 1 31
0 √6560 - 1
0 √ 1961
Retain H0 .
(f) (5 pts) Test the hypothesis: H0 : 礻x,y = 0 versus H1 : 礻x,y 0 at the 5% level of significance.
H0 : 礻x,y = 0 versus H1 : 礻x,y 0
t* = = 3 √345 3 2 √ 11
Reject H0 , 礻x,y is significantly different from 0.
(g) (5 pts) Construct the 95% confidence interval for 礻x,y .
之r = ln( ) = 0 √ 7414.
The 95% confidence interval for 之ρ is 0 √ 7414 士 = (0 √ 2514< 1 √ 2314).
The 95% confidence interval for 礻x,y is (tanh(0 √ 2514)< tanh(1 √ 2314)) = (0 √ 246< 0 √843).
(h) (2 pts) If y is regressed on z instead, what would the R2 be?
The value of R2 remains the same, because no matter regressing y on z or regressing z on y, R2 = rx(2),y which is order invariant.
3. (30 points) Emily took a random sample of 20 facilities where anthrax vaccine is being kept. Then she visits each facility in the sample in order to record how many vials of vaccine are shown on the inventory
(x) and to count the number of vials actually available (y) in the storage refrigerator. Her data and partial work are as follows:
Facility |
a |
b |
c |
d |
e |
f |
g |
h |
i |
j |
k |
l |
m |
n |
o |
p |
q |
r |
s |
t |
Sum |
x |
36 |
78 |
101 |
65 |
21 |
84 |
10 |
13 |
31 |
26 |
25 |
2022-12-09