= 22pt
Math 3310 --- Exam II

1a. 1 Way ANOVA:
SSTrt =
é
ê
ê
ë
(119.0)2
7
+
(142.8)2
7
+
(175.0)2
7
ù
ú
ú
û		 -
(436.8)2
21
,
  =
[ 2023 + 2913.12 + 4375 ] - 9085.44,
  = 225.68,
SSE = 317.34 - 225.68 = 91.66

Source df SS MS F
Treatment 2 225.68 112.84 22.16
Error 18 91.66 5.09  
Total 20 317.34    


1b. 90% Confidence Interval For µ2:
x 2 ± t
 
a
2
,ERR
MSE
n2
We have t0.15, 18 = 1.734, hence:
20.4 ±
(1.734)
5.09
7
,
20.4 ± (1.734)(0.8527),
18.9 < µ2 < 21.9.


1c. 90% Confidence Interval For (µ1 - µ2):
( x1 - x 2) ± t
 
a
2
,ERR
MSE
n1
+
MSE
n2

We have t0.15, 18 = 1.734, hence:
(17.0 - 20.4) ±
(1.734)
5.09
7
+
5.09
7
,
-3.4 ± (1.734)(1.0259),
-5.5 < (µ1 - µ2) < -1.3

2. We know that n=200, x = 160, so p = x/n, or p = 160/200 = 0.80.
  1. H0: p ³ 0.90 vs. H1: p < 0.90.
  2. Test Statistic:
    Z* =
    p - p
    p(1-p)
    n
    		,
      =
    0.80-0.90
    (0.9)(0.1)
    200
    		,
      =
    -0.1
    0.0212
    		= -4.71.
  3. Rejection Region: One tail, large sample, Use 0.5 - a for Z = 1.64
  4. Reject H0.

3. 90% Confidence Interval for µd:
d ± t
 
a
2
,n-1
s
n

We have t.05, 9 = 1.833, hence:
0.88 ±
(1.833)
1.05
10
,
0.88 ± (1.833)(0.332),
0.27 < µd < 1.49.


4. We know that TRT df = 3, BLK df = 4, ERR df = 12. Thus, we have:
MSE =
SSE
(t-1)(b-1)
,
3.0 =
SSE
12
,
SSE = 36.0,
SSB = 160.8 - 102 - 36 = 22.8

Source df SS MS F
Treatment 3 102.0 34.0 11.3
Block 4 22.8 5.7 1.9
Error 12 36.0 3.0  
Total 19 160.8    


  1. H0: b1 = b2 = b3 = b4 = b5 vs. H1: Not all bi are equal.
  2. Test Statistic: 1.9
  3. Rejection Region: Find Fa, BLK, ERR = F.05, 4,12 = 3.26.
  4. Fail to Reject H0.
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