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294 CHAPTER 6
Section 6.7
1. (a) X= 3.05, sX= 0.34157, nX= 4, Y= 1.8, sY= 0.90921, nY= 4.
The number of degrees of freedom is
4+0.909212
42
(b) X= 3.05, sX= 0.34157, nX= 4, Y= 1.8, sY= 0.90921, nY= 4.
The number of degrees of freedom is
4+0.909212
42
2. X= 255.8, sX= 8.216325, nX= 6, Y= 270.7375, sY= 11.902933, nY= 8.
Page 294
SECTION 6.7 295
The number of degrees of freedom is
6+11.9029332
82
We can conclude that the dissolve times differ between the two shapes.
3. X= 2.31, sX= 0.89, nX= 15, Y= 2.8, sY= 1.1, nY= 15.
The number of degrees of freedom is
15 +1.12
15 2
4. (a) X= 47.79, sX= 2.19, nX= 14, Y= 47.15, sY= 2.65, nY= 40.
The number of degrees of freedom is
14 +2.652
40 2
Page 295
296 CHAPTER 6
(b) X= 69.33, sX= 6.26, nX= 12, Y= 58.5, sY= 5.59, nY= 24.
(c) X= 109.71, sX= 17.02, nX= 17, Y= 84.52, sY= 13.51, nY= 29.
The number of degrees of freedom is
17 +13.512
29 2
5. X= 24.8, sX= 0.91043, nX= 10, Y= 19.5, sY= 1.3106, nY= 10.
The number of degrees of freedom is
SECTION 6.7 297
10 +1.31062
10 2
6. X= 66.1667, sX= 19.630758, nX= 6, Y= 28.2, sY= 17.767949, nY= 5.
Since the population standard deviations are assumed to be equal, estimate their common value with
the pooled standard deviation
X+ (nY−1)s2
Y
7. X= 53.0, sX= 1.41421, nX= 6, Y= 54.5, sY= 3.88587, nY= 6.
The number of degrees of freedom is
6+3.885872
62
Page 297
298 CHAPTER 6
We cannot conclude that the mean cost of the new method is less than that of the old method.
8. (a) No. Even though the sample standard deviations are equal, the population variances may be quite
different.
9. X= 8.38, sX= 0.96, nX= 15, Y= 9.83, sY= 1.02, nY= 15.
The number of degrees of freedom is
15 +1.022
15 2
10. X= 9.30, sX= 2.71, nX= 14, Y= 9.47, sY= 2.22, nY= 7.
The number of degrees of freedom is
SECTION 6.7 299
14 +2.222
72
11. X= 36.893, sX= 3.2211, nX= 13, Y= 33.000, sY= 2.5449, nY= 9.
The number of degrees of freedom is
13 +2.54492
92
12. X= 6.1, sX= 4.1, nX= 10, Y= 6.6, sY= 4.3, nY= 10.
The number of degrees of freedom is
10 +4.32
10 2
Page 299
300 CHAPTER 6
We cannot conclude that the mean reductions differ between the two models.
13. (a) X= 27.3, sX= 5.2, nX= 12, Y= 32.7, sY= 4.1, nY= 14.
The number of degrees of freedom is
12 +4.12
14 2
14. X= 36.893077, sX= 3.2211, nX= 13, Y= 33.000000, sY= 2.5449, nY= 9.
The number of degrees of freedom is
13 +2.54492
92
15. X= 62.714, sX= 3.8607, nX= 7, Y= 60.4, sY= 5.461, nY= 10.
(3.86072/7)2
7−1+(5.4612/10)2
10 −1
Page 300
SECTION 6.8 301
16. (a) One-tailed. The alternative hypothesis is of the form µX−µY>∆.
17. (a) SE Mean = StDev/√N = 0.482/√6 = 0.197.
Section 6.8
1. D= 0.17214, sD= 0.44316, n= 14. There are 14 −1 = 13 degrees of freedom.
The null and alternate hypotheses are H0:µD= 0 versus H1:µD6= 0.
2. D= 1.2667, sD= 6.215, n= 6. There are 6 −1 = 5 degrees of freedom.
Page 301
302 CHAPTER 6
3. The differences are 4,−1,0,7,3,4,−1,5,−1,5.
4. The differences are 0.0195,−0.0120,0.0131,0.1041,0.0211,0.0434,0.0991,0.0458,−0.0249,0.0539.
5. The differences are −7,−21,4,−16,2,−9,−20,−13.
6. The differences are 3.1,−1.2,0.4,−0.4,−3.1,−7.7.
Page 302
SECTION 6.8 303
7. The differences are 35,57,14,29,31.
8. The differences are 0.56,0.61,0.64,0.67.
9. The differences are 1,2,−1,0,2,−1,2.
Page 303
10. (a) The differences are 68, 93, 87, 72, 70, 63, 101, 81. D= 79.375, sD= 13.383759, n= 8.
(b) The differences are 68, 93, 87, 72, 70, 63, 101, 81. D= 79.375, sD= 13.383759, n= 8.
11. (a) The differences are 5.0,4.6,1.9,2.6,4.4,3.2,3.2,2.8,1.6,2.8.
(b) The appropriate null and alternate hypotheses are H0:µR−µB≤2 vs. H1:µR−µB>2.
12. (a) One-tailed. The alternate hypothesis is of the form µD> µ0.
Page 304
SECTION 6.9 305
The confidence interval is 33.6316 ±2.718(17.1794), or (−13.062,80.325).
13. (a) SE Mean = StDev/√N = 2.9235/√7 = 1.1050.
Section 6.9
1. (a) The signed ranks are Signed
x x −14 Rank
Page 305
306 CHAPTER 6
(b) The signed ranks are Signed
x x −30 Rank
(c) The signed ranks are Signed
x x −18 Rank
Page 306
2. (a) The signed ranks are Signed
x x −41 Rank
(b) The signed ranks are Signed
x x −41.8 Rank
Page 307
308 CHAPTER 6
(c) The signed ranks are Signed
x x −42 Rank
3. (a) The signed ranks are Signed
x x −45 Rank
Page 308
SECTION 6.9 309
We cannot conclude that the mean conversion is less than 45.
(b) The signed ranks are Signed
x x −30 Rank
30.1 0.1 1
30.3 0.3 2
28.8−1.2−3
Page 309
310 CHAPTER 6
(c) The signed ranks are Signed
x x −55 Rank
52.3−2.7−1
The null and alternate hypotheses are H0:µ= 55 versus H1:µ6= 55.
The sum of the positive signed ranks is S+= 70.5. n= 24.
Page 310
4. (a) The signed ranks are Signed
x x −41 Rank
40.76 −0.24 −1
We can conclude that the mean thickness is greater than 41 mm.
(b) The signed ranks are Signed
x x −43 Rank
Page 311
312 CHAPTER 6
(c) The signed ranks are Signed
x x −44 Rank
43.76 −0.24 −1
We cannot conclude that the mean thickness differs from 44 mm.
5. The signed ranks are Signed
x x −0 Rank
0.01 0.01 2
Page 312
6. The ranks of the combined samples are Value Rank Sample
0.43 1 Y
The null and alternate hypotheses are H0:µX−µY= 0 versus H1:µX−µY6= 0.
7. The ranks of the combined samples are Value Rank Sample
12 1 X
The null and alternate hypotheses are H0:µX−µY= 0 versus H1:µX−µY6= 0.
Page 313
