b)
7
.
2
08
.
96
X
5
.
1
±
²
i
; X
i
= +1.5 (2.7) + 96.08 = 4.05 + 96.08 = 100.13
;
7
.
2
08
.
96
X
8
.
±
±
i
;
X
i
= .8 (2.7) + 96.08 = 2.16 + 96.08 =
93.92
2. a) area above z = .4 = .1544 + .5 = .6544, so 65.44
area above z = .93 = .1762, so
%;
17.62
%
b) area beyond z = .4 is .1554, so 15.54
%
Chapter 5
1.
67
.
4
54
.
52
.
2
25
7
.
2
6
.
98
08
.
96
70
.
2
,
08
.
96
X
±
±
±
z
V
This z is larger in magnitude than the critical value (1.96) so the result is
significant at the .05 level (twotailed).
2.
33
.
9
100
7
.
2
52
.
2
±
±
z
For
D
= .01, z
crit
= 2.33 (onetailed), or 2.58 (twotailed). In either case these
results are easily significant at the .01 level. Because the sample size was 4 times
larger for part b than part a, the zscore from part a could have been simply multiplied
by 2 (i.e., the squareroot of 4).
Chapter 6
1.
a)
25
7
.
2
x
V
;
x
crit
z
X
V
P
r
= 96.08 ± 1.96 (.54); so,
P
= 96.08 ± 1.06
Therefore, the 95% CI goes from 95.02 to 97.14 (note, as a check, that the sum of these two limits is
192.16, which is exactly twice as large as the sample mean, as will always be the case). Because the
proposed population mean of 98.6 is not in this 95% CI, we know that the sample mean would differ
significantly from this population mean at the .05 level.
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 Fall '12
 WallaceSherman
 Statistics, twotailed tcrit=, Younger vs

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