35
= 1.74(B) (a) se = s / √n → Solve for n, so n = se = (s / se)2. For the current problem, n = (10.3 / 1)2 = 106.09 → round up to next integer for the reason explain in lecture → use n = 107
1. Blood pressure
(A) sem = 10.3 / 35
= 1.74
(B) (a)
se = s
/ √n → Solve for n, so
n = se
= (s / se)2. For the current problem, n = (10.3 / 1)2
= 106.09 → round up to next integer for the reason explain in lecture →
use n = 107
3. t curve
(A) Sketch curve (Figure 13.3 in textbook.)
(B) t9,.90 = 1.38; Shade region to right = Pr(t9 >= 1.38) = 0.10
(C) t9,.10 = -1.38. Shade region to left = Pr(t9 <= -1.38) = 0.10.
(D) Area in both tails combined = 0.20
Comments:
Sketching the curve and labeling the horizontal axis with relevant landmarks helps bring home the distinction between the t quantiles (e.g., t9, .90 = 1.38) and areas under the curve (e.g., 0.90). It also makes clear that fact that the right-tail probability is the complement of the cumulative probability.
T tables do not include negative t values because it knows readers can use the symmetry of the curve to determine values to the left of center. Visualization the curve, for instance, makes it clear that the 10th percentile is the mirror image of the 90th percentile.
5. Approximating the areas beyond a t quantile.
(A) Sketch of the curve with landmarks.
(B) The right tail region beyond 2.65 on t8 is a little bigger than 0.01 and a little smaller than 0.025.
(C) The precise probability computed with StaTable = �(0.029256) = is 0.014628.
7. t score for a confidence interval. Use t28-1,1-.05/2 = t27, .975 = 2.05.
Comment: The notation is compact, with subscripts showing the df and the cumulative probability associated with the t score. Recall that df = n - 1 and the cumulative probability is the proportion of the distribution (area under the curve) less than the current score. The reason we use a cumulative probability point that is .975 is because this point has a right-tail of .025, representing half the alpha level.
9.
Red wine and polyphenol levels. 
= 5.5 s = 2.517 n = 9, sem = 2.517 / 9
= 0.839, df = 9 - 1 = 8. Use t8,.975 = 2.31. The
95% confidence interval for � is 5.5 � (2.31)(0.839) = 5.5 � 1.9 = (3.6,
7.4)
11. Boy
height.
sem = 3.1 / 26 =
0.608, df = 26 - 1 = 25, t25, .975 =
2.05 and the
95% CI for � is 63.8 � (2.06)(0.608) = 63.8 �
1.3 = (62.5, 65.1).
13. Respiratory
function in furniture workers.
n = 7,
= 2.969,
s = 0.9876, sem = 0.9876 / 7 =
0.373. Notice that a = .10 for 90% confidence.
The 90% CI for � is 2.969 � (t7-1,1-.10/2)(.373) = 2.969 � (t6,.95)(.373) = 2.969 � (1.94)(.373) = 2.969 � 0.724 = (2.245, 3.693)
15.
Body weight in high school girls.
(A) Stemplot
9|4
9|6
10|034
10|55
11|44
�10 (lbs.)
No major departures from Normality
(B) x-bar = 103.888, s
= 6.918, se = 6.918 / √9 = 2.306, df = 8,
95% CI for � = 103.888 � (2.306)(2.306) = 103.888 � 5.318 =
(98.570 to 109.206)
(C) m = 5.318
(D) n = (1.96
∙ 6.918 / 3)2 = 20.43 → 21
Significance Tests
17. P-value from tstat. (A) 15 (B) 2.13 and 2.60 (C)
0.025 and 0.01, respectively.
(D) 0.01 < P < 0.025 (E) 0.02 < P < 0.05
19. Beware a
= 0.05.
(A) No, you would not reject H0 since P > a.
(B) Yes, you would reject H0 since P
< a.
(C) It is not reasonable to come to a different conclusion. The results
are nearly identical.
21. Menstrual cycle length.
(A):
H0: � = 29.5 days vs. H1: � 
29.5 days
B: Skip
Step C: n = 9,
= 27.78, s = 2.906, sem = 0.9687,
tstat = (27.78 - 29.5) / 0.9687 = -1.77; df = 9 - 1 =
8
Step D:
To determine the p value note the following landmarks: t8,.90 =
1.40 and t8,.95 = 1.86. The area in the tail is
between .05 and .10. Double this for the two-sided p value: 10 < p < .20.
The precise p (calculated with StaTable) is .11. This p value
suggests the evidence is not significant; we would retain H0
for want of evidence.
23.
Cholesterol levels in Asian immigrants
Step A: H0: � = 190 vs. H1: � 
190
Step B: Skip
Step C: tstat = (181.52 - 190) / (40 /
100) = -2.12; df = 100 - 1 = 99
Step D: We will use the t distribution with df = 100 as the best approximation to
t99. Landmarks on t100
are t100,.975
= 1.98 and t100,.99
= 2.36. Thus, the one-sided p value is between .025 and
.01. The two-sided p value is less than .05 and more than .02.
25. SIDS sample