n=25, so use t percentage point in interval (24 degrees of
freedom). The formula:
m =
_ X
±t0.025 * s/Ön
t0.025,df = 24 = 2.06, so
m = 148,000 ±2.06 * 62,000 / 5,
and the interval is [122456, 173544].
The confidence interval in part a) gives a range of
plausible values for the mean of the distribution of home
sale prices. Look, instead, at the distribution of sample sale
prices: sample mean is $148,000 and sample standard deviation is
$62,000. The friend paid $206,000 for a home, this is less than
one standard deviation above the sample mean ($210,000), not
unusual at all.
Confidence interval for a difference in population
means using unmatched data.
sample mean for women, [`X]W, is 11; sample mean for
men, [`X]M, is 16. sample variance for women, s2W=10;
sample variance for men, s2M=21.5. This is an unmatched
problem, and the 2 population variances are unknown, use formula
8-20. Degrees of freedom for the t value are, using 8-22,
10-2=8. For a 95% confidence interval t0.025,df = 8 = 2.31.
Using 8-21,
s2p =
4*s2W + 4*s2M
4 + 4
= 15.75.
The confidence interval for the difference in women's and men's
salaries, mW - mM, is
mW - mM =
_ X
W
-
_ X
M
±t0.025* sp*
æ Ö
1
nW
+
1
nM
where nW = nM = 5, so
mW - mM = -5 ±2.31 * 3.97 * 0.632,
hence
mW - mM = -5 ±5.8.
This only provides very weak evidence that, on average,
women earn less than men. Indeed, the confidence interval contains
0 and hence is consistent with no discrimination. Furthermore, this
analysis does
not take into account possible explanations for this difference
(e.g. experience, training, etc.) and so, while a policy of
discrimination would likely result in a difference, a difference is
not, in itself, proof of discrimination.
Confidence interval for a difference in population
means using matched data. The data is matched by litter.
Use formula 8-12. Start by forming the differences between
treatment and control, D = Treatment - Control. The
sample mean difference, [`D], is 3 and the sample
variance, s2D, of D is 3.56. n = 10, so d.f. = 9 and
t0.025,df = 9 = 2.26.
D =
_ D
±t0.025,df = 9 * sd/
__ Ö10
,
D = 3 ±2.26 * 0.596,
D = 3 ±1.35.
The experiment provides evidence that an interesting
environment leads to greater brain development,as measured by
brain weight.
Confidence interval for a proportion. In a sample of
size n tires, a proportion, P = 0.10, failed to meet standards.
Expected number of 'successes' is 0.10*10 = 1 < 5, so
need to use the graphical method. Find P = 0.10 on the horizontal
axis and read off the 95% interval from the vertical axis (n=10):
0 £ p £ 0.45 .
Expected number of 'successes' is 0.10*25 = 2.5 < 5, so
need to use the graphical method. Find P = 0.10 on the horizontal
axis and read off the 95% interval from the vertical axis (half way
between n=20 and n=30):
0.01 £ p £ 0.30.
Expected number of 'successes' is 0.10*50 = 5, so
can use either method. Using the graphical method, find P = 0.10 on
the horizontal axis and read off the 95% interval from the vertical
axis (n=50):
0.04 £ p £ 0.22.
Using formula 8.27,
p = P ±1.96 *
æ Ö
P(1-P)
n
,
p = 0.1 ±1.96 *
æ Ö
0.1(0.9)
50
,
p = 0.1 ±0.08.
Use formula 8.27,
p = P ±1.96 *
æ Ö
P(1-P)
n
,
p = 0.1 ±1.96 *
æ Ö
0.1(0.9)
200
,
p = 0.1 ±0.04.
Confidence intervals for differences in proportions.
Use formula 8-29 for parts a and b:
nus = njapan = 300. Pus = 0.40 and Pjapan = 0.33
pus - pjapan = Pus - Pjapan ±1.96 *
æ Ö
Pus(1-Pus)
nus
+
Pjapan(1-Pjapan)
njapan
,
pus - pjapan = 0.07 ±1.96 * 0.039,
pus - pjapan = 0.07 ±0.077,
nus = nla = 300. Pus = 0.40 and Pla = 0.57
pus - pla = Pus - Pla ±1.96 *
æ Ö
Pus(1-Pus)
nus
+
Pla(1-Pla)
nla
,
pus - pla = -0.17 ±0.079,
File translated from TEX by TTH, version 2.00. On 24 Mar 1999, 15:30.