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Unformatted text preview: 160 Chapter 4 Applications of Statistical Inference CONFIDENCE
INTERVALS
USING x2, F,
AND T EXAMPLE In this section. we assume that all samples arise from normal distribution
However. in actual practice, we must always question the assumption of normaif 1' and make certain that those underlying distributions are at least approximnlci'
normal before using this theory. We use the fact (Theorem 4.53} that W 2 (n — 1)SZ/a2 is X2(n—1) to ﬁnd 
conﬁdence interval for (72. From Table IV in the Appendix with n — 1 degrees 1.5
freedom select a and b such that __ 2
P<rl§w<b)=1~a, 1 _.
a where P(W 5 u) = P(W 3 a} = uiz so that n = xliwzm—i) and b = Xian—1' . (Note that X30) is a number that cuts offp probability to the right of it for a x29
random variable.) Then. solving the inequalities. we have a 1 _ b
1.a/ZP<(n‘— US: 5 02 E (11—1)?) zp(wgozgw> b n Thus the probability that the random interval [(n — 1)SZ/b, (n — '1)S2/a] ' 1'
the unknown 02 is 1 — (1. Once the values of X1,X2,. . . ,Xn are observed to .r,,.t2......x,, and :2 computed, then the interval [(n — 1)52/b.(n  1).92/a] is .
100(1 — a)% conﬁdence interval for 02. It follows that a 100(1 — oz)% confian
interval for o, the standard deviation. is given by l(n—1)sz (ii—ml n—l
\, b ’V a Vbs’i "—1 Assume that the time in days required for maturation of seeds of a species or
Guardiola, a ﬂowering plant found in Mexico, is N(/1,02). A. random samp}
of n = 13 seeds. both parents having narrow leaves, yielded f = [8.97 (la .._
and 13
1252 : Z (.r, — if = 12844]. {:1
A 90% conﬁdence interval for 02 is [128.41 [28.41 21.03 ' 5.226 1 Z [6'11‘24'57] EXAMPLE Section 4.2 Confidence Intervals Uslng X2, F, and T 161 because 5.226 = x§I95(12) and 21.03 = x&05(12) from Table IV in the Appendix.
The corresponding 90% conﬁdence interval for a is [#611, x/24.57] = [247,496]. _— There are occasions when it is of interest to compare the variances of two
normal distributions. We do this by ﬁnding a conﬁdence interval for rig/0,2,
using the ratio of Sign)? and Shag, where 55’; and Si, are the two sample
variances based on two independent samples of sizes n and m from N(ux,a§)
and (Munch. respectively. However the reciprocal of that ratio can be rewritten as follows:
2 . _ 2
5—: [(m 21)SY:/(m‘ 1)
UY _ UY [L 1;”il/wv ' Since (m q US$40]? and (n — 1)S§/a§ are independent chisquare variables with
(m — l) and (n — 1) degrees of freedom, respectively, we know from Example 4.15
that the distribution of this ratio is F(m—1,n—1). That is, (m —1)S§, 5—7,;
_ “iii _ “i
(n ~1)S§ i can — 1) a; has an F distribution with n = m — 1 and r2 = n — 1 degrees of freedom.
This is the ratio that we want to use to ﬁnd a conﬁdence interval for org/0Y2. To form the conﬁdence interval, select constants c and d from Table VII in the
Appendix so that Because of the limitations of Table VII, we generally let c = F1_a/2(m—1,n—1) =
1/Fa/2(n—1,m—1) and d = Fa/2(m—.l,n—1). (See Exercise 4..l7(d).) If sﬁ and s;
are the observed values DEE; and 5%. respectively. then 1 53 52
“‘—_ hf“ —_1, _1 A
[Fa/2(n—lJn—1) 5% /2(m " )5? is a 100(1 — a)% conﬁdence interval for 013/03. By taking square roots of both
endpoints, we would obtain a 100(1 — a)% conﬁdence interval for ax/ay. In Example 4.21, denote oz by 0%. There ()1 — Us? = 1252 = 128.41. Assume that
the time in days required for maturation of seeds of a species of Guardiola, both
parents having broad leaves, is N(p.y,a$). A random sample of size m = 9 seeds 162 Chapter 4 Applications of Slallstlcal Inference yielded y = 23.20 and 9
8sf, = 2 (y, — y)2 = 36.72.
[:1 A 98% conﬁdence interval for 0.3/03 is given by (HQW 5.67 (36.72)/8 ’ (36.72)/8 ]:[0‘41’10'49] because F0,01(12, 8) = 5.67 and F0.01(8, 12) = 4.50. It follows that a 98% conﬁdence interval for aX/ay is
[V041, V1049] = [064,324].
__ Although we are able to formally ﬁnd a conﬁdence interval for the ratio of two
distribution variances and/or standard deviatiOns, we should point out that these
intervals are generally not too useful because they are often very wide. Moreover.
these intervals are not very robust. That is, the conﬁdence coefﬁcients are not very
accurate ifwe deviate much from underlying normal distributions, because, in those
instances, the distribution of (n ‘— 1}52,f02 could deviate greatly from x2(n—1). We have found a conﬁdence interval for [he mean [L of a normal distribu
tion, assuming that the value of the standard deviation 0 is known or when a
is unknown but the sample size is large. However, in many applications, the
sample sizes are small and we do not know the value of the standard devia
tion, although in some cases we might have a very good idea about its value.
For illustration, a manufacturer of light bulbs probably has a good notion from
past experience of the value of the standard deviation of the length of life
of different types of light bulbs. But certainly, most of the time, the investid
gator will not have any more idea about the standard deviation than about
the mean—and frequently less. Let us consider how to proceed under these
circumstances. If the random sample arises from a normal distribution, we use the fact
(Example 4.1—4) that 3—11
T: a/ﬁ _7u s _ S/ﬁ
(n 1)52(n_1) 1 02 hasatdistribution withr = n—l degrees Offreedom,becauseZ = (Y—M/(a/jr—r)
is N(0,1), W = (n — 1)52/02 is X2(n—1), and Z and W are independent. Select
ta/2(n—1) so that P[T Eta/2(n—1)]= 01/2. Then Y_
l—a =P:—ta/2(n—1) 3 SA; Eta/2(n—1)] manna I If.
U
I
.
1:.
I
'3
i.
I
r. E EXAMPLE EXAMPLE Section 4.2 Confidence Intervals Using x2, F, and T 163 = P[—Y— ta/2(n—1)<%> s w : —Y+t = P[Y — la/2(n—1)<%> g n 3 7+ ta/2(n—1)<%>]. The observations of a random sample provide computed values of J? and 52 and is a 100(1 — a)% conﬁdence interval for a. Let X equal the amount of butterfat in pounds produced by a typical cow during a
305day milk production period between her ﬁrst and second calves. Assume that
the distribution of X is NUmIz). To estimate a a farmer measured the butterfat
production for n = 20 cows yielding the following data: 481 537 513 583 453 510 570 500 457 555
618 327 350 643 499 421 505 637 599 392 For these data, X = 507.50 and s = 89.75. Thus a point estimate of a is f = 507.50.
Since t0.05(19) = 1.729. a 90% conﬁdence interval for a is 9.75
507.50 :l: 1.729(8——> ¢2_0
507.50 :l: 34.70, or equivalently, [472.80, 542.20]. Let T have a : distribution with n — 1 degrees of freedom. Then ta/2(n—1) > Za/z,
Consequently, we would expect the interval if i anew/ﬂ) to be shorter than
the interval 2 :l: raﬂ(n~1)(r{.ﬁi). After all, we have more information, namely the
value of a, in constructing the ﬁrst interval. However, the length of the second
interval is very much dependent on the value of s. If the observed 5 is smaller
than a, a shorter conﬁdence interval could result by the second procedure. But on
the average, 3:1: Zelda/ﬂ) is the shorter of the two conﬁdence intervals. (See Exercise 4.210.) To compare confidence intervals when cr is known or when a is unknown, 50
samples of n = 5 observations were simulated from a N(50, 16) distribution. For
each sample of size 5, a 90% conﬁdence interval was calculated using the known
or :2 4, namely, If :l: 1.645(4/«/5). Those 50 conﬁdence intervals are depicted in
Figure 4.21(a). For those same data a 90% confidence interval was calculated for
it assuming that a was unknown and using f :l: 2.132(s/\/5). These are depicted in
Figure 4.2l(b). Note the different lengths of the latter intervals while the length
ot' the z intervals are all equal. Some of the [ intervals are longer and some are
shorter than the corresponding z intervals. The average length of these intervals
is 7.399, while the length of a z interval is 5.885. It can be shown that the expected
length of a t interval with the given characteristics is 7.169. For the z intervals.
43 (86%] contain the mean a = 50, while 45 (90%) of these I intervals contain
the mean. If this simulation were repeated, the results will be different but it 164 Chapter 4 Applications of Statistical Inference should always be true that approximately 90% of each set of intervals contain the mean.
_._._ 424446485052545658 424446485052545658
(u) z Intervals, 0 known (/3) z Intervals. 0 unknown Figure 4.21 90% conﬁdence intervals for u If we are not able to assume that the underlying distribution is normal but
,u and a are both unknown, approximate conﬁdence intervals for n can still be
constructed using Y—M TIS/ﬁ’ which now only has an approximate r distribution. Generally. this approxi
mation is quite good for many nonnormal distributions (i.c.. it is robust). in
particular, if the underlying distribution is symmetric, unimodal, and of the
continuous type. However. if the distribution is highly skewed. there is great
danger using this approximation {See Exercise 5.311) In such a situation,
it would be safer to use certain nonparametric methods for ﬁnding a canﬁv
dence interval for the median of the distribution, one of which is given in
Section 4.9. There is one other aspect of conﬁdence intervals that should be mentioned. So
far we have created only what are called twosided conﬁdence intervals for the
mean a. Sometimes it happens that you might want only a lower (or upper) bound
on n. We proceed as follows. Say 7 is the mean of a random sample of size n from the normal distribution
N(,u, 02), where say for the moment that 52 is known. Then X—u
P(0/\/ﬁ Szu) 1—01. P[Y—z.(%> Ell] =1—oz. Once 7 is observed to be equal to i, then [It — Zu(O/\/l_1),00) is a 100(1 — a)%
onesided conﬁdence interval for it. That is, with the conﬁdence coefﬁcient of
1 — (1,} — za(a/ﬁ) is a lower bound for ,u. Similarly, (—oo, )7 + axe/J)? )] is a or equivalently, Section 4.2 Confidence Intervals Using x2, F, and T 165 onesided conﬁdence interval for u and E + quG/x/ﬁ) PTOVidCS an upper bound
for n with conﬁdence coefﬁcient 1 ~ 02. When a is unknown. We would use T = (17 — MAS/ﬂ?) to ﬁnd the cor
responding lower or upper bounds for H, namely E — ta(n—1)(s/\/71) and E +
ran—new). Now consider the problem of constructing conﬁdence intervals for the difference
of the means of two normal distributions when the variances are unknown
but the sample sizes are small. Let X1,X2....,X" and Y1.Y2,...,Ym be two
independent random samples from the distributions MI“ 1.0.3.) and N(uy,o$),
respectively. If the sample sizes are not large {say considerably smaller than 30),
this problem can be a difﬁcult one. However. even in these cases, if we can assume
common, but unknown, variances. say of = r13 = 02, there is a way out of our
difﬁculty. We know that q
:7? “ y *(#X "15”) Jazm+azfm is N (0. 1). Moreover. since the random samples are independent, 2: U 2 (ll—I)S§, +(m—1)S¥, 7 ‘l
O" 0‘ is the sum of two independent chi—square random variables; and thus the distri—
bution of U is x2(ﬁt+Yn2}. In. addition. the independence of the sample means and sample variances implies that Z and U are independent. According to the
deﬁnition of a T random variable, 4
r/U/(n +171 —2) has a I distribution with n + m ~ 2 degrees of freedom. That is, T: X ‘ Y “ (HX ' Hi”) 1/03,”: +03fm ﬂu: — its; + (m ~21)s$,]/(” + m _ 2)
0'“ 0' _ X ‘ T; _ (flit — Mr) J[(n—l).5‘§+(ml)5§,][l 1]
+
n + m — 2 u m has a I distribution with r = n +m — 2 degrees of freedom. Thus, with to =
Iii/ztnntm—Z). T._ P(—f()ETEIU)=1—a Solvng the inequality for tax — [,Ly yields _. I! 1 — ﬁ {'1 1
P(X"?“'(:Sr _+_SILV—MYSX—Y'i[OSP “‘) — +
IT 111 \ )1 HI 166 Chapter4 Applications of Statistical Inference EXAMPLE where the pooled estimator of the common standard deviation is S 2 (n — 1)S§, + (m — 1)S§,
P n+m—2 ' 1ch}, and sp are the observed values of 17, V, and Sp, then _ _ [ 1+1 _+t 1+1
._ _ S /_ __ _ /_ _
A y 0 P\ n m’x y 05” n m is a 100(1 — 00% conﬁdence interval for [ix — My. Suppose that scores on a standardized last in mathematics taken by students from:
large and small high schools are Numb?) and N(uy,02), respectively, where 02'
is unknown. If a random sample of n = 9 students from large high schools yielded
f = 81.31, s? = 60.76 and a random sample of m = 15 students from small high
schools yielded y = 7861, 53 = 48.24, the endpoints for a 95% conﬁdence interval
for ,LLX — My are given by ' 81.31 — 78.61i2.074\/W\/g + 11.5 22 because [0025(22) = 2.074. The 95% conﬁdence interval is [—3.65, 9.05].
__ In the case that the variances 0A2, and a; are known or the sample sizes large so that 53 m of, and Si % :13, we would use the facts that 7—7—(ptx1ly) 7_?_(MX_MY) Joﬁfn +031": ,lSﬁ/ni—Sﬁ/m have N(0, 1) and approximate N(0, 1) distributions, respectively, to ﬁnd conﬁdence
intervals for [LX — ﬂy. The respective intervals are and 2 2 2 2
a a _ _ s S,
—X+—1 and .r—yizu/z —"+—). l  y i Za/l
r1 "'1 H m Remark 'I't is'h1'tereaﬁl_13 to! consider ﬂit. two“mph .T in mm detail. II ii. T—‘F‘—lurm Section 4.2 Confidence Intervals Using X2, F, and T 167 Mowﬁnee {rt—1ij n: l. (m~1}.lm a: Land [n+m}t{n+mn1] A: have have that 2"?“(M1 "1!!!) 53431
in ll The: In this term we hate that each variance is divided by the wrung sample site!
That is. as stated [allowing Example 4.sz if the sample sizes are large or the
variances ltnuwn. we wuld like if .52 tr] e! D' .5
ll “I ll m in the denominator; so T seems to use the wrong sample sizes Thus. using this
1" is particularly had when the sample sizes and the variances are unequal; and
thus cauticm must be taken it: using that T in constructing a eonﬁdenee interval
for “I  luy. That is, it'n c m and er} 4: a}. then T dues not have a distributien
which is close to that at a Student r—dlstrlbutlun with n + m —r 2 degrees of
freedom: its spread is much teas than the Student t‘e as the term aim in the
denominator is much larger than it shuuld be. Du the other hand1 it at c n
and a}. 4: uﬁ. then Jim + sign isgenerelly matter the it should be and the
distributinn of T is spread nut more than that ﬂf the Student t. There in way out of this difficulty. homer. When the underlying distribu
tions are close to nermal, but the sample sizes and the variances are seething“.r
much different1 We suggest the use 0! where it has been proved that H has an approximate rdistrlhutinn with Le]
degrees 01' treedem, with n~l(n)1+;l:l(g)r Hera Lu] is the "ﬂoor" at greatest integer in use the number of degrees of
freedom equals a rounded down. This type at appreraimatim has ﬁiat suggested
by E. L. Weleh. EXAMPLE To help understand the above remark, a simulation was done using Maple. In order to obtain a qq plot of the quantiles of a t—distribution, a CAS or some type
of computer program is very important because of the challenge in ﬁnding these
quantiles. l 168 Chapter4 Applications of Statisticailnferenoe _ EXERCISES 4.2 Maple was used to simulate N = 500 observations of T and N = 500 observations
of U. In Figure 4.22, :1 = 6, m = 18, the X observations were generated from
the N{0,1} distribution, and the Y observations were generated from the N (0,36)
distribution. For the value of u for Weieh’s approximate 1 distribution, we used the
distribution variances rather than the sample variances. For the simutation results shown in Figure 4.23. n = 18, m = 6, the X observa
tions were generated from the N(0.1) distribution, and the Y observations were
generated from the N(0, 36) distribution. Remember that in the development of the T statistic, it was assumed that the variances of the two distributions are equal.
__ 2 U ObSBrvations. T{19] p.d.f. Superimposed T(19) Quantiles Versus U Order StatisticI Figure 4.22 Observations of T and of U, n = 6, m : 18, a; = 1, 0,2,, = 36 4.21 Let X equal the length (in centimeters} ofa certain species of ﬁsh when caught in the spring. Assume that the distribution of X is N[u,a
observations ofX are 13.1 5.1 18.0 8.? 16.5 '18 6.8
12.0 17.3 25.4 19.2 15.8 23.0 (3) Give a point estimate of the standard deviation 0 of this species of ﬁsh.
(b) Find a 95% Conﬁdence interval for o. 4.22 A student who works in a blood lab tested 25 men for cholesterol levels and found
the following values: 164 272 261 248 235
230 242
335 297 }. A random sample of n = 13 192 203 278 268
305 286 310 345 289 326
328 400 228 1 94 338 252 Section 4.2, Confidence Intervals Using )8, F, and T 169 r‘NbJRU'IO‘DJOO —6—5—4—3~2—1 01
TObservations. T02) p.d.[. Superimposed UUbsen(alions, T[S)_p.d.lf. Stiperimposed T(5) Quantiles Versus U Order Statistics Figure 4.23 Ubsel‘vations of T and of U, n : 18, m = 6, 01% = 1, 0,2, = 36 Assume that these values represent observations of a random sample taken from
Nluﬁz). (a: Calculate the sample mean and sample variance for these data. (b) Find :1 90% conﬁdence interval for «I. {c} Find a 90% conﬁdence interval for tr. (d) Does the trasumption of normality seem to be valid? Let X1.X2,X3,....Xﬂ be a random sample from N(u..02), with known mean it.
Describe how you would construct a conﬁdence interval for the unknown variance
02.1[in'r1Usethc fact that Z" (Xi — #32,}:52 is x207). int
Let X and Y equal the weights of a phosphorusfree laundry detergent in a “6—
pound" box and a “12—pound" box. respectively. Assume that the distributions of
X and Y are Nu”.ch and N(py.e%). respectively. A random sample of n = 10
observations of X yielded a sample mean of? =: 6.10 pounds with a sample variance
of 5% = 0.0040, while an independent random sample of m = 9 observations of Y
yielded a sample mean of? = 12.10 pounds with a sample variance ofs? = 0.0076. (a) Give a point estimate of HIE/0'3.
(It) Find a 95% conﬁdence interval for 05,103.. Let X and Y equal the concentratiOn in parts per billion of chromium in the blood
for healthy persons and for persons with a suspected disease. respectively. Assume that the distributions of X and Y are t‘leX.a§) and N(uy.a$), respectively. Using
it = 8 observations of X: 15 231218 9 281110 170 Chapter 4 Applications of Statistical Inference and m = 10 observations of Y:
25 20 35 15 40 16 10 22 18 32
(a) Give a point estimate of 034092,.
(b) Find a onesided 95% conﬁdence interval which is an upper bound for oﬁ/dﬁ. Let X1 , X2, . . . .X,, be a random sample ofsize n from a normal distribution, N(u,a1J.
Select a and b so that P(05M5b>=1—a. 02 So a 100(1 —01)% conﬁdence interval fora is [if (n —— tub s, .f[n — 1)/a s]. Find values of a and b that minimize the length of this conﬁdence interval. That is, minimize ewee) under the restriction b
G‘{b}— 6(a) =f g(u)du 21—01. where C(11) and ﬂu} are the distribution function and p.d.f. ofa X 2(12—1) distribution.
respectively. HINT: Due to the restriction, b is a function of a. In particular, by taking . . . . . . db a
derivatives of the reslriclJng equation Wlth respect to a, show that — = ﬂ da ng1'
. dk . dk _
Determine d—. By setting a— : 0, show that a and b must satisfy
a a all/Ze—ﬂ/l __ bn/ZQ—bﬂ = 0 NOTE: It is possible to solve for the values of a and b. See Exercise 5.2—1. Thirteen tons of cheese are stored in some old gypsum mines, including “22pound"
wheels (label weight). A random sample of n = 9 of these wheels yielded the
following weights in pounds: 21.50 18.95 18.55 I940 19.15
22.35 22.90 22.20 23.10 Assuming that the distribution of the weights of the wheels of cheese is N (12,02),
ﬁnd a 95% conﬁdence interval for it. In a stud of maximal aerobic capacity (Journal of Applied Physiology 65. 6 [Decem
ber 1983?, pp. 269ti—2708], 12 women were used as subjects. and one measurement
that was made was blood plasma volume. The following data give their blood plasma
volumes in liters: 3.15 2.99 2.77 3.12 2.45 3.85
2.99 3.87 4.06 2.94 3.53 3.20 Assume that these are observations of a normally distributed random variable X
that has mean u and standard deviation 0. (a) Give the value of a point estimate of ,Lt. (b) Determine point estimates of a: and o. (c) Find a 90% conﬁdence interval for it. A leakage test was conducted to determine the effectiveness of a seal designed to
keep the inside of a plug airtight. An air needle was inserted in the plug and this was
placed under water. The pressure was then increased until leakage was observed.
Let X equal the pressure in pounds per square inch. Assume that the distribution of Section 4.2 Confidence Intervals Using f, F, and T 171 X is N{u.nz). Using the following 11 = 10 observations of X:
3.1 3.3 4.5 2.8 3.5 3.5 3.7 4,2 3.9 3.3 (a) Find a point estimate of u.
(It) Find a point estimate of a. (c) Find a 95% onelsided Conﬁdence interval for u that provides an upper bound
for ii. Let X[.X1,....XN be a random sample of size n from the normal distribution
N{u.al]. Calculate the expected length of a 95% conﬁdence interval for u assuming
that n = 5 and the variance is (a) known. (b) unknown. HINT? To ﬁnd JETS), ﬁrst determine E[‘/(n ~ 1)S2/a2 recalling that
{n —— 1).?13'02 is xzbt—l). An interim automotive supplier places several electrical wires in a harness. A pull
test measures the force required to pull spliced wires apart. A customer requires
that each wire that is spliced into the harness must withstand a pull force of 20
pounds. Let X equal the pull force required to pull 20 gauge wires apart. Assume
that the distribution of X is Ntu,n2). The following data give 20 observations
of X. 28.8 24.4 30.1 25.6 26.4 23.9 22.1 22.5 27.6 28.1
20.8 27.7 24.4 25.1 24.6 26.3 28.2 22.2 26.3 24.4 (a) Find point estimates for u and a. (b) Find a 99% onesided conﬁdence interval for u that provides a lower bound
for u. Independent random samples of the heights of adult males living in two countries
yielded the following results: n =. 12. E = 65.7 inches, 5,; z: 4 inches and m = 15,
i = 68.2 inches, 5,. = 3 inches. Find an approximate 98% conﬁdence interval for
the difference .ux — try of the means of the populations of heights. Assume that 2 ., 2
a“, — 0).. Consider the butterfat production [in pounds) for a COW during a 305day milk
production period following the birth of a calf. Let X and 1" equal the butterfat
production for such cows on a farm in Wisconsin and a farm in Michigan. Twelve
observations of X are: 649 657 714 877 975 468
567 849 721 791 874 405 Sixteen observations of Y are: 699 891 632 315 589 764 524 727
597 368 652 978 479 733 549 790 (3) Assuming that X is [Wig/.02) and Y is N04 hag), ﬁnd a 95% conﬁdence interval
for ux — try. (b) Construct boxandwhisker diagrams for these two sets of data on the same
graph. (c) Does there seem to be a signiﬁcant difference in butterfat production for cows
on these two farms? A test was conducted to determine if a wedge on the end of a plug ﬁtting designed
to hold a seal onto that plug was doing its job. The data taken were in the form
of measurements of the force required to remrwe a seal from the ping ﬁrst with
the wedge in place, say X. and the force required without the plug, say Y. Assume
that the distributions of X and Y are bimbo!) and N(/.ty,oz). Ten independent
observations of X are: 3.26 2.26 2.62 2.62 2.36 3.00 2.62 2.40 2.30 2.40 172 Chapter 4 Applications of Statistical Inference CONFIDENCE
INTERVALS
AND TESTS OF
HYPOTHESES Ten independent observations of Y are: 1.80 1.46 1.54 1.42 1.32 1.56 1.36 1.64 2.00 1.54 (a) Find a 95% conﬁdence interval for MK — [.t y.
(b) Construct boxandwhisker diagrams of these data on the same ﬁgure.
(c) Is the wedge necessary? Let 7, 7, $2 , and 8%, be the respective sample means and unbiased estimates of the
variances using independent samples of sizes n and m from the normal distributions
N([lx,0§) and N(uy,a$), where ax, ny. aﬁ, and 0% are unknown. If, however. (xi/0,2, = d. a known constant. argue that a) (17—?) antwe . 15 NtO, 1).
«dais/u +034": _ (II1)S_2X {m —1)52y
b ———~— + —
( ) do)? a?
(c) The two random variables in {a} and (b) are independent. ((1) With these results. construct a random variable (not depending upon 0,2,) that _
has a 1 distribution and can be used to construct a conﬁdence interval for #X — My.
Let 7 denote the mean of a random sample of size n from a distribution that has _ mean u and variance a2 i: it}. Findﬂso that the probability is approximately 0.95
that the random interval X — 1,12 to X + l/2 includes a. ( is x2{n+m—2). Let 7&7 and 7 be the means of two independent random samples,each of size n, from
the respective distributions Nursing] and Non/472), where the common variance
02 is known. Find n. such that P(X—Y—U/5</.Lx—My<Y—?+U/5)=0.90‘. The ﬁrst major area of statistical inference involves the estimation of parameters.
We have introduced both point estimation through maximum likelihood estimation
and interval estimation with conﬁdence intervals. We now consider a second major
area of statistical inference, namely tests of statistical hypotheses, in which many
such tests are closely related to conﬁdence intervals for parameters. To see this, let
us begin with an illustration. Many statisticians are involved in a reform movement in teaching introductory
statistics. This concerns mainly those statistics courses that do not have calculus as
a prerequisite. This reform requires students to be more actively involved through
projects. diacussions, computer work' and analyses of statistical reports found in
the media. There is much less lecturing by the instructor and hopefully more.
thinking about the use of good statistical methods by the student. Initially, we ﬁnd
that in such a course both the instructor and the student really work harder, but
there seems to be more satisfaction for both in doing so. That is, the claim is the
students learn statistics better. To be honest, there has not been a good assessment
of that claim. How might We investigate such a situation? Suppose all persons involved, some of whom might be for the reform and some
against it, agree on some type of ﬁnal test to be given to all students taking a course
using the reform method. Moreover, let us assume that many students, who have
taken the traditional statistics course that uses the lecture system, have taken this
type of test with scores that average about 75 with a standard deviation of 10. Later
on we consider the possibility of comparing two groups of students, one using the 350 Appendix c Answers to OddNumbered Exercises 3.75 0.6915 using normal approximation, 0.7030 using binomial. 3.7—? 0.3085. 3.79 0.6247 using normal approximation, 0.6148 115ng Poisson.
3.7” 0.5548; (b) 0.3823.
3.7]3 0.6813 using normal approximation. 0.6788 using binomial.
3.715 0.3802; (b) 0.7571. 3.717 0.4734 using normal approximation, 0.4749 using Poisson approximation with A = 50, 0.4769 using
b(5000, 0.01). 3.719 0.6455 using normal approximation, 0.6449 using Poisson.
3.72! (a) 0.8289 using normal approximation, 0.8294 using Poisson.
(b) 0.0261 using tables in book, 0.0218 using Maple. CHAPTER 4 4.1—3 (3) 0.890; (b) 0.05.
4.1—3 0.8644.
4.15 (3) d=2.131‘, 11(15): 3 — 2.1313/4, no.5) = 3 + 2.1315/4.
(a) d = 1.746; _ _ /7S§+9S§. /1 1_
X—Yi1.746 T\§+1—0, (c) [—8.5517. 10.3517];
c = 1/482, d = 4.20; S2 2 4.20 2
P( Y2 <a—g< 25Y>=0.95;
4.82SX 0X SK
(0 [0.1365, 2.7626]. (3) s = 6.144; (b) [440610.142] or [41079521].
[ ;;1<X.— 102 ZL1(X1— 102] X37201) , Mag/201)
0.4987; (b) [0,1835].
[1947,2233].
)7 = 3.580;
5 = 0.512;
[0, 3.877].
f = 25.475, S = 2.4935; (b) [24.059, [—115.480, 129.105]; (C) no. (9) (n —1)s§/d+(m—1)s§(a’ 1)
+
11+ m — 2 2 ~ y :i: ta/72(n+m 2)\/ nm' Appendix C Answers to OddNumbered Exercises 351 4.21? )7 = 135 or n = 136.
4.3] (a) 1.4 < 1.645, do not reject H0;
(b) 1.4 > 1.282, reject H0.
4.33 (a) a = 0.0478; (b) E : 24.1225 > 22.5, do not reject H0.
4.35 (a) n = 25, c = 1.6.
4.3—? c = 19.5, n = 164 or 165.
g 4.4] (a) t: 3.0 > 1.753, reject H0;
(b) Since tomSUS} 2 2.947, the approximate pvalue of this test is 0.005.
4.43 (a) r: (35 47)/(s/m) g —1.729;
(b) —1.789 < —1.729, reject H0;
(c) 0.025 c pvalue < 0.05, pvalue % 0.045.
4.45 1.477 c 1.833, do not reject H0.
4.47 (a) X2 = 8.895 < 10.12 so the company was successful.
(b) Since $9,509) = 8.907,pvalue ~ 0.025.
4.49 (a) 0.3032 using b(100,0.08), 0.313 using Poisson approximation, 0.2902 using normal approximation; (b) 0.1064 using b(100,0.04), 0111 using Poisson approximation, 0.1010 using normal approximation.
4.41] (a) a = 0.1056, 01 = 0.1040 using b(192,0.75);
(b) ,6 = 0.3524, [3 = 0.3467 using b(192,0.8). 4.41} With 71 = 130, C: 8.5, a R 0,055,1‘3 % 0.094.
 )7 — 7
4.5.1 (a) z: ——————— 3 10mm): 2473;
155% + 125; 1 1
27 (E + E) (b) t: 5.570 > 2.473, reject H0. (c) The critical region is 2 S2]
i; 3 F0,025(15,12) = 3.18 or 4; 3 F0,025(12, .15) = 2.96.
Sy S}
52 1356 75 s2
s A = ' = 1.96 3.18 d 1 =0.51 2.96, tH.
11106 5% < an S; < accep 0 4.53 (a) t < —1.706; (b) —1.714 < —1.706, reject H0;
(c) 0,025 < pvalue < 0.05; (e) 1.836 < 3.28, do not reject. 4.88 4.56 (a) m = 0.84 < 2.53 = F0.01(24, 28),
5.81
— = 1.19 2. 1 = 2 4
4.88 < 9 Fund 33 ), 2.
Y' _ (b) 3.402 2 2.326 = 20.01. reject pt, = My.
4.5:: [0.007, 0071], yes. do not reject a}, = a 4.59 (a) z = ——p1"’2— 3 1.645; v50 —IA9)(1/"1+1/n2) ...
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This note was uploaded on 12/28/2010 for the course ECON 41 taught by Professor Guggenberger during the Fall '07 term at UCLA.
 Fall '07
 Guggenberger

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