160-172 Chapter 4.2 Confidence Intervals Using Chi Square, F, and T

160-172 Chapter 4.2 Confidence Intervals Using Chi Square, F, and T

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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 4.5-3} 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/w-v ' Since (m q US$40]? and (n — 1)S§/a§ are independent chi-square variables with (m — l) and (n — 1) degrees of freedom, respectively, we know from Example 4.1-5 that the distribution of this ratio is F(m—1,n—1). That is, (m —1)S§, 5—7,; _ “iii _ “i (n ~1)S§ i can — 1) a; has an F distribution with n = m — 1 and r2 = n — 1 degrees of freedom. This is the ratio that we want to use to find a confidence interval for org/0Y2. To form the confidence interval, select constants c and d from Table VII in the Appendix so that Because of the limitations of Table VII, we generally let c = F1_a/2(m—1,n—1) = 1/Fa/2(n—1,m—1) and d = Fa/2(m—.l,n—1). (See Exercise 4..l-7(d).) If sfi and s; are the observed values DEE; and 5%. respectively. then 1 53- 52 “‘—_ hf“ —_1, _1 A [Fa/2(n—lJn—1) 5% /2(m " )5? is a 100(1 — a)% confidence interval for 013/03. By taking square roots of both endpoints, we would obtain a 100(1 — a)% confidence interval for ax/ay. In Example 4.2-1, denote oz by 0%. There ()1 — Us? = 1252 = 128.41. Assume that the time in days required for maturation of seeds of a species of Guardiola, both parents having broad leaves, is N(p.y,a$). 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 _7-u s _ S/fi (n 1)52(n_1) 1 0-2 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 305-day milk production period between her first and second calves. Assume that the distribution of X is NUm-Iz). 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 4.2-10.) 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 4.2-1(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 4.2-l(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 4.2-1 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 5.3-11-) 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 two-sided 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)% one-sided 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 one-sided 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$), respectively. If the sample sizes are not large {say considerably smaller than 30), this problem can be a difficult one. However. even in these cases, if we can assume common, but unknown, variances. say of = r13 = 02, there is a way out of our difficulty. We know that q :7? “ y *(#X "15”) Jazm+azfm is N (0. 1). Moreover. since the random samples are independent, 2: U 2 (ll—I)S§, +(m—1)S¥, 7 ‘l O" 0‘ is the sum of two independent chi—square random variables; and thus the distri— bution of U is x2(fit+Yn-2}. In. addition. the independence of the sample means and sample variances implies that Z and U are independent. According to the definition of a T random variable, 4 r/U/(n +171 —2) has a I distribution with n + m ~ 2 degrees of freedom. That is, T: X ‘ Y “ (HX ' Hi”) 1/03,”: +03fm flu: — its; + (m ~21)s$,]/(” + m _ 2) 0'“ 0' _ X ‘ T; _ (flit — Mr) J[(n-—l).5‘§+(m-l)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—(ptx-1ly) 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 r-distrlhutinn 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 q-q 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 4.2-2, :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 4.2-3. 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 Statistic-I Figure 4.2-2 Observations of T and of U, n = 6, m : 18, a; = 1, 0,2,, = 36 4.2-1 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. 4.2-2 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-‘NbJ-RU'IO‘D-JOO —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 4.2-3 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 phosphorus-free 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 one-sided 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 ewe-e) 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 “22-pound" 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% one-sided 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 305-day 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 box-and-whisker 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 box-and-whisker 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—?) ant-we . 15 NtO, 1). «dais/u +034": _ (II-1)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 Odd-Numbered Exercises 3.7-5 0.6915 using normal approximation, 0.7030 using binomial. 3.7—? 0.3085. 3.7-9 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. 3.7-17 0.4734 using normal approximation, 0.4749 using Poisson approximation with A = 50, 0.4769 using b(5000, 0.01). 3.7-19 0.6455 using normal approximation, 0.6449 using Poisson. 3.7-2! (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. 4.1-5 (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 Odd-Numbered Exercises 351 4.2-1? )7 = 135 or n = 136. 4.3-] (a) 1.4 < 1.645, do not reject H0; (b) 1.4 > 1.282, reject H0. 4.3-3 (a) a = 0.0478; (b) E : 24.1225 > 22.5, do not reject H0. 4.3-5 (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 p-value of this test is 0.005. 4.4-3 (a) r: (35- 47)/(s/m) g —1.729; (b) —1.789 < —1.729, reject H0; (c) 0.025 -c p-value < 0.05, p-value % 0.045. 4.4-5 1.477 c 1.833, do not reject H0. 4.4-7 (a) X2 = 8.895 < 10.12 so the company was successful. (b) Since $9,509) = 8.907,p-value ~ 0.025. 4.4-9 (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. 4.4-1] (a) a = 0.1056, 01 = 0.1040 using b(192,0.75); (b) ,6 = 0.3524, [3 = 0.3467 using b(192,0.8). 4.4-1} 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 4.5-3 (a) t < —1.706; (b) —1.714 < —1.706, reject H0; (c) 0,025 < p-value < 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 4.5-9 (a) z = ——p1"’2— 3 1.645; v50 —IA9)(1/"1+1/n2) ...
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160-172 Chapter 4.2 Confidence Intervals Using Chi Square, F, and T

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