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Unformatted text preview: 2. Alternative hypothesis: Ha : pd > O (Upper—Tailed Test)
Ha : #d < 0 (LowerTailed Test)
Ha : Md #— 0 (TwoTailed Test:) 3. Test Statistic: _ _
_ d— 0 d t‘ sd/ﬁ: sd/ﬁ where
n = the number of paired differences d = the mean of the sample differences
sd = the standard deviation of the sample differences 4. Rejection Region: Reject H0 when t > ta for Ha : ,ud > O (Upper—Tailed Test)
t < —ta for Ha : pd < 0 (Lower—Tailed Test) t > tat/2 or t < —ta/2 for Ha : Md 7é 0 (Two—Tailed Test:) or when p—value < (1. Here the critical values of t, ta and ta/g, are based on n — 1 degrees of
freedom from the Student’s t probability table. ~ tulZ Assumptions: The experiment is designed as a paireddifference test so that the n dif—
ferences represent a random sample from a" normal population. SmallSample (1 — oz)100% Conﬁdence Interval for Md 2 ,ul — #2, Based on a
Paired—Difference Experiment __ 3 d
d :l: ta ——
/2 (W)
where 305/ ﬂ is the estimated standard error of 3.. Example 10. To compare the wearing qualities of two types of automobile tires, A
and B, a tire of type A and one of type B are randomly assigned and mounted on the rear
wheels of each of ﬁve automobiles. The automobiles are then operated for a speciﬁc number
of miles, and the amount of wear is recorded for each tire, These measurements appear in
table below. Do the data present sufficient evidence to indicate a difference in the average
wear for the two tire types? The sample means and variances for Tire A and Tire B are calculated as follows:
351 = 10.24 i 31 = 1316 $2 = 976 82 = 1328 13 Automobile Tire A Tire B d = A — B 1 10.6 10.2 0.4
2 9.8 9.4 0.4
3 12.3 11.8 0.5
4 9.7 9.1 0.6
5 8.8 8.3 0.5 Solution. Notice that this is a pairedsample. The sample mean of differences, d1, d2,
. , dn, is calculated as __ 1 n
d = — d, = 0.48
which is exactly the same as the difference of the sample means:
E1 — 52 = 10.24 — 9.76 = 0.48. Also, the sample standard deviation of differences, d1, d2, , d", is calculated as 7L Let Md = M1 — M2
1. We need to test the null hypothesis Hozud=0 or ngul—u220
2. An appropriate two—tailed alternative
Haz/idyéO or Hazul—M27EO 3. The test statistic is t— d‘ 0 —0'4§———— 212.8. “ sd/ﬁ = 0.0837/x/5 4. The p—Value is
p—Value = P{t > 12.8} .. From Table 4 of Appendix I with df : 4, we read to 005 = 4.604; that is P{t > 4.604} = 0.005. Thus,
p—~Value = P {t > 12.8} > P{t > 4.604} = 0005. Hence, the results are highly signiﬁcant and we conclude that there is a difference in the
mean wear for tire types A and B .I 14 Example 1.1,, Find a 95% conﬁdence interval for ad : a1  #2 using the data in Example
1 0,. Solution. From Table 4 of Appendix I with df '= 4, we read to 005 = 2.776. A 95%
conﬁdence interval for the difference between the mean wear is we 0.48 :: 2.776 (W)
x/E 01.48 i 0.10 or
0.38 < #1 — ,u2 < 058, Notice that the conﬁdence interval does not contain 0. This reconﬁrms that we must reject
H0 : ,ad 2 0 in Example 10,, HOME‘W‘ORK: pp414 — 417 101,35, 10,37, 10,39, 10141, 104.45, 101.41 10.6 Inferences Concerning a Population Variance The sample variances
n 32: 1 Zea—a)? — 1
n i=1 is used as an unbiased estimator of the population variance (72. This means that, in repeated sampling, the average of all our sample estimates will equal the target parameter, 02. Deﬁnition 2. The statistic
(n — 1) 52 X2: 2 a
is called a chi—square variable and its sampling distribution is called the chisquare probability
distribution with n — 1 degrees of freedom. ﬂxz) Certain critical values of the chi—square statistic, which are used for making inferences
about the population variance, have been tabulated in Table 5 of Appendix I, 15 Example 12,, The probability X2; based on n = 16 measurements (df = 15), exceeds 249958 is 0.05. That is,
x?) 05 = 24.9958, For a sample of n = 6 measurements, 95% of the area under the X2 distribution lies to the right of 14.145476. That is,
X395 2 1145476, Similarly, for a sample of n = 24 measurements, 975% of the area under the X2 distribution
lies to the right of 116885. That is, X3975 = 11,6885, The statistical test of a null hypothesis concerning a population variance
H0 2 02 = 03 uses the test statistic
2 (n " 1) 32 X =_‘——2
‘70 Test of Hypothesis C0ncerning a Population Variance
l. Null hypothesis: H0 : a2 = 03
2.. Alternative hypothesis: Ha : 02 > 03 (Upper—Tailed Test)
Ha : 02 < 03 (Lower—Tailed Test)
Ha : 02 7E 08 (Two—Tailed Test:)
3. Test Statistic:
2 (n ‘“ 1) 32
X = *3—
Go
4. Rejection Region: Reject H0 when X2 > xi for Ha : 02 > 03 (Upper—Tailed Test)
X2 < xia for Ha : 02 < 03 (Lower—Tailed Test)
X2 > X242 or X2 < xia/g for Ha : 02 75 0(2) (Two—Tailed Test:) or when p—Value < a. Here the critical values of X2: xi, X342: and xia/Q, are based on
n —— 1 degrees of freedom from the chi—square probability table. Assumptions: The sample of size n is selected at random from a normally distributed
population. (1 — a) 100% Conﬁdence Interval for a2 (n—1)s2 <02< (71—1)52
2 Xa/Z XT—a/Z where X342 and XT—a/Z are the upper and lower Xg—values, which locate one—half of a in each
tail of the chi—square distribution. Example 13. A cement manufacturer claims that concrete prepared from his product
has a relatively stable compressive strength and that the strength measured in kilograms
per square centimeter (kg/c7712) lies within a range of 40 kg/cm2. A sample of n = 10
measurements produced a mean and variance equal to, respectively, 5 =312 and 52: 195 Do these data present sufﬁcient evidence to reject the manufacturer’s claim? Solution. 1. The null hypothesis is
[10:02 = 102 = 100 2. An appropriate upper—tailed alternative is
Ha ; 5—2 > 1001. 3. The test statistic is 2 (n — 1) s2 (10 — 1) (195) 1755
=___=——=—=17.55.
X 03 102 100 4. The critical value approach: Use oz 2 01,05 and df = n —— 1 = 9, the critical value from Table 5 in Appendix I is
X: = X305 2 16.9190. Thus, the rejection region is : Reject H0 if X2 > 169190.. Since the observed value of the
test statistic is X2 = 1755 > 169190, we reject H0 and conclude that the range of concrete
strength measurements exceeds the manufacturer’s claim. 5. The p—value approach: The p—value is
p—value = P {X2 > 17155} H
From Table 5 in Appendix I with df = n — 1 = 9, we read X305 = 1519190 and 17 That is
P {X2 > 169190} = 0105,, and
P {X2 > 19,0228} 2 0025
Since "
" P {X2 > 190228} < P {X2 > 1755} < P {X2 > 169190},
we see that 0,025 < p—value < 0.051, f(x2) 0 1619190 x2 5 190228
l————> Reject H0 Again, we reject H0 and conclude that the range of concrete strength measurements exceeds
the manufacturer’s claim] Example 14. Using the data in Example 13, Construct a 95% conﬁdence interval to
estimate the true value of the population variance. Solution. The 95% conﬁdence interval for 02 is (n—1)52 <02 < (71—1)s2
X342 Xi—a/z From Table 5 in Appendix I with df = n — 1 = 9, we read Xi—a/2 = X?) 975 = 270039
and
X342 2 x3025 = 1910228,. Substituting these values into the formula for the interval estimate, we get (10—1)(195) 2 (10—1)(195)
191.0228 <0 < 2170039 01‘
92.258 < 02 < 6491906, Thus, we can estimate the population variance to fall into the interval 92.258 to 649,906. Example 15. An experimenter is convinced that her measuring instrument had a vari—
ability measured by standard deviation 0 = 2‘, During an experiment, she recorded the 18 measurements 41, 52, and 10.2. Do these data conﬁrm or disprove her assertion? Test the
appropriate hypothesis. Solution. 1. The null hypothesis is
H0 : 02 = 4 2. An appropriate twotailed alternative is
Ha : 02 % 41, 3.. The test statistic is 2 (n — 1) s2 (3 — 1) (10,57) 2114
X 03 22 4 5 9 4. The critical value approach: Use a = 0.10 and df = n — 1 = 2, the critical values from Table 5 in Appendix I are
X342 2 X505 2 599147 and
x1042 = x3095 = 0.102587, Thus, the rejection region is : Reject H0 if X2 > 599147 or X2 < 01102587. Since the
observed value of the test statistic is 01102587 < X2 = 5.29 < 599147, we cannot reject H0
and these data conﬁrm her assertion a = 2.. 5. The p—value approach: Since this two—tailed test, we have % (p—value) = P {X2 > 5,29} 1. From Table 5 in Appendix I with df = n — 1 = 2, we read x310 = 460517 and
X3005 = 5.99147 That is P {X2 > 4.60517} 2 0.10
and P {X2 > 599147} = 0105,
Since P {X2 > 5.99147} < P {x2 > 5129} < P {X2 > 460517} we have 1 005 < 5 (p—value) < 0.10.
01‘ 010 < p—value < 0.20. 19 Hence, the results are not statistically signiﬁcant and we reject H0. There is insufﬁcient
evidence to reject the null hypothesis H0 : 02 = 4.. Example 16. Using the data in Example 15, construct a 95% conﬁdence interval to
estimate the true value of the population variance. Solution. The corresponding 90% conﬁdence interval for a2 is (n—1)82 2 (n—1)82 2 < 0' <
Xa/2 XT—a/Q
From Table 5 in Appendix I with df = n _ 1 = 2, we read
Xi_a/2 = X395 = 0102587 and
xiv/2 = x305 = 5.99147. Substituting these values into the formula for the interval estimate, we get (3 — 1) (10.57) 2 (3 — 1) (10.57)
5.99147 <0 < 0.102587 01‘
3.528 < 02 < 206.069.. Thus, we can estimate the population variance to fall into the interval 3.528 to 206.069. This
interval includes the hypothesized value a2 = 4. There is insufﬁcient evidence to reject the null hypothesis H0 : 02 = 4. HOMEW'ORK: 1219.423 — 424
10.49, 10.51, 10.53, 10.55 10.7 Comparing two Population Variances One way to compare two population variances, a? and 0%, is to use the ratio of the sample
variances, 8% /33 .. If 3% /53 is nearly equal to 1, we will ﬁnd little evidence to indicate that a?
and 0% are unequal. On the other hand, a very large or very small value for 5% /53 provides
evidence of a difference in the population variances. How large or how small must Sf /33 be for sufficient evidence to exist to reject the
following null hypothesis?
H0 : a? = 0%..
When independent random samples are drawn from two normal populations with equal
variances — that is, a? = 0% — then 3% /33 has a probability distribution in repeated sampling
that is known as an F distribution, shown in the ﬁgure below. 20 f0”) In Table 6 of Appendix 1, critical values of F for right—tailed areas corresponding to
or = 0.10, 0.05, 0,025, 0,01, and 0005 are tabulated for various combinations of dfl numerator degrees of freedom and dfg denominator degrees of freedom. Example 17.
1. The value of F with area 0.05 to its right for dfl = 6 and dfg = 9 is 3.37. 2., The value of F with area 0.05 to its right for dfl = 5 and dfg = 10 is 3.33.
3. The value of F with area 0.01 to its right for dfl = 6 and df2 : 9 is 580.. The statistical test of the null hypothesis uses the test statistic If we can identify the larger population variance as of, the alternative hypothesis implies a
onetailed test — that is,
Ha : a? > a; we can ﬁnd the right—tailed critical value for rejecting H0 directly from Table 6' in Appendix
I. However, when the alternative hypothesis requires a two—tailed test — that is, Hazo§7é0§ the rejection region is divided between the upper and lower tail of the F distribution. How—
ever, if we always call the population with the larger sample variance “Population 1”, then
the observed value of our test statistic will always be in the right tail of the F distribution.
Even though half of the rejection region, the area a / 2 to its left, will be in the lower tail of
the distribution, we will never need to use it! That is, for a two—tailed test, * The area in the right tail of the rejection region is only or / 2. 21 * The area to the right of the observed test statistic is only (p—value) / 2. Test of Hypothesis Concerning a Population Variance
1. Null hypothesis: H0 : a? = a;
2. Alternative hypothesis: Ha : a? > a; (One—Tailed Test)
Ha : a? 750% (Two—Tailed Test:) 3. Test Statistic: where 3% is the larger sample variance.
4. Rejection Region: Reject H0 when F > Fa for Ha : of > a; (OneTailed Test)
F > Fez/2 for Ha : 0% 7S 0% (TwoTailed Test:) or when pvalue < a. Here the critical values of F, Fa, and Fa/g, are based on dfl 2
n1 — 1 numerator degrees of freedom and dfz : n2 — 1 denominator degrees from the F probability table. 0 Assumptions: The samples are randomly and independently selected from normally
distributed populations. (1 — a) 100% Conﬁdence Interval for 0% /03 2 2 2
51 1 01 <51)
_ . < _ < _ Fd ,d‘
(33) Fdf1,df2 0% 83 f2 f1
where l‘i‘dfbdf2 and Fdf2,df'1 are based on dfl 2 n1 — 1 numerator degrees of freedom and
dfg 2 n2 — 1 denominator degrees from the F probability table, with area a / 2 to its right. Example 18. An experimenter is concerned that the variability of responses using two
different experimental procedures may not be the same. Before conducting his research, he
conducts a prestudy with random samples of 10 and 8 responses and gets 5% = 7.14 and
3% z 3.21, respectively. Do the sample variances present suﬂicient evidence to indicate that the population variances are unequal? Solution. 22 1. We wish to test the null hypothesis
H0 : a? = a; 2. An appropriate two—tailed alternative is Ha : 0% 7E 03
3. The test statistic is 2 7 14
s .
F = —1 = — = 2.22.
8% 3.21 4. Use Table 6 in Appendix I for 07/2 2 0.025 with dfl = m — 1 = 9 and dfg 2 n2— 1 = 7,
Fez/2 = 4.82. So we can reject H0 if F > 4.82 with a z 0.05. Because the test statistic does
not fall into the rejection region, we cannot reject H0 : a? = 0%. Thus, there is insufﬁcient
evidence to indicate a difference in the population variances.. Example 19. Refer to Example 18 and ﬁnd a 90% conﬁdence interval for a? /03. Solution. The 90% conﬁdence interval for a? /03 is 2 2 2
8 1 a 8
(—3) F < 1 < ( A) Fdf2df1
32 df:,df'2 ‘72 32 where 81: 7 14 and 82“ — 3 21 with df1—— n1 — 1— — 9 and df2_ — n2 — 1 = 7. Use Table 6 1n
Appendlrc I, we found F917 2 368 and F719 = 329 Substituting these values into the formula for the conﬁdence interval, we get 7.14 1 Z: 7.14
— — — 3.29
(3,21) 3 68 < < (3.— 21)( ) 02
060<——<7..32 ‘72 The calculated interval 0.60 to 7.32 includes 10, the value hypothesized in H0. This indi—
cates that it is quite possible that 0% = 03 and therefore agrees with the test conclusions. Do not reject H0: 01 — 02. I 01' Example 20.. The variability in the amount of impurities present in a batch of chemical
used for a particular process depends on the length of time the process is in operation. A
manufacturer using two production lines 1 and 2 has made a slight adjustment to line 2,
hoping to reduce the variability as well as the average amount of impurities in the chemical.
Samples of 721 = 25 and 77.2 = 25 measurements from the two batches yield these means and variances:
El 2 3.2 8% = 1.04 32 = 3.0 s; = 0.51 Do the data present sufﬁcient evidence to indicate that the process variability is less for line
2? 23 Solution. for) Distribution for
production line 2 Distribution for
production line 1 Level of impurities 1. We wish to test the null hypothesis , 2_ 2
H0.01—0—2 2. An appropriate one—tailed alternative is , 2. 2
Ha.al>02 3. The test statistic is s2 1.04
F = *1 = — = 2.04.
8% 0.51 4. The p—Value is
p—Value = P {F > 2.04} Using> Table 6 in Appendix I with dfl = dfg = 25 — 1 = 24, we get F0 05 = 198 and F0 025 = 2.27. That is,
P{F > 1.98} = 0.05
and
P{F > 2.27} = 0025..
Since
P{F > 2.27} < P{F > 2.04} < P{F > 1.98},
we obtain 0025 < p—Value < 0.05. The results are judged signiﬁcant at the 5% level, and H0 is rejected. We cannot conclude
that the variability of line 2 is less than that of line 1.. HOMEW'ORK: 1911.430 — 432 10.58, 10.59, 10.61, 10.63, 10.65 24 ...
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 Spring '08
 Cheng
 Statistics

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