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stats 363 Chapter-10b notes

stats 363 Chapter-10b notes - 2 Alternative hypothesis Ha...

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Unformatted text preview: 2. Alternative hypothesis: Ha : pd > O (Upper—Tailed Test) Ha : #d < 0 (Lower-Tailed Test) Ha : Md #— 0 (Two-Tailed 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 paired-difference test so that the n dif— ferences represent a random sample from a" normal population. Small-Sample (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 paired-sample. 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 chi-square 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 two-tailed 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; (One-Tailed Test) F > Fez/2 for Ha : 0% 7S 0% (Two-Tailed 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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