Statistics for Business (Env) Chapter 10 Statistical Inferences Based on Two Samples 1 Statistical Inferences Based on Two Samples 10.1 Comparing Two Population Means by Using Independent Samples: Variances Known 10.2 Comparing Two Population Means by Using Independent Samples: Variances Unknown 10.3 Paired Difference Experiments 10.4 Comparing Two Population Proportions by Using Large, Independent Samples 2 Comparing Two Population Means by Using Independent Samples: Variances Known • Suppose a random sample has been taken from each of two different populations • Suppose that the populations are independent of each other – Then the random samples are independent of each other • Then the sampling distribution of the difference in sample means is normally distributed 3 Do the achievement scores for children taught by method A differ from the scores for children taught by method B? 4 A research design that uses a separate sample for each treatment condition (or for each population) is called an independent-measures research design or a between-subjects design. The goal of an independent-measures research study is to evaluate the mean difference between two populations (or between two treatment conditions). 5 Sampling Distribution of the Difference of Two Sample Means #1 • Suppose population 1 has mean µ1 and variance σ12 – From population 1, a random sample of size n1 is selected which has mean and variance s12 • Suppose population 2 has mean µ2 and variance σ22 – From population 2, a random sample of size n2 is selected which has mean and variance s22 • Then the sample distribution of the difference of two sample means… 6 Sampling Distribution of the Difference of Two Sample Means #2 • Is normal, if each of the sampled populations is normal – Approximately normal if the sample sizes n1 and n2 are large • Has mean = µ1 – µ2 • Has standard deviation x x 1 2 2 1 n1 2 2 n2 7 If you select one score from each of these two populations, the closest two values are X1 =50 and X2 =30. The two values that are farthest apart are X1 =70 and X2 =20. µ2 µ1 8 Sampling Distribution of the Difference of Two Sample Means #3 9 z-Based Confidence Interval for the Difference in Means (Variances Known) • Let be the mean of a sample of size n1 that has been randomly selected from a population with mean m1 and standard deviation 1 • Let be the mean of a sample of size n2 that has been randomly selected from a population with m2 and 2 • Suppose each sampled population is normally distributed or that the samples sizes n1 and n2 are large • Suppose the samples are independent of each other, then … 10 z-Based Confidence Interval for the Difference in Means Continued • A 100(1 – ) percent confidence interval for the difference in populations µ1–µ2 is 12 22 x1 x2 z 2 n1 n2 11 Example 10.1 The Bank Customer Waiting Time Case #1 • A random sample of size 100 waiting times observed under the current system of serving customers has a sample mean of 8.79 – Call this population 1 – Assume population 1 is normal or sample size is large – The variance is 4.7 • A random sample of size 100 waiting times observed under the new system of time of 5.14 – Call this population 2 – Assume population 2 is normal or sample size is large – The variance is 1.9 • Then if the samples are independent … 12 Example 10.1 The Bank Customer Waiting Time Case #2 • At 95% confidence, z/2 = z0.025 = 1.96, and x1 x2 z 2 12 22 4.7 1.9 8.79 5.14 1.96 n1 n2 100 100 3.65 0.5035 3.15 , 4.15 • According to the calculated interval, the bank manager can be 95% confident that the new system reduces the mean waiting time by between 3.15 and 4.15 minutes 13 z-Based Test About the Difference in Means (Variances Known) • Test the null hypothesis about H0: µ1 – µ2 = D0 – D0 = µ1 – µ2 is the claimed difference between the population means – D0 is a number whose value varies depending on the situation – Often D0 = 0, and the null means that there is no difference between the population means 14 z-Based Test About the Difference in Means (Variances Known) • Use the notation from the confidence interval statement on a prior slide • Assume that each sampled population is normal or that the samples sizes n1 and n2 are large 15 Test Statistic (Variances Known) • The test statistic is x1 x2 D0 z 2 1 n1 2 2 n2 • The sampling distribution of this statistic is a standard normal distribution • If the populations are normal and the samples are independent ... 16 z-Based Test About the Difference in Means (Variances Known) • Reject H0: µ1 – µ2 = D0 in favor of a particular alternative hypothesis at a level of significance if the appropriate rejection point rule holds (i.e. calculated z is in the rejection region). • Rules are on the next slide… 17 Hypothesis Tests for Two Population Means Two Population Means, Known Population Variances Lower-tail test: Upper-tail test: Two-tail test: H0: μ1 μ2 H1: μ1 < μ2 H0: μ1 ≤ μ2 H1: μ1 > μ2 H0: μ1 = μ2 H1: μ1 ≠ μ2 i.e., i.e., i.e., H0: μ1 – μ2 0 H1: μ1 – μ2 < 0 H0: μ1 – μ2 ≤ 0 H1: μ1 – μ2 > 0 H0: μ1 – μ2 = 0 H1: μ1 – μ2 ≠ 0 18 Hypothesis tests for μ1 – μ2 Two Population Means, Known Population Variances Lower-tail test: Upper-tail test: Two-tail test: H0: μ1 – μ2 0 H1: μ1 – μ2 < 0 H0: μ1 – μ2 ≤ 0 H1: μ1 – μ2 > 0 H0: μ1 – μ2 = 0 H1: μ1 – μ2 ≠ 0 -z Reject H0 if Z < -Z z Reject H0 if Z > Z /2 -z/2 /2 z/2 Reject H0 if Z < -Z/2 or Z > Z/2 19 EXAMPLE Two cities, Boston and Kingston are both in Massachusetts. The mean household income in Boston is $38,000. The population s.d. is known to be $6,000 for a sample of 40 households. The mean income in Kingston is $35,000 for a sample of 35 households. The population s.d. is known to be $7,000. At the .01 significance level can we conclude the mean income in Boston is more? EXAMPLE Step 1 State the null and alternate hypotheses. H0: µB < µK H1: µB > µK Step 4 State the decision rule. The null hypothesis is rejected if t is greater than 2.326 or p < .01. Step 2 Select the level of significance. The .01 significance level is stated in the problem. Step 3 Find the appropriate test statistic. Since both samples are more than 30, we can use z as the test statistic. 21 EXAMPLE Step 5: Compute the value of z and make a decision. t $38,000 $35,000 1.98 ($6,000) 2 ($7,000) 2 40 35 Because the computed Z of 1.98 < critical Z of 2.26, the p-value of .0239 > .01 (), the decision is not to reject the H0. We cannot conclude that the mean household income in Boston is larger. 22 Comparing Two Population Means by Using Independent Samples: Variances Unknown • In general, the true values of the population variances σ12 and σ22 are not known • They have to be estimated from the sample variances s12 and s22, respectively 23 Comparing Two Population Means by Using Independent Samples: Variances Unknown #2 • • Also need to estimate the standard deviation of the sampling distribution of the difference between sample means Two approaches: 1. If it can be assumed that σ12 = σ22 = σ2, then calculate the “pooled estimate” of σ2 2. If σ12 ≠ σ22, then use approximate methods 24 Pooled Estimate of σ2 • Assume that σ12 = σ22 = σ2 • The pooled estimate of σ2 is the weighted averages of the two sample variances, s12 and s22 • The pooled estimate of σ2 is denoted by sp2 s 2p n1 1s12 n2 1s22 n1 n 2 2 • The estimate of the population standard deviation of the sampling distribution is x1 x2 2 s p 1 1 n1 n2 25 One sample compared with 2 samples statistics Assume that σ12 = σ22 = σ2 Mean SS2 df2 26 t-Based Confidence Interval for the Difference in Means (Variances Unknown) • Select independent random samples from two normal populations with equal variances • A 100(1 – ) percent confidence interval for the difference in populations µ1 – µ2 is 1 1 x1 x2 t 2 s 2p n n 2 1 • where s 2p n1 1s12 n2 1s22 n1 n 2 2 • and t/2 is based on (n1+n2-2) degrees of freedom (df) 27 Finding the value of the test statistic requires two steps: Step One: Pool the sample standard deviations. s 2p 2 (n1 1) s1 2 (n2 1) s2 n1 n2 2 Step Two: Determine the value of t from the following formula. t X1 X 2 1 1 s p n1 n 2 28 Hypothesis tests for μ1 – μ2 Two Population Means, Unknown Population Variances Lower-tail test: Upper-tail test: Two-tail test: H0: μ1 – μ2 0 H1: μ1 – μ2 < 0 H0: μ1 – μ2 ≤ 0 H1: μ1 – μ2 > 0 H0: μ1 – μ2 = 0 H1: μ1 – μ2 ≠ 0 -t Reject H0 if t < -t t Reject H0 if t > t /2 -t/2 /2 t/2 Reject H0 if t < -t/2 or t > t/2 29 Example: A recent EPA study compared the highway fuel economy of domestic and imported passenger cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a sample standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a sample standard deviation of 3.9. At the .05 significance level can the EPA conclude that the mpg is higher on the imported cars? 30 Example: (continued) Step 1 State the null and alternate hypotheses. H0: µD > µI H1: µD < µI Step 2 State the level of significance. The .05 significance level is stated in the problem. Step 3 Find the appropriate test statistic. Both samples are less than 30, so we use the t distribution. 31 Example: (continued) Step 4 The decision rule is to reject H0 if t<-1.708. There are n1 + n2 – 2 or 25 degrees of freedom. Step 5 We compute the pooled variance. 2 2 ( n 1 )( s ) ( n 1 )( s 2 1 1 2 2) sp n1 n 2 2 (15 1)( 2.4) 2 (12 1)(3.9) 2 9.918 15 12 2 32 Example: (continued) We compute the value of t as follows. t X1 X 2 1 1 s n n 2 1 33.7 35.7 2 p 1 1 9.918 15 12 1.640 33 Example: (continued) Since a computed z of –1.64 > critical z of –1.71, H0 can not be rejected. There is insufficient sample evidence to claim a higher mpg on the imported cars. -1.71 -1.64 34 Example: Comparing Mean weights To show if boys are heavier than girls of the same age, a survey is conducted in which a sample of 15 boys shows a mean weight of 41Kg and a standard deviation of 3Kg. A group of 10 girls of the same age shows a mean weight of 38Kg and a standard deviation of 2Kg. Assuming both the weights of boys and girls follow the normal distribution. At the level of significant 0.05, test if the average weight of boys is greater than the average weight of girls of the same age. Step 1 State the null and alternate hypotheses. H0: µg > µb H1: µg < µb 35 Example: (continued) Step 2 State the level of significance. The .05 significance level is stated in the problem. Step 3 Find the appropriate test statistic. Both samples are less than 30, so we use the t distribution. Step 4 The decision rule is to reject H0 if t > t0.05 =1.714. There are n1 + n2 – 2 or 23 degrees of freedom. 36 Example: (continued) Step 5 Compute the pooled variance and t. S2p = [(15-1)*32 + (10-1)*22]/ (15+10-2) = 7.04 Sp = 2.65 t = (41-38) / sqrt(7.04*(1/15 + 1/10)) = 2.77 =1.714 Since t =2.77 > t0.05 =1.714, we reject H0 . So the mean weight of boys is larger than the mean weight of girls of the same age. 37 Example: Directed reading activities in the classroom A class of 21 third-graders participates in these activities for 8 weeks while a control classroom of 23 third-graders follows the same curriculum without the activities. After the 8 weeks, all children take a reading test (scores in table). At a level of significance 0.05, can we conclude directed reading activities help improve reading ability? Step 1 State the null and alternate hypotheses. H0: µ1 = µ2 H1: µ1 = µ2 38 Example: Directed reading activities (continued) Step 2 State the level of significance. The .05 significance level is stated in the problem. Step 3 Find the appropriate test statistic. Both samples are less than 30, so we use the t distribution. Step 4 The decision rule is to reject H0 if t > t0.025 =1.97 or t < -t0.025 . There are n1 + n2 – 2 or 42 degrees of freedom. 39 Example: Directed reading activities (continued) Step 5 Compute the pooled variance and t. S2p = [(21-1)*11.012 + (23-1)*17.152]/ (21+23-2) = 211.79 t = (51.48-41.52) / sqrt(211.79*(1/21 + 1/23)) = 9.96/4.39=2.27 Since t =2.27 > t0.025 =1.97, we reject H0 . So there are significant difference between the 2 group. 40 Example: Directed reading activities (continued) Step 1 State the null and alternate hypotheses. H0: µ2 > µ1 H1: µ2 < µ1 Step 5 Compute the pooled variance and t. S2p = [(21-1)*11.012 + (23-1)*17.152]/ (21+23-2) = 211.79 t = (51.48-41.52) / sqrt(211.79*(1/21 + 1/23)) = 9.96/4.39=2.27 There are n1 + n2 – 2 or 42 degrees of freedom. The rule is to reject H0 if t > t0.05 =1.65. 41 Pooled Variance t Test: Example You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data: NYSE NASDAQ Number 21 25 Sample mean 3.27 2.53 Sample std dev 1.30 1.16 Assuming both populations are approximately normal with equal variances, is there a difference in average yield ( = 0.05)? Chap 9-42 Calculating the Test Statistic The test statistic is: X X μ μ t 1 2 1 1 1 S n1 n2 2 p 2 3.27 2.53 0 1 1 1.5021 21 25 2 2 n 1 S n 1 S 21 11.30 2 25 11.16 2 2 1 1 2 2 S p (n1 1) (n2 1) (21 - 1) (25 1) 2.040 1.5021 43 Solution H0: μ1 - μ2 = 0 i.e. (μ1 = μ2) H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2) = 0.05 df = 21 + 25 - 2 = 44 Critical Values: t = ± 1.96 Reject H0 Reject H0 .025 -1.96 .025 0 1.96 t 2.040 Test Statistic: Decision: 3.27 2.53 t 2.040 Reject H0 at = 0.05 1 1 Conclusion: 1.5021 21 25 There is evidence of a difference in means. 44 Two kinds of studies So far, we have studied : Independent samples two sets of sample data that come from two independent populations (e.g. women and men, or students from program A and from program B). However, sometimes we want to study Paired samples two sets of sample data that come from related populations (e.g. “before treatment” and “after treatment”). 45 Paired/Dependent Samples Independent samples are samples that are not related in any way. Dependent samples are samples that are paired or related in some fashion. *The same subjects measured at two different points in time (repeated-measures). *Matched or paired observations *Hypothesis test proceeds just as in the one sample case. 46 Paired-Sample t Test: Example Assume you work in the finance department. Is the new financial package faster (=0.05 level)? You collect the following processing times for same set of jobs: Existing System (1) 9.98 Seconds 9.88 9.84 9.99 9.94 9.84 9.86 10.12 9.90 9.91 New Software (2) Difference Di 9.88 Seconds .10 9.86 .02 9.75 .09 9.80 .19 9.87 .07 9.84 .00 9.87 - .01 9.98 .14 9.83 .07 9.86 .05 D D i n SD .072 D D i n 1 .06215 47 2 Paired-Sample t Test: Example Is the new financial package faster ( 0.05 level)? H0: mD 0 H1: mD > 0 Reject .05 .05 D = .072 Critical Value=1.8331 df = n - 1 = 9 Test Statistic t D mD .072 0 3.66 SD / n .06215/ 10 t 1.8331 Decision: Reject H0 t Stat. in the rejection zone. 3.66 Conclusion: The new software package is faster. 48 Paired-Sample: Example-twins Suppose we collect 8 pairs of twins. The first twin in the pair is healthy; the second is not. For each twin, we measure grey matter density (gmd). Is grey matter density in the populations significantly different ? Processed data from the 8 pairs is shown below (units not given). Consider the population differences, D = X1 - X2, Hypothesis Testing Involving Paired Observations (continued) If σD is unknown, we can estimate the unknown population standard deviation with a sample standard S D deviation: where D is the mean of the differences sd is the (sample) s.d. of the differences n is the number of pairs (differences) The test statistic for D is now a t statistic, with n-1 d.f. n 2 (D D ) i i 1 n 1 D μD t SD n 50 Confidence Interval of Paired Observations, σD Unknown (continued) The confidence interval for μD is SD D t n1 n n where SD is: SD 2 (D D ) i i 1 n 1 51 Hypothesis Testing for Mean Difference, σD Unknown Paired Samples Lower-tail test: Upper-tail test: Two-tail test: H0: μD 0 H1: μD < 0 H0: μD ≤ 0 H1: μD > 0 H0: μD = 0 H1: μD ≠ 0 -t Reject H0 if t < -t t Reject H0 if t > t Where t has n - 1 d.f. Chap 9-52 /2 -t/2 /2 t/2 Reject H0 if t < -t/2 or t > t/2 Paired Samples Example • Assume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data: Salesperson Chen Li Zhang Wang Wan Number of Complaints: Before (1) After (2) 6 20 3 0 4 4 6 2 0 0 (2) - (1) Difference, Di - 2 -14 - 1 0 - 4 -21 D = D i n = -4.2 SD (D 2 D ) i n 1 5.67 Chap 9-53 Paired Samples: Solution • Has the training made a difference in the number of complaints (at the 0.01 level)? H0: μD = 0 H1: μD 0 = .01 d.f. = n - 1 = 4 Critical Value = ± 4.604 D = -4.2 Test Statistic: D μ D 4 .2 0 t 1 .6 6 S D / n 5 .6 7 / 5 Reject Reject /2 /2 - 4.604 4.604 - 1.66 Decision: Do not reject H0 (t stat is not in the reject region) Conclusion: There is not a significant change in the number of complaints. Chap 9-54 An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random sample of eight cities revealed the following information. At the .05 significance level can the testing agency conclude that there is a difference in the rental charged? City Atlanta Chicago Cleveland Denver Honolulu Kansas City Miami Seattle Hertz ($) 42 56 45 48 37 45 41 46 Avis ($) 40 52 43 48 32 48 39 50 EXAMPLE 4 Step 1 Ho: md = 0 H1: md 0 Step 4 H0 is rejected if t < -2.365 or t > 2.365; or if p-value < .05. We use the t distribution with n-1 or 7 degrees of freedom. Step 2 The stated significance level is .05. Step 3 The appropriate test statistic is the paired ttest. Step 5 Perform the calculations and make a decision. 56 City Hertz Avis d d2 Atlanta 42 40 2 4 Chicago 56 52 4 16 Cleveland 45 43 2 4 Denver 48 48 0 0 Honolulu 37 32 5 25 Kansas City 45 48 -3 9 Miami 41 39 2 4 Seattle 46 50 -4 16 57 d 8.0 d 1.00 n 8 2 d d 2 n sd n 1 d t sd n 82 78 8 3.1623 8 1 1.00 3.1623 0.894 8 58 P(t>.894) = .20 for a one-tailed t-test at 7 degrees of freedom. Because 0.894 is less than the critical value, the p-value of .20 > a of .05, do not reject the null hypothesis. There is no difference in the mean amount charged by Hertz and Avis. 59 Comparing Two Population Proportions Goal: test a hypothesis or form a confidence interval for the difference between two population proportions (p1 – p2). Assumptions: two independent samples from two populations n1p1 5 , n1(1-p1) 5 n2p2 5 , n2(1-p2) 5 The point estimate for the difference is p s1 p s2 60 Two Population Proportions Since we begin by assuming the null hypothesis is true, we assume p1 = p2 and pool the two ps estimates The pooled estimate for the overall proportion is: X1 X2 p n1 n2 where X1 and X2 are the numbers from samples 1 and 2 with the characteristic of interest 61 Two Population Proportions (continued) The test statistic for p1 – p2 is a Z statistic: p Z where p s1 p s2 p1 p 2 1 1 p (1 p) n1 n2 X1 X2 X X , p s1 1 , p s2 2 n1 n2 n1 n2 62 Confidence Interval for Two Population Proportions Population proportions p s1 The confidence interval for p1 – p2 is: p s2 Z p s1 (1 p s1 ) n1 p s2 (1 p s2 ) n2 63 Example Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than 5 days last year, while a sample of 300 unmarried workers showed 35 missed more than five days. Use a .05 significance level. 64 The null and the alternate hypotheses H0: pU < pM H1: pU > pM The null hypothesis is rejected if the computed value of z is greater than 1.65 or the p-value < .05. The pooled proportion 35 22 pc = .1036 300 250 65 z 35 22 300 250 1.10 .1036 (1 .1036 ) .1036 (1 .1036 ) 300 250 Because the calculated z of 1.10 < a critical z of 1.65 ( of .05), the null hypothesis is not rejected. We cannot conclude that a higher proportion of unmarried workers miss more days in a year than the married workers. 66 Chapter Ten Two-Sample Tests of Hypothesis ONE- Conduct a test of hypothesis about the difference between two independent population means with Known/ Unknown Variances TWO- Conduct a test of hypothesis regarding the difference in two population proportions with Known/ Unknown Variances THREE- Understand the difference between dependent and independent samples. FOUR- Conduct a test of hypothesis about the mean difference between paired or dependent observations. 67 Table of the Standard Normal Distribution z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0.00 0.0000 0.0398 0.0793 0.1179 0.1554 0.1915 0.2257 0.2580 0.2881 0.3159 0.3413 0.3643 0.3849 0.4032 0.4192 0.4332 0.4452 0.4554 0.4641 0.4713 0.4772 0.4821 0.4861 0.4893 0.4918 0.4938 0.4953 0.4965 0.4974 0.4981 0.4987 0.01 0.0040 0.0438 0.0832 0.1217 0.1591 0.1950 0.2291 0.2611 0.2910 0.3186 0.3438 0.3665 0.3869 0.4049 0.4207 0.4345 0.4463 0.4564 0.4649 0.4719 0.4778 0.4826 0.4864 0.4896 0.4920 0.4940 0.4955 0.4966 0.4975 0.4982 0.4987 0.02 0.0080 0.0478 0.0871 0.1255 0.1628 0.1985 0.2324 0.2642 0.2939 0.3212 0.3461 0.3686 0.3888 0.4066 0.4222 0.4357 0.4474 0.4573 0.4656 0.4726 0.4783 0.4830 0.4868 0.4898 0.4922 0.4941 0.4956 0.4967 0.4976 0.4982 0.4987 0.03 0.0120 0.0517 0.0910 0.1293 0.1664 0.2019 0.2357 0.2673 0.2967 0.3238 0.3485 0.3708 0.3907 0.4082 0.4236 0.4370 0.4484 0.4582 0.4664 0.4732 0.4788 0.4834 0.4871 0.4901 0.4925 0.4943 0.4957 0.4968 0.4977 0.4983 0.4988 0.04 0.0160 0.0557 0.0948 0.1331 0.1700 0.2054 0.2389 0.2704 0.2995 0.3264 0.3508 0.3729 0.3925 0.4099 0.4251 0.4382 0.4495 0.4591 0.4671 0.4738 0.4793 0.4838 0.4875 0.4904 0.4927 0.4945 0.4959 0.4969 0.4977 0.4984 0.4988 0.05 0.0199 0.0596 0.0987 0.1368 0.1736 0.2088 0.2422 0.2734 0.3023 0.3289 0.3531 0.3749 0.3944 0.4115 0.4265 0.4394 0.4505 0.4599 0.4678 0.4744 0.4798 0.4842 0.4878 0.4906 0.4929 0.4946 0.4960 0.4970 0.4978 0.4984 0.4989 0.06 0.0239 0.0636 0.1026 0.1406 0.1772 0.2123 0.2454 0.2764 0.3051 0.3315 0.3554 0.3770 0.3962 0.4131 0.4279 0.4406 0.4515 0.4608 0.4686 0.4750 0.4803 0.4846 0.4881 0.4909 0.4931 0.4948 0.4961 0.4971 0.4979 0.4985 0.4989 0.07 0.0279 0.0675 0.1064 0.1443 0.1808 0.2157 0.2486 0.2794 0.3078 0.3340 0.3577 0.3790 0.3980 0.4147 0.4292 0.4418 0.4525 0.4616 0.4693 0.4756 0.4808 0.4850 0.4884 0.4911 0.4932 0.4949 0.4962 0.4972 0.4979 0.4985 0.4989 0.08 0.0319 0.0714 0.1103 0.1480 0.1844 0.2190 0.2517 0.2823 0.3106 0.3365 0.3599 0.3810 0.3997 0.4162 0.4306 0.4429 0.4535 0.4625 0.4699 0.4761 0.4812 0.4854 0.4887 0.4913 0.4934 0.4951 0.4963 0.4973 0.4980 0.4986 0.4990 0.09 0.0359 0.0753 0.1141 0.1517 0.1879 0.2224 0.2549 0.2852 0.3133 0.3389 0.3621 0.3830 0.4015 0.4177 0.4319 0.4441 0.4545 0.4633 0.4706 0.4767 0.4817 0.4857 0.4890 0.4916 0.4936 0.4952 0.4964 0.4974 0.4981 68 0.4986 0.4990 69