Calculation of the range

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IEGR 351 CHAPTER 10: INFERENCES CONCERNING PROPORTIONS Sections 10.1-10.3 Agenda ■ The Estimation of Proportions ■ Hypothesis Concerning One Proportion ■ Hypothesis Concerning Several Proportions The Estimation of Proportions ■ This chapter will discuss how to do hypothesis testing on the proportions. ■ Some examples we can use to think about this are acceptance sampling or life testing of a component. ■ The information we need to use to estimate a proportion is the number of times X, that an appropriate event occurs in n trials, occasions, or observations. ■ The point estimator is denoted as the sample proportion or 𝑋 𝑛 ■ If the n trials in our problems satisfies the assumptions in the binomial distribution we can use the mean and standard deviation we can find mean and standard deviation of the proportion of successes given by the following equations 𝑛𝑝 =p 𝑛 or 𝑛𝑝(1 − 𝑝) = 𝑛 𝑝(1 − 𝑝) 𝑛 Estimation of Proportions ■ The confidence interval associated with the proportion is as follows ■ 𝑥 𝑛 − 𝑧𝛼Τ2 𝑥 𝑛 𝑥 1−𝑛 𝑥 𝑛 < 𝑝 < + 𝑧𝛼Τ2 𝑛 𝑥 𝑛 𝑥 1−𝑛 𝑛 ■ The error equation associated with proportions is as follows ■ 𝐸 = 𝑧𝛼Τ2 𝑝(1−𝑝) 𝑛 ■ This is the equation to find the sample size if we know the proportion ■ 𝑛 = 𝑝(1 − 𝑝) 𝑧𝛼Τ2 2 𝐸 ■ This is the equation to find the sample size if we do not know the proportion ■ 𝑛= 1 𝑧𝛼Τ2 2 4 𝐸 ■ Let’s take a look at some examples Estimation of Proportions ■ If x=36 of n=100 persons interviewed are familiar with the tax incentives for installing certain energy-saving devices, construct a 95% confidence interval for the corresponding true proportion. ■ 𝑥 𝑛 = 36 100 = 0.36 ■ And 𝑧𝛼Τ2 = 1.96 ■ 0.36 − 1.96 0.36∗0.64 100 < 𝑝 < 0.36 + 1.96 0.36∗0.64 100 = 0.266 < 𝑝 < 0.454 ■ We are 95% confident that the population proportion of the person familiar with the tax incentive, p, is contained in the interval from 0.266 to 0.454. Estimation of Proportions ■ In a sample survey conducted in a large city, 136 of 400 persons answered yes to the question of whether their city’s public transportation is adequate. With 99% confidence what can we say about the maximum error, if ■ 𝑥 𝑛 = 136 400 = 0.34 is used as an estimate of the corresponding true proportion? ■ And 𝑧𝛼Τ2 = 2.575 ■ 𝐸 = 2.575 0.34∗0.66 400 = 0.061 Estimation of Proportions ■ Suppose that we want to estimate the true proportion of defectives in a very large shipment of adobe bricks, and that we want to be at least 95% confident that the error is at most 0.04. How large a sample will we need if ■ We have no idea what the true proportion might be; ■ 𝑛= 1 1.96 2 4 0.04 = 600.25 ~601𝑟𝑜𝑢𝑛𝑑 𝑢𝑝 𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 ■ We know that the true proportion does not exceed 0.12? ■ 𝑛 = 0.12 0.88 1.96 2 0.04 = 253.55~254 Hypothesis Concerning One Proportion ■ Many methods used in sampling inspection, quality control, and reliability verification are based on tests of null hypothesis that a proportion(percentage or probability) equals some specified constant. ■ Based on the table below we will look at tests performed with approximate large sample tests based on the normal approximation to the binomial distribution. ■ Null hypothesis: 𝑝 = 𝑝0 ■ Alternative hypothesis: 𝑝 < 𝑝0 , 𝑝 > 𝑝0 , 𝑝 ≠ 𝑝0 ■ This test is used when the desire to control the uniformity of a product or operation. ■ The test statistic used for this is below ■ 𝑍= 𝑋−𝑛𝑝0 𝑛𝑝0 (1−𝑝0 ) Critical Regions for Testing 𝐩 = 𝑝0 (large sample) Alternative Hypothesis Reject null hypothesis if: 𝑝 < 𝑝0 𝑍 < −𝑧𝛼Τ2 𝑝 > 𝑝0 𝑍 > 𝑧𝛼Τ2 𝑝 ≠ 𝑝0 𝑍 < −𝑧𝛼Τ2 or 𝑍 > 𝑧𝛼Τ2 Hypotheses Concerning One Proportion ■ Transceivers provide wireless communications among electronic components of consumer products. Responding to a need for a fast, low-cost test of Bluetoothcapable transceivers engineers developed a product test at the wafer level. IN one set of trial with 60 devices selected from different wafer lots , 48 devices passed. Test the null hypothesis p>0.070 at the 0.95 level of significance. ■ Parameter of interest: Testing transceivers ■ Null hypothesis: p = 0.70 ■ Alternative hypothesis: p > 0.70 ■ Level of significance: 𝛼 = 0.05 ■ Test Statistic: 𝑍 = 𝑋−𝑛𝑝0 𝑛𝑝0 (1−𝑝0 ) Hypothesis Concerning One Proportion ■ Criterion: Reject the null hypothesis if Z> 1.645 ■ Calculation: ■ 𝑧= 48−60(0.70) 60(0.70) (0.30) = 1.69 ■ Decision: Since 𝑧 = 1.69 is greater than 1.645, we reject the null hypothesis at level 0.05. In other words, there is sufficient evidence to conclude that the proportion of good transceivers that would be produced is greater than 0.70. The P-value, (P(Z>1.69)=0.0455, somewhat strengthens this conclusion. Hypotheses Concerning Several Proportions ■ In this case we are testing whether two or more binomial populations have the same parameter p. We are interested in the null hypothesis being 𝑝1 = 𝑝2 … . . 𝑝𝑛 and the alternative hypothesis similar to we have seen before. ■ Test Statistic for test concerning the difference between two proportions is below ■ 𝑍= 𝑋1 𝑋2 −𝑛 𝑛 1 1 ො ො 𝑝(1− 𝑝) + 𝑛1 𝑛2 𝑤𝑖𝑡ℎ 𝑝Ƹ = 𝑋1 +𝑋2 𝑛1 +𝑛2 ■ The large sample confidence interval for the difference between two proportions is below ■ 𝑥2 𝑛1 𝑥 − 𝑛2 2 ± 𝑧𝛼Τ2 𝑥1 𝑥 (1−𝑛1 ) 𝑛1 1 𝑛1 + 𝑥2 𝑥 (1−𝑛2 ) 𝑛2 2 𝑛2 Hypotheses Concerning Several Proportions ■ A study shows that 16 of 200 tractors produced on one assembly line required extensive adjustments before they could be shipped, while the same was true for 14 out of 400 tractors produced on another assembly line. At the 0.01 level of significance, does this support the claim that the second production line does superior? ■ Parameter of interest: The quality of tractors from the production line ■ Null hypothesis: 𝑝1 = 𝑝2 this can also be written as 𝑝1 − 𝑝2 = 𝛿0 ■ Alternative hypothesis: 𝑝1 > 𝑝2 ■ Level of significance: 𝛼 = 0.01 ■ Test Statistic: 𝑍 = 𝑋1 𝑋2 − 𝑛 𝑛 1 1 ො ො 𝑝(1− 𝑝) + 𝑛1 𝑛2 𝑤𝑖𝑡ℎ 𝑝Ƹ = 𝑋1 +𝑋2 𝑛1 +𝑛2 ■ Criterion: Reject null hypothesis if Z>2,33, where Z is given by the formula Hypotheses Concerning Several Proportions ■ Calculations ■ 𝑍= 16 14 − 200 400 1 1 0.05 (0.95) + 200 400 = 2.38 𝑤𝑖𝑡ℎ 𝑝Ƹ = 16+14 200+400 = 0.05 ■ Decision: Since Z=2.38 exceed 2.33, the null hypothesis must be rejected; we conclude that the true proportion of tractors requiring extensive adjustments is greater for the first assembly line that for the send. The P-value is 0.0087 Hypothesis Concerning Several Proportions ■ With reference to the example on the previous slide let’s find the 95% confidence interval. ■ 0.08 − 0.035 ± 1.96 0.08 0.92 200 + 0.035 0.965 400 = 0.003 < 𝑝1 − 𝑝2 < 0.087 ■ The first assembly line has a rate of extensive adjustment between 3 and 87 out of 1,000 higher than the rate for the second assembly IEGR 351 CHAPTER 9: INFERENCES CONCERNING VARIANCES Part 1: Sections 9.1-9.3 Agenda ■ The Estimation of Variances ■ Hypothesis Concerning One Variance ■ Hypothesis Concerning Two Variances The Estimation of Variances ■ The sample variance is an unbiased estimator of 𝜎 2 , it does not follow that the sample standard deviation is also unbiased estimator of 𝜎. ■ 𝑆2 = ത 2 σ𝑛 𝑖=1(𝑋𝑖 −𝑋) 𝑛−1 is an unbiased estimator of 𝜎 2 ■ For large samples it is common practice to estimate 𝜎 with 𝑠 ■ Population standard deviations can also be estimated in terms of sample range 𝑅 ■ 𝑅 𝑑2 this is a good estimate to use when the sample size is less than or equal to 5 for larger samples it is more appropriate to use 𝑠 𝑎𝑠 𝑎𝑛 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝜎. Below is a sampling distribution of the range for different n. n 2 3 4 5 6 7 8 9 10 𝑑2 1.128 1.693 2.059 2.326 2.534 2.704 2.847 2.970 3.078 𝑑3 0.853 0.888 0.880 0.864 0.848 0.833 0.820 0.808 0.797 Estimation of Variances ■ Let’s use data in the previous chapter to illustrate using the range estimates. ■ Mine 1: 8,260 8,130 8,350 8,070 8,340 ■ Mine 2: 7,950 7,890 7,900 8,140 7,920 7,840 ■ Let’s look at the first sample and you can try the second sample to get practice. ■ To take the range we take the difference of the smallest and largest value of the sample then we have to find the value 𝑑2 of the sample ■ 𝑅 𝑑2 = 8350−8070 2.326 = 120.4 Estimation of Variances ■ In practical applications, interval estimates 𝜎 𝑜𝑟 𝜎 2 are based on the sample standard deviation or the sample variance. For random samples from normal populations, we make use of Theorem 6.5 according to which is a random variable having the chi square distribution with n-1 degrees of freedom. ■ (𝑛−1)𝑆 2 𝜎2 ■ We can assert with probability 1-𝛼 that the inequality will be satisfied. ■ 2 χ1−𝛼/2 < (𝑛−1)𝑆 2 𝜎2 < χ2𝛼/2 x^2 = (𝑛−1)𝑆 2 𝜎2 ■ Solving this inequality for 𝜎 2 we can get the confidence interval, ( take the square root we get the confidence interval for 𝜎 Estimation of Variances ■ (𝑛−1)𝑠 2 χ2𝛼/2 ■ The confidence intervals for 𝜎 𝑎𝑛𝑑 𝜎 2 obtained by taking equal tails as in the above formula, do not actually give the narrowest confidence interval, because the chi square distribution is not symmetrical. ■ Let’s look at an example to use these equations ■ Suppose the refractive indices of 20 pieces of glass (randomly selected from a large shipment purchased by the optical firm) have a variance of 1.20 ∗ 10−4 . Construct a 95% confidence interval for 𝜎, the standard deviation of the population sampled. ■ Solution: For 20-1=19 degrees of freedom, χ20.975 = 8.907 and χ20.025 = 32.852 according to the Table 5 for the chi squared table we can substitute the values into the equation ■ 𝜎02 𝜎 2 ≠ 𝜎02 ■ This test is used when the desire to control the uniformity of a product or operation. ■ The test statistic used for this is below 2 ■ χ = (𝑛−1)𝑆 2 𝜎02 with n-1 degrees of freedom ■ Below are the critical regions Critical Regions for Testing 𝜎 2 = 𝜎02 (normal population) Alternative Hypothesis Reject null hypothesis if: 𝜎 2 < 𝜎02 2 χ2 < χ1−𝛼 𝜎 2 > 𝜎02 χ2 > χ2𝛼 𝜎 2 ≠ 𝜎02 2 χ2 < χ1−𝛼/2 or χ2 < χ2𝛼/2 Hypothesis Concerning One Variance ■ Remember since the chi squared distribution is not symmetrical. For moderate tests of degrees of freedom the two tests are nearly the same. ■ Let’s take a look at an example we will still be following the same 8 step procedure ■ The lapping process which is used to grind certain silicon wafers to the proper thickness is acceptable only if 𝜎, the population standard deviation of the thickness of dice cut from the wafers is at most 0.50 mil. Use the 0.05 level of significance to test the null hypothesis 𝜎 = 0.50 against the alternative hypothesis 𝜎 > 0.50, if the thicknesses of 15 dice cut from such wafers have a standard deviation of 0.064 mil. ■ Parameter of interest: Lapping process to grind silicon wafers ■ Null hypothesis: 𝜎 = 0.50 ■ Alternative hypothesis: 𝜎 > 0.50 ■ Level of significance: 𝛼 = 0.05 ■ Test Statistic: χ2 = (𝑛−1)𝑆 2 𝜎02 Hypotheses Concerning One Variance ■ Criterion: Reject the null hypothesis if Χ 2 > 23.685, the value of 2 Χ0.05 for 14 degrees of freedom ■ Calculation: 2 ■ χ = (15−1)(0.64)2 (0.50)2 = 22.94 ■ Decision: Since χ2 =22.94 does not exceed 23.685, the null hypothesis cannot be rejected; even though the sample standard deviation exceeds 0.50, there is not sufficient evidence to conclude that the lapping process is unsatisfactory. Hypotheses Concerning Two Variances ■ Back in chapter 8 we performed a two sampled t test where the variances of the two populations sampled are equal. This section of the book test the null hypothesis 𝜎12 = 𝜎22 , which applies to independent random samples from two normal populations. ■ Statistic for test of equality of two variances (normal populations) ■ 𝐹= 𝑆12 𝑆22 ■ The F distribution with 𝑛1 − 1 and 𝑛2 − 1 degrees freedom ■ If the null hypothesis 𝜎12 = 𝜎22 is true, the ratio of the sample variances 𝑆12 𝑎𝑛𝑑 𝑆22 provides a statistic on which tests of the null hypothesis can be based. ■ The critical region is as follows on the next page. Since the tables we have only show the alpha values for the right hand tails you can use the reciprocal of the original test statistic and make use of the relation ■ 𝐹1−𝛼 ν1 , ν2 = 1 𝐹𝛼 (ν2 ,ν1 ) Hypotheses Concerning Two Variances Critical regions for testing 𝝈𝟐𝟏 = 𝝈𝟐𝟐 normal population Alternative hypothesis Size Mean 𝜎12 < 𝜎22 𝑆12 𝐹= 2 𝑆2 𝐹 > 𝐹𝛼 (𝑛2 − 1, 𝑛1 − 1) 𝜎12 > 𝜎22 𝑆12 𝐹= 2 𝑆2 𝐹 > 𝐹𝛼 (𝑛2 − 1, 𝑛1 − 1) 𝜎12 ≠ 𝜎22 𝑆12 𝐹= 2 𝑆2 𝐹 > 𝐹𝛼/2 (𝑛𝑀 − 1, 𝑛𝑚 − 1) Note: Similar to the chi squared test, equal tails are used in the two-tailed test as a matter of mathematical convenience, even though the F distribution is not symmetrical ■ Confidence Interval ■ 𝐹1−𝛼Τ2 𝑛1 − 1, 𝑛2 − 1 𝑠22 𝑠12 < 𝜎22 𝜎12 < 𝐹𝛼/2 𝑛1 − 1, 𝑛2 − 1 Hypotheses Concerning Two Variances ■ It is desired to determine whether there is less variability in the silver plating done by Company 1 than in that done by Company 2. If independent random samples of size 12 of the two computers’ work yield 𝑠1 = 0.035 mil and 𝑠2 = 0.062 mil , test the null hypothesis 𝜎12 = 𝜎22 against the alternative hypothesis 𝜎12 < 𝜎22 at the 0.05 level of significance. ■ Parameter of interest: variability in silver plating ■ Null hypothesis: 𝜎12 = 𝜎22 ■ Alternative hypothesis: 𝜎12 < 𝜎22 ■ Level of Significance:𝛼 = 0.05 ■ Test Statistic: 𝐹 = 𝑆12 𝑆22 Hypotheses Concerning Two Variances ■ Criterion: Reject null hypothesis if F>2.82, the value of 𝐹0.05 for 11 and 11 degrees of freedom ■ Calculations : 𝐹 = (0.062)2 (0.035)2 = 3.14 ■ Decision: Since F=3.14 exceeds 2.82, the null hypothesis must be rejected in other words, the data support the contention that the plating done by Company 1 is less variable than that done by Company 2. Hypotheses Concerning Two Variances ■ ■ ■ ■ ■ ■ ■ ■ Let’s use this data again Mine 1: 8,260 8,130 8,350 8,070 8,340 Mine 2: 7,950 7,890 7,900 8,140 7,920 7,840 Use a level of significance 0.02 to test whether it is reasonable to assume that the variances of the two population sampled are equal. Parameter of interest: variability in silver plating Null hypothesis: 𝜎12 = 𝜎22 Alternative hypothesis: 𝜎12 ≠ 𝜎22 Level of Significance:𝛼 = 0.02 ■ Test Statistic: 𝐹 = 𝑆12 𝑆22 Hypotheses Concerning Two Variances ■ Criterion: Reject null hypothesis if F>11.4, the value of 𝐹0.01 for 4 and 5 degrees of freedom ■ Calculations : 𝐹 = 15750 10920 = 1.44 ■ Decision: Since F=1.44 does not exceeds 11.4, the null hypothesis cannot be rejected; there is no real statistical reason to doubt the equality of the variances of the two populations.. Hypotheses Concerning Two Variances ■ Let’s say we have the following data from an earlier problem based on making green gasoline from sucrose. The equal sample sizes2are 𝑛1 = 𝑛2 = 9, 𝑠12 = 0.4548, and 𝑠22 = 𝜎 0.1089. Obtain a 98% confidence interval for 22 . 𝜎1 ■ Since the degrees of freedom for the F are (𝑛2 −1, 𝑛1 − 1) = (8,8) and 𝛼 Τ2 = 20.01, we 1 𝜎 find 𝐹0.01 = 6.03 and 𝐹0.99 = = 1/6.03. The 98% confidence interval for 22 𝐹0.01 𝜎1 becomes ■ 1 0.1089 0.10889 , 6.03 6.03 0.4548 0.4548 or (0.04,1.44) ■ The wideness of the interval illustrates the large amounts of variability in variances when sample sizes are small. The second variance 𝜎22 could be as small as one-twenty-fifth of 𝜎12 or it could be larger than 𝜎12 . ■ The procedures used in this chapter depend strongly on the assumption that the underlying population is normal. The sample variance can change when it depart from normality. IEGR 351 Directions: Please review the slides on Inference and the new slides I will put up on proportions. Remember to show all 8 steps if doing testing to get full points. Problem 1. Using the data below estimate the 𝜎 for the Brinell hardness of Alloy 1 in terms of the sample range. Alloy1 Alloy2 66.3 71.3 63.5 60.4 64.9 62.6 61.8 63.9 64.3 68.8 64.7 70.1 65.1 64.8 64.5 68.9 68.4 65.8 63.2 66.2 Problem 2. Using the data below construct a 99% confidence interval for the variance of the yield of carbon chain lengths n=9. 0.63 2.64 1.85 1.68 1.09 1.67 0.73 1.04 0.68 Problem 3. If 12 determinations of the specific heat of iron have a standard deviation of 0.0086, test the null hypothesis that 𝜎 = 0.010 for such determinations. Use the alternative hypothesis 𝜎 ≠ 0.010 and the level of significance 𝛼 = 0.01. Problem 4. Pull strength tests on 10 soldered leads for a semiconductor device yield the following results in poundsforce required to rupture the bond: 15.8 12.7 13.2 16.9 10.6 18.8 11.1 14.3 17.0 12.5 Another set of 8 leads was tested after encapsulation to determine whether to pull strength has been increased by encapsulation of the device, with the following result: 24.9 23.6 19.8 22.1 20.4 21.6 21.8 22.5 Use 0.02 level of significance to test whether it is reasonable to assume that the two samples come from populations of equal variances. Problem 5. In a random sample of 200 claims filed against an insurance company writing collision insurance for cars, 84 exceeded $3,500. Construct a 95% confidence interval for the true proportion of claims filed against this insurance company that exceed $3,500 using the large sample confidence interval formula. Problem 6. In a recent study, 69 of 120 meteorites were observed to enter the earth’s atmosphere with a velocity of 69 less than 26 miles per second. If we estimated the corresponding true proportion as 120 = 0.575 What can we say with 95% confidence about the maximum error? Problem 7. What is the size of a smallest sample required to estimate an unknown proportion of customers who would pay for an additional service, to within a maximum error of 0.06 with at least 95% confidence? Problem 8. A manufacturer of submersible pumps claims that at most 30% of the pumps require repairs within the first 5 years of operation. If a random sample of 120 of these pumps includes 47 which require repairs within the first 5 years, test the null hypothesis p=0.03 against the null hypothesis p>0.30 at the 0.05 level of significance.
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Kindly see attached files with the step by step solution of the different problems. I've included both a word and a pdf versions so that you can use any of them in case you find a problem when trying to read the equations used

Problem 1.

Calculation of the range:
Minimum (alloy 1) = 61.8
Maximum (alloy 1) = 68.4
Range = Maximum – Minimum = 68.4 – 61.8 = 6.6
On the other hand, the corresponding d2 statistic for a sample size of 10 is 3.078. Taking this into
account, the estimate of the population standard deviation would be:
𝜎=

𝑅𝑎𝑛𝑔𝑒
6.6
=
~2.144
𝑑2
3.078

Problem 2.

Taking into account the small sample size and that the population standard deviation is unknown, the
confidence interval should be constructed by applying the t statistic, such that:
𝜇 = 𝑥̅ ± 𝑡 ∗

𝑠
√𝑛

Where x is the sample mean, s the sample standard deviation, n the sample size and t is the two-side t
statistic corresponding to 8 (n-1) degrees of freedom and a 99% confidence level. Having a look at a t
student distribution, the corresponding t statistic would be of 3.355. On the other hand, the sample
mean and sample standard deviation would be:
𝑥̅ =

∑𝑖 𝑥𝑖 0.63 + 2.64 + 1.85 + ⋯ + 1.04 + 0.68
=
= 1.33
𝑛
9

∑𝑖(𝑥𝑖 − 𝑥̅ )2
(0.63 − 1.33)2 + (2.64 − 1.33)2 + ⋯ + (0.68 − 1.33)2
𝑠=√
=√
= 0.67
𝑛−1
9−1
Substituting these values into the equation for the 99% confidence interval,
𝜇 = 𝑥̅ ± 𝑡 ∗

𝑠
√𝑛

= 1.33 ± 3.355 ∗

0.67
√9

= 1.33 ± 0.75 = (0.58, 2.58)

Meaning that the 99% confidence interval for the yield of carbon chain lengths ranges between 0.58 and
2.58.

Problem 3.

Taking the provided information into account, the hypotheses being tested would be:



Null hypothesis: σ = 0.010
Alternative hypothesis: σ ≠ 0.010

Calculation of the corresponding chi statistic:
𝛸2 =

(𝑛 − 1)𝑠 2 (12 − 1)0.00862
=
= 8.1356
𝜎2
0.0102

On the other hand, the corresponding critical statistics considering 11 degrees of freedom and a
significance level of α = 0.01 would be 2.60 and 26.8. Taking into account that the computed chi statistic
(8.14) is outside the critical region, there is not enough evidence to support the rejection of the null
hypothesis. From this point of view, the computed sample standard deviation (0.0086) is not
significantly different from the population standard deviation (0.0100), and the small difference
observed shall be attributed to a random effect. As a result, the researcher would conclude that the
population standard deviation (σ) is 0.010 with a 99% confidence level.

Problem 4.

Taking the provided information into account, the hypotheses being tested would be:



Null hypothesis: σ1 = σ2
Alternative hypothesis: σ1 ≠ σ2

Calculation of the respective sample sta...


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