The sample mean and sample variance are given by: x sample standard deviation Excel: STDEV(values)

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1 Unless we have made a very large number of measurements, we don't have an accurate estimate of the mean or standard deviation of a data set. If we assume the values are normally distributed, we can estimate the mean and standard deviation from the data. The sample mean and sample variance are given by: N = i (4.4a) N Ecel: AVERAGE(values) i= S ( ) (4.4b) N 2 2 i N i Variance 2 S S sample standard deviation Ecel: STDEV(values) How close are these values to the true mean and standard deviation? That depends on how many samples we have. FiniteStatistics.doc 9/26/2008 9:37 AM Page

2 p() 4.4 Finite Statistics Normal Distribution Function Sigma = 0.5 Sigma =.0 Sigma = Sigma = X FiniteStatistics.doc 9/26/2008 9:37 AM Page 2

3 For a normally distributed data set, we can say that the probability of a sample, i, differing from the data set mean value,, is given by i t, PS (P%) (4.5) t,p is referred to as the t estimator. Eample: TABLE 4. Sample of Variable i i Find the sample mean, standard deviation, 95% precision interval within which one should epect any measured value to fall, standard deviation of the means, and the 95% estimate of the true mean value Mean: Standard Deviation: t 9,95% : % Precision Interval: Standard Deviation of the means: % Precision interval about the true mean value: FiniteStatistics.doc 9/26/2008 9:37 AM Page 3

4 Table 4.4 Student-t Distribution t 50 t 90 t 95 t FiniteStatistics.doc 9/26/2008 9:37 AM Page 4

5 Standard Deviation of the Means If we take a set of N measurements of the same variable, then repeat this process M times, the mean of each data set will differ somewhat from the others. It can be shown that the mean values themselves will follow a normal distribution even if the original distribution is not normal. The standard deviation of the means is given by: S N (4.6) Notice that the standard deviation of the mean decreases as the sample size increases. We can now say with a certainty of P% that the mean of a S / 2 FiniteStatistics.doc 9/26/2008 9:37 AM Page 5

6 sample of N values differs from the true mean of the distribution by an amount = t,p S (P%) (4.7) PROBLEM 4.6 Consider a process in which the applied measured load has a known true mean of 00 N and a variance of 400 N 2. An engineer takes 6 measurements at random. What is the probability that this sample will have a mean value between 90 and 0? KNOWN: ' = 00 N 2 =400 N 2 (so, = 20 N) FIND: For N = 6, P(90 0)? FiniteStatistics.doc 9/26/2008 9:37 AM Page 6

7 SOLUTION Begin by finding the z value for a corresponding z / N For = 90 N, z / 6 = -2.0 For = 0 N, z / 6 = 2.0 So, P(90 0) P( 2.0 z 2.0) = So, there is about a 95% chance. FiniteStatistics.doc 9/26/2008 9:37 AM Page 7

8 Table 4.3 Probability Values for Normal Error Function One-Sided Integral Solution for p( z ) (2 ) / 2 z 0 e 2 /2d z FiniteStatistics.doc 9/26/2008 9:37 AM Page 8

9 4.7 Data Outlier Detection How do you handle spurious data points? The most common and simplest approach is to label points that lie outside the range of 99.8% probability of occurrence,, as outliers. This three-sigma test works well with data set of 0 or more points. Eample 4. Given 0 data points with a mean of 27 psi and a standard deviation of 3.8 test for data outliers. Three-sigma test for small data sets gives a range of 27 ± t 9,99.8% S = 27 ± 4.3*3.8,.25 < > There are no data points outside this range. For large data sets, the modified three-sigma test for outliers can be used. A modified z variable is computed with the data set mean and standard deviation. z 0 for each data point is calculated and the corresponding probability value for the Normal Error Function is found. If the probability value is less than 0.% then the data point is considered an outlier. There is one data point ( i = 8) that is outside this range. 4.8 Number of Measurements Required Some sample statistics must be known to estimate the variation in the data set and therefore estimate a confidence interval in the data yet to be acquired. = t S,P FiniteStatistics.doc 9/26/2008 9:37 AM Page 9

10 The 95% confidence interval is therefore S CI t, 95% S t, 95% N (95%) The one-sided precision value d is d CI 2 Therefore if follows that N t, 95% S t, 95% d (95%) Problem 4.4 Estimate the number of measurements of a time-dependent acceleration signal obtained from a vibrating vehicle that would lead to an acceptable confidence interval about the mean of 0. g, if the standard deviation of the signal is epected to be 2 g. 2 S N KNOWN: CI = 0. g S = 2 g FIND: N SOLUTION Let d = CI/2 = 0.05 g. We are looking for the number of measurements required to keep ts 0.05g at 95%. N = ( ts / d) 2 If we select a large number of measurements, such that t N,95 =.96, then N 650 For this value, the t value remains unchanged. Thus, a large number of measurements are required due to the close restriction on CI. FiniteStatistics.doc 9/26/2008 9:37 AM Page 0

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