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MECH 373 Instrumentation and Measurements

MECH 373 Instrumentation and Measurements. Lecture 13/A. Statistical Analysis of Experimental Data (Chapter 6). • Introduction • General Concepts and Definitions • Probability • Probability Distribution Function • Parameter Estimation. PDF with Engineering Applications.

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MECH 373 Instrumentation and Measurements

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  1. MECH 373Instrumentation and Measurements Lecture 13/A Statistical Analysis of Experimental Data (Chapter 6) • Introduction • General Concepts and Definitions • Probability • Probability Distribution Function • Parameter Estimation

  2. PDF with Engineering Applications A number of distribution functions are used in engineering applications. We will discuss briefly about some common distribution functions. ►Binomial (values either true/false) ►Poisson ►Normal (Gaussian) ►Student’s t ► χ2 (Chi-squared) ►Weibull ►Exponential ►Uniform

  3. Probability Distribution Function Binomial Distribution • The binomial distribution is a distribution which describes discrete random variables that can have only two possible outcomes: “success” or “failure”. • This distribution has applications in production quality control, when the quality of a product is either acceptable or unacceptable. • The binomial distribution provides the probability (P) of finding exactly r successes in a total of n trials and is expressed as where, p is the probability of success which remains constant throughout the experiment, and is called as n combination r, which is the number of ways that we can choose r identical items from n items. • The expected number of successes in n trials for binomial distribution is • The standard deviation of the binomial distribution is

  4. Probability Distribution Function Poisson Distribution • The Poisson distribution is used to estimate the number of random occurrences of an event in a specified interval of time or space if the average number of occurrences is already known. • The probability (P) of occurrence of x events is given by where, λ is the expected or mean number of occurrences during the interval of interest. • The expected value of x for the Poisson distribution, the same as the mean (μ), is given by • The standard deviation is given by • Probability that the number of occurrences is less than or equal to k is given by

  5. -1 0 1 2 3 4 5 Probability Distribution Function Normal (Gaussian) Distribution • The Normal or Gaussian distribution is a simple distribution function that is useful for a large number of common problems involving continuous random variables. • Symmetric about μ • Bell-shaped • Mean μ: the peak of the density occurs • Standard deviation σ: indicates the spread of the bell curve. m = 2

  6. Standard Normal Distribution(mean=0, standard deviation=1) For a given population, the probability of having a single value of x between a lower limit of x1 and upper limit of x2 is Since f(x) is in the form of an error function, the integration must be performed numerically. To simplify the integration process, the integrand is modified with a change of variable. A non-dimensional variable z is defined as: A function f(z) which is called the standard normal density function is defined as This function represents the normal probability density function for a random variable z, with mean μ = 0, and standard deviation σ = 1. The function f(z) is plotted below versus z.

  7. Standard Normal Distribution(mean=0, standard deviation=1) The probability that x is between x1 and x2 is the same as the probability that the transformed variable z is between z1 and z2. That is

  8. Confidence Intervals in Sample Population It is also of interest to determine the probability that a measurement will fall within one or more standard deviations of the mean. The values are listed in the following table: • This implies that if a measurement is performed in an environment where the deviation from the mean value is totally influenced by random variables, we are • • 68.26% confident that the value of the random variable will fall within one standard deviation (1σ) from the mean. • • 95.44% confident that the value of the random variable will fall within two standard deviation (2σ) from the mean. • • 99.74% confident that the value of the random variable will fall within three standard deviations (3σ) from the mean. • • 99.96% confident that the value of the random variable will fall within 3.5 standard deviations (3.5σ) from the mean.

  9. 3s Z [-3,3] 1s Z[-1,1] 2s Z [-2,2] 68% 95% 99.7% Standard Normal Distribution (mean=0, standard deviation=1)

  10. 95% 99.7% 68% Normal Distribution Example 1 • The distribution of heights of American women aged 18 to 24 is approximately normally distributed with mean 65.5 inches and standard deviation 2.5 inches. • 68% of these American women have heights between 65.5 – 1(2.5) and 65.5 + 1(2.5) inches, or between 63 and 68 inches. • 95% of these American women have heights between 65.5 - 2(2.5) and 65.5 + 2(2.5) inches, or between 60.5 and 70.5 inches. • 99.7% of these American women have heights between 65.5 - 3(2.5) and 65.5 + 3(2.5) inches, or between 58 and 73 inches.

  11. Normal distribution mean value u = 10.0, Standard deviation σ = 1.0 A, Probability a single value between 9 and 12? Z1 = (x – u)/σ = (9 - 10) / 1 = -1 P(-1 ≤ Z ≤ 0) = 0.3413 (Area from 0 to -1) Z2 = (12 – 10) / 1 = 2 P(0 ≤ Z ≤ 2) = 0.4772 ( area from 0 to 2) P(-1 ≤ Z ≤ 2) = 0.3413 + 0.4772 = 0.8185 (82%). B, Probability less than 9? Z1 = -1 P(-1 ≤ Z ≤ 0) = 0.3413 (Area from 0 to -1) P (-∞ ≤ Z ≤ -1) = 0.5 – 0.3413 = 0.1587 (from -1 to negative infinity) Normal Distribution Example 2

  12. Diameter tolerance of cylinders – normal distribution. Average diameter u = 4in, Standard deviation = 0.002. A, Probability of D ≤ 4.002. Z = (D – u) /σ = (4.002 – 4) / 0.002 = 1 P(0 ≤ Z ≤ 1) = 0.3413 (area from 0 to 1) P (-∞ ≤ Z ≤ 0) = 0.5 (area from 0 to negative infinity) P(-∞ ≤ Z ≤ 1) = 0.5 + 0.3413 = 0.8413 (84%) B, D > 2 σ rejected? P(0 ≤ Z ≤ 2) = 0.4772, also P(-2 ≤ Z ≤ 0) = 0.4772 P(-2 ≥ Z ≥ 2) = 1 – 2*0.4772 = 0.0456 (about 5%). Normal Distribution Example 3

  13. Parameter Estimation In most experiments the sample size is small relative to the population, yet the principle intention from the statistical aspect is to estimate the parameters describing the entire population. The most common estimated parameter is the population mean and in some cases it is also necessary to estimate the population standard deviation. • Estimate population mean using sample mean • Estimate population standard deviation using sample standard deviation • However, different samples from the sample population will generate different values. Therefore, it is necessary to determine the uncertainty intervals of estimated parameters.

  14. Interval Estimation of Population Mean • Estimate of population mean • Confidence interval • Confidence level (degree of confidence) • Level of significance Confidence level = 1- a where α is the probability that the mean will fall outside the confidence interval.

  15. From Sample to Population • Choose many samples from population • Find the mean of each sample • Determine the uncertainly range of the means (Sample mean is a variable !!!) How to estimate the statistics of a population from only one single sample? The central limit theorem makes it possible to make an estimate of the confidence interval with a suitable confidence level.

  16. Central Limit Theorem (a) • Consider a population of the random variable x with a mean of μ and standard deviation of σ • Several different samples, each of size n, mean of • If sample size n is sufficiently large (>30), then (standard error of the mean) • For the central limit theorem to apply, the sample size n must be large (typically greater than 30) • If original population is normal, the distribution for the is normal. We can use the following equation • If original population is not normal and n is large (n > 30) the distribution for the is normal • If original population is not normal and n < 30 the distribution for the is only approximately normal

  17. Confidence Interval Interval Estimation of Population with confidence level 1-a

  18. Interval Estimation of Population • • If z is zero, this means that • • However, we only expect the true value of μ to lie somewhere within the confidence interval, which is • • If the sample size is less than 30, the assumption that the population standard deviation can be represented by the sample standard deviation may not be accurate. • • In case of small samples, a statistic called Student’s t distribution is used. That is

  19. Normal Distribution Table

  20. Example 6.11 We would like to determine the confidence interval of the mean of a batch of resistor made in a certain process. Based on 36 measurements, the average resistance is 25  and the sample stanadard deviation is 0.5 . Determine the 90% confidence interval of the mean resistance of the batch. • n = 36 • Average = 25  • Sample standard deviation S = 0.5  • Determine 90% confidence interval of the mean • Solution: • 1-α=90%, -> α = 0.1 • 0.5- α/2=0.5-0.05=0.45 • Table 6.3 -> Z α/2=1.645

  21. Student’s t-distribution (n<30) In case of small samples, a statistic called Student’s t distribution is used. That is • Student’s t, degree of freedom ν = n-1

  22. Confidence Interval Interval Estimation of Population Mean (n<30) with confidence level 1-a

  23. Student’s t-Distribution Table

  24. Example 6.12 • n=6 • 1250, 1320, 1542, 1464, 1275, and 1383 h • Estimate population mean and 95% confidence interval on the mean • Solution: • Mean=(1250+1320+1542+1464+1275+1383)/6=1372 h • S=[(1250-1372)2+(1320-1372)2+(1542-1372)2+(1464-1372)2+(1275-1372)2+(1383-1372)2]/(6-1)]1/2 = 114 h • ν=n-1=6-1=5, α=1-95%=0.05, α/2=0.05/2=0.025 • Table 6.6 -> t α/2=2.571

  25. Example 6.13 • Reduce the 95% confidence interval to ±50 h from ±120 h • Determine how many more systems should be tested • Solution: Assume n>30, • α=1-95%=0.05, 0.5-α/2=0.5-0.05/2=0.475 • Table 6.3 -> z α/2=1.96 • n = (1.96x114/50)2=20 <30 => t-distribution • ν=n-1=20-1=19, α/2=0.05/2=0.025 • Table 6.6 -> t α/2=2.093 • n = (2.093x114/50)2=23

  26. ν = 4 Area=α/2 ν = 10 Area=α/2 Chi-squared Distribution Function • Relates the sample variance to the population variance

  27. ν = 4 Area=α/2 ν = 10 Area=α/2 Interval Estimation of Population Variance With confidence level 1-a

  28. Chi-squared Distribution Table

  29. Example 6.14 • n = 20, mean = 0.32500 in, S = 0.00010 in • obtain 95% confidence interval for the standard deviation • Solution: • ν=n-1=20-1=19, α=1-95%=0.05, α/2=0.05/2=0.025, 1- α/2=1-0.05/2=0.975 • Table 6.7 -> χ2ν, α/2=32.825, χ2ν, 1-α/2=8.9065

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