Category: statistics

Statistics – Inverse Gamma Distribution ”; Previous Next Inverse Gamma Distribution is a reciprocal of gamma probability density function with positive shape parameters $ {alpha, beta } $ and location parameter $ { mu } $. $ {alpha } $ controls the height. Higher the $ {alpha } $, taller is the probability density function (PDF). $ {beta } $ controls the speed. It is defined by following formula. Formula ${ f(x) = frac{x^{-(alpha+1)}e^{frac{-1}{beta x}}}{ Gamma(alpha) beta^alpha} \[7pt] , where x gt 0 }$ Where − ${alpha}$ = positive shape parameter. ${beta}$ = positive shape parameter. ${x}$ = random variable. Following diagram shows the probability density function with different parameter combinations. Print Page Previous Next Advertisements ”;

Statistics – Relative Standard Deviation ”; Previous Next In probability theory and statistics, the coefficient of variation (CV), also known as relative standard deviation (RSD), is a standardized measure of dispersion of a probability distribution or frequency distribution. Relative Standard Deviation, RSD is defined and given by the following probability function: Formula ${100 times frac{s}{bar x}}$ Where − ${s}$ = the sample standard deviation ${bar x}$ = sample mean Example Problem Statement: Find the RSD for the following set of numbers: 49, 51.3, 52.7, 55.8 and the standard deviation are 2.8437065. Solution: Step 1 – Standard deviation of sample: 2.8437065 (or 2.84 rounded to 2 decimal places). Step 2 – Multiply Step 1 by 100. Set this number aside for a moment. ${2.84 times 100 = 284}$ Step 3 – Find the sample mean, ${bar x}$. The sample mean is: ${frac{(49 + 51.3 + 52.7 + 55.8)}{4} = frac{208.8}{4} = 52.2.}$ Step 4Divide Step 2 by the absolute value of Step 3. ${frac{284}{|52.2|} = 5.44.}$ The RSD is: ${52.2 pm 5.4}$% Note that the RSD is expressed as a percentage. Print Page Previous Next Advertisements ”;

Statistics – Scatterplots ”; Previous Next A scatterplot is a graphical way to display the relationship between two quantitative sample variables. It consists of an X axis, a Y axis and a series of dots where each dot represents one observation from a data set. The position of the dot refers to its X and Y values. Patterns of Data in Scatterplots Scatterplots are used to analyze patterns which generally varies on the basis of linearity, slope, and strength. Linearity – data pattern is either linear/straight or nonlinear/curved. Slope – direction of change in variable Y with respect to increase in value of variable X. If Y increases with increase in X, slope is positive otherwise slope is negative. Strength – Degree of spreadness of scatter in the plot. If dots are widely dispersed, the relationship is consider weak. If dot are densed around a line then the relationship is said to be strong. Print Page Previous Next Advertisements ”;

Statistics – Process Sigma ”; Previous Next Process sigma can be defined using following four steps: Measure opportunities, Measure defects, Calculate yield, Look-up process sigma. Formulae Used ${DPMO = frac{Total defect}{Total Opportunities} times 1000000}$ ${Defect (%) = frac{Total defect}{Total Opportunities} times 100}$ ${Yield (%) = 100 – Defect (%) }$ ${Process Sigma = 0.8406+sqrt{29.37}-2.221 times (log (DPMO)) }$ Where − ${Opportunities}$ = Lowest defect noticeable by customer. ${DPMO}$ = Defects per Million Opportunities. Example Problem Statement: In equipment organization hard plate produced is 10000 and the defects is 5. Discover the process sigma. Solution: Given: Opportunities = 10000 and Defects = 5. Substitute the given qualities in the recipe, Step 1: Compute DPMO $ {DPMO = frac{Total defect}{Total Opportunities} times 1000000 \[7pt] , = (10000/5) times 1000000 , \[7pt] , = 500}$ Step 2: Compute Defect(%) $ {Defect (%) = frac{Total defect}{Total Opportunities} times 100 \[7pt] , = frac{10000}{5} times 100 , \[7pt] , = 0.05}$ Step 3: Compute Yield(%) $ {Yield (%) = 100 – Defect (%) \[7pt] , = 100 – 0.05 , \[7pt] , = 99.95}$ Step 3: Compute Process Sigma $ {Process Sigma = 0.8406+sqrt{29.37}-2.221 times (log (DPMO)) \[7pt] , = 0.8406 + sqrt {29.37} – 2.221 times (log (DPMO)) , \[7pt] , = 0.8406+sqrt(29.37) – 2.221*(log (500)) , \[7pt] , = 4.79 }$ Print Page Previous Next Advertisements ”;

Statistics – Harmonic Resonance Frequency ”; Previous Next Harmonic Resonance Frequency represents a signal or wave whose frequency is an integral multiple of the frequency of a reference signal or wave. Formula ${ f = frac{1}{2 pi sqrt{LC}} } $ Where − ${f}$ = Harmonic resonance frequency. ${L}$ = inductance of the load. ${C}$ = capacitanc of the load. Example Calculate the harmonic resonance frequency of a power system with the capcitance 5F, Inductance 6H and frequency 200Hz. Solution: Here capacitance, C is 5F. Inductance, L is 6H. Frequency, f is 200Hz. Using harmonic resonance frequency formula, let”s compute the resonance frequency as: ${ f = frac{1}{2 pi sqrt{LC}} \[7pt] implies f = frac{1}{2 pi sqrt{6 times 5}} \[7pt] , = frac{1}{2 times 3.14 times sqrt{30}} \[7pt] , = frac{1}{ 6.28 times 5.4772 } \[7pt] , = frac{1}{ 34.3968 } \[7pt] , = 0.0291 }$ Thus harmonic resonance frequency is $ { 0.0291 }$. Print Page Previous Next Advertisements ”;

Statistics – Grand Mean ”; Previous Next When sample sizes are equal, in other words, there could be five values in each sample, or n values in each sample. The grand mean is the same as the mean of sample means. Formula ${X_{GM} = frac{sum x}{N}}$ Where − ${N}$ = Total number of sets. ${sum x}$ = sum of the mean of all sets. Example Problem Statement: Determine the mean of each group or set”s samples. Use the following data as a sample to determine the mean and grand mean. Jackson 1 6 7 10 4 Thomas 5 2 8 14 6 Garrard 8 2 9 12 7 Solution: Step 1: Compute all means $ {M_1 = frac{1+6+7+10+4}{5} = frac{28}{5} = 5.6 \[7pt] , M_2 = frac{5+2+8+14+6}{5} = frac{35}{5} = 7 \[7pt] , M_3 = frac{8+2+9+12+7}{5} = frac{38}{5} = 7.6 }$ Step 2: Divide the total by the number of groups to determine the grand mean. In the sample, there are three groups. $ {X_{GM} = frac{5.6+7+7.6}{3} = frac{20.2}{3} \[7pt] , = 6.73 }$ Print Page Previous Next Advertisements ”;

Statistics – Signal to Noise Ratio ”; Previous Next Sign to-commotion proportion (contracted SNR) is a measure utilized as a part of science and designing that analyzes the level of a coveted sign to the level of foundation clamor. It is characterized as the proportion of sign energy to the clamor power, regularly communicated in decibels. A proportion higher than 1:1 (more prominent than 0 dB) shows more flag than clamor. While SNR is regularly cited for electrical signs, it can be connected to any type of sign, (for example, isotope levels in an ice center or biochemical motioning between cells). Signal-to-noise ratio is defined as the ratio of the power of a signal (meaningful information) and the power of background noise (unwanted signal): ${SNR = frac{P_{signal}}{P_{noise}}}$ If the variance of the signal and noise are known, and the signal is zero: ${SNR = frac{sigma^2_{signal}}{sigma^2_{noise}}}$ If the signal and the noise are measured across the same impedance, then the SNR can be obtained by calculating the square of the amplitude ratio: ${SNR = frac{P_{signal}}{P_{noise}} = {(frac{A_{signal}}{A_{noise}})}^2} $ Where A is root mean square (RMS) amplitude (for example, RMS voltage). Decibels Because many signals have a very wide dynamic range, signals are often expressed using the logarithmic decibel scale. Based upon the definition of decibel, signal and noise may be expressed in decibels (dB) as ${P_{signal,dB} = 10log_{10}(P_{signal})} $ and ${P_{noise,dB} = 10log_{10}(P_{noise})} $ In a similar manner, SNR may be expressed in decibels as ${SNR_{dB} = 10log_{10}(SNR)} $ Using the definition of SNR ${SNR_{dB} = 10log_{10}(frac{P_{signal}}{P_{noise}})} $ Using the quotient rule for logarithms ${10log_{10}(frac{P_{signal}}{P_{noise}}) = 10log_{10}(P_{signal}) – 10log_{10}(P_{noise})} $ Substituting the definitions of SNR, signal, and noise in decibels into the above equation results in an important formula for calculating the signal to noise ratio in decibels, when the signal and noise are also in decibels: ${SNR_{dB} = P_{signal,dB} – P_{noise,dB}} $ In the above formula, P is measured in units of power, such as Watts or mill watts, and signal-to-noise ratio is a pure number. However, when the signal and noise are measured in Volts or Amperes, which are measures of amplitudes, they must be squared to be proportionate to power as shown below: ${SNR_{dB} = 10log_{10}[{(frac{A_{signal}}{A_{noise}})}^2] \[7pt] = 20log_{10}(frac{A_{signal}}{A_{noise}}) \[7pt] = A_{signal,dB} – A_{noise,dB}} $ Example Problem Statement: Compute the SNR of a 2.5 kHz sinusoid sampled at 48 kHz. Add white noise with standard deviation 0.001. Set the random number generator to the default settings for reproducible results. Solution: ${ F_i = 2500; F_s = 48e3; N = 1024; \[7pt] x = sin(2 times pi times frac{F_i}{F_s} times (1:N)) + 0.001 times randn(1,N); \[7pt] SNR = snr(x,Fs) \[7pt] SNR = 57.7103}$ Print Page Previous Next Advertisements ”;

Statistics – Hypothesis testing ”; Previous Next A statistical hypothesis is an assumption about a population which may or may not be true. Hypothesis testing is a set of formal procedures used by statisticians to either accept or reject statistical hypotheses. Statistical hypotheses are of two types: Null hypothesis, ${H_0}$ – represents a hypothesis of chance basis. Alternative hypothesis, ${H_a}$ – represents a hypothesis of observations which are influenced by some non-random cause. Example suppose we wanted to check whether a coin was fair and balanced. A null hypothesis might say, that half flips will be of head and half will of tails whereas alternative hypothesis might say that flips of head and tail may be very different. $ H_0: P = 0.5 \[7pt] H_a: P ne 0.5 $ For example if we flipped the coin 50 times, in which 40 Heads and 10 Tails results. Using result, we need to reject the null hypothesis and would conclude, based on the evidence, that the coin was probably not fair and balanced. Hypothesis Tests Following formal process is used by statistican to determine whether to reject a null hypothesis, based on sample data. This process is called hypothesis testing and is consists of following four steps: State the hypotheses – This step involves stating both null and alternative hypotheses. The hypotheses should be stated in such a way that they are mutually exclusive. If one is true then other must be false. Formulate an analysis plan – The analysis plan is to describe how to use the sample data to evaluate the null hypothesis. The evaluation process focuses around a single test statistic. Analyze sample data – Find the value of the test statistic (using properties like mean score, proportion, t statistic, z-score, etc.) stated in the analysis plan. Interpret results – Apply the decisions stated in the analysis plan. If the value of the test statistic is very unlikely based on the null hypothesis, then reject the null hypothesis. Print Page Previous Next Advertisements ”;

Statistics – Harmonic Mean ”; Previous Next What is Harmonic Mean? Harmonic Mean is also a mathematical average but is limited in its application. It is generally used to find average of variables that are expressed as a ratio of two different measuring units e. g. speed is measured in km/hr or miles/sec etc. Weighted Harmonic Mean Formula $H.M. = frac{W}{sum (frac{W}{X})}$ Where − ${H.M.}$ = Harmonic Mean ${W}$ = Weight ${X}$ = Variable value Example Problem Statement: Find the weighted H.M. of the items 4, 7,12,19,25 with weights 1, 2,1,1,1 respectively. Solution: ${X}$ ${W}$ $frac{W}{X}$ 4 1 0.2500 7 2 0.2857 12 1 0.0833 19 1 0.0526 25 1 0.0400   $sum W$ $sum frac{W}{X}$= 0.7116 Based on the above mentioned formula, Harmonic Mean $G.M.$ will be: $H.M. = frac{W}{sum (frac{W}{X})} \[7pt] , = frac{6}{0.7116} \[7pt] , = 8.4317 $ ∴ Weighted H.M = 8.4317 We”re going to discuss methods to compute the Harmonic Mean for three types of series: Individual Data Series Discrete Data Series Continuous Data Series Individual Data Series When data is given on individual basis. Following is an example of individual series: Items 5 10 20 30 40 50 60 70 Discrete Data Series When data is given alongwith their frequencies. Following is an example of discrete series: Items 5 10 20 30 40 50 60 70 Frequency 2 5 1 3 12 0 5 7 Continuous Data Series When data is given based on ranges alongwith their frequencies. Following is an example of continous series: Items 0-5 5-10 10-20 20-30 30-40 Frequency 2 5 1 3 12 Print Page Previous Next Advertisements ”;

Statistics – Individual Series Arithmetic Mode ”; Previous Next When data is given on individual basis. Following is an example of individual series − Items 5 10 20 30 40 50 60 70 In case of individual items, the number of times each value occurs is counted and the value which is repeated maximum number of times is the modal value. Example Problem Statement − Calculate Arithmetic Mode for the following individual data − Items 14 36 45 36 105 36 Solution − The Arithmetic Mode of the given numbers is 36 as it is repeated maximum number of times,3. Calculator Print Page Previous Next Advertisements ”;