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 – Goodness of Fit ”; Previous Next The Goodness of Fit test is used to check the sample data whether it fits from a distribution of a population. Population may have normal distribution or Weibull distribution. In simple words, it signifies that sample data represents the data correctly that we are expecting to find from actual population. Following tests are generally used by statisticians: Chi-square Kolmogorov-Smirnov Anderson-Darling Shipiro-Wilk Chi-square Test The chi-square test is the most commonly used to test the goodness of fit tests and is used for discrete distributions like the binomial distribution and the Poisson distribution, whereas The Kolmogorov-Smirnov and Anderson-Darling goodness of fit tests are used for continuous distributions. Formula ${ X^2 = sum {[ frac{(O_i – E_i)^2}{E_i}]} }$ Where − ${O_i}$ = observed value of i th level of variable. ${E_i}$ = expected value of i th level of variable. ${X^2}$ = chi-squared random variable. Example A toy company builts football player toys. It claims that 30% of the cards are mid-fielders, 60% defenders, and 10% are forwards. Considering a random sample of 100 toys has 50 mid-fielders, 45 defenders, and 5 forwards. Given 0.05 level of significance, can you justify company”s claim? Solution: Determine Hypotheses Null hypothesis $ H_0 $ – The proportion of mid-fielders, defenders, and forwards is 30%, 60% and 10%, respectively. Alternative hypothesis $ H_1 $ – At least one of the proportions in the null hypothesis is false. Determine Degree of Freedom The degrees of freedom, DF is equal to the number of levels (k) of the categorical variable minus 1: DF = k – 1. Here levels are 3. Thus ${ DF = k – 1 \[7pt] , = 3 -1 = 2 }$ Determine chi-square test statistic ${ X^2 = sum {[ frac{(O_i – E_i)^2}{E_i}]} \[7pt] , = [frac{(50-30)^2}{30}] + [frac{(45-60)^2}{60}] + [frac{(5-10)^2}{10}] \[7pt] , = frac{400}{30} + frac{225}{60} + frac{25}{10} \[7pt] , = 13.33 + 3.75 + 2.50 \[7pt] , = 19.58 }$ Determine p-value P-value is the probability that a chi-square statistic,$ X^2 $ having 2 degrees of freedom is more extreme than 19.58. Use the Chi-Square Distribution Calculator to find $ { P(X^2 gt 19.58) = 0.0001 } $. Interpret results As the P-value (0.0001) is quite less than the significance level (0.05), the null hypothesis can not be accepted. Thus company claim is invalid. Print Page Previous Next Advertisements ”;

Statistics – Frequency Distribution ”; Previous Next Frequency distribution is a table that displays the frequency of various outcomes in a sample. Each entry in the table contains the frequency or count of the occurrences of values within a particular group or interval, and in this way, the table summarizes the distribution of values in the sample. Example Problem Statement: Constructing a frequency distribution table of a survey was taken on Maple Avenue. In each of 20 homes, people were asked how many cars were registered to their households. The results were recorded as follows: 1 2 1 0 3 4 0 1 1 1 2 2 3 2 3 2 1 4 0 0 Solution: Steps to be followed for present this data in a frequency distribution table. Divide the results (x) into intervals, and then count the number of results in each interval. In this case, the intervals would be the number of households with no car (0), one car (1), two cars (2) and so forth. Make a table with separate columns for the interval numbers (the number of cars per household), the tallied results, and the frequency of results in each interval. Label these columns Number of cars, Tally and Frequency. Read the list of data from left to right and place a tally mark in the appropriate row. For example, the first result is a 1, so place a tally mark in the row beside where 1 appears in the interval column (Number of cars). The next result is a 2, so place a tally mark in the row beside the 2, and so on. When you reach your fifth tally mark, draw a tally line through the preceding four marks to make your final frequency calculations easier to read. Add up the number of tally marks in each row and record them in the final column entitled Frequency. Your frequency distribution table for this exercise should look like this: Frequency table for the number of cars registered in each household Number of cars (x) Tally Frequency (f) 0 ${lvertlvertlvertlvert}$ 4 1 ${require{cancel} cancel{lvertlvertlvertlvert} lvert}$ 6 2 ${cancel{lvertlvertlvertlvert}}$ 5 3 ${lvertlvertlvert}$ 3 4 ${lvertlvert}$ 3 By looking at this frequency distribution table quickly, we can see that out of 20 households surveyed, 4 households had no cars, 6 households had 1 car. Print Page Previous Next Advertisements ”;

Statistics – Quartile Deviation ”; Previous Next It depends on the lower quartile ${Q_1}$ and the upper quartile ${Q_3}$. The difference ${Q_3 – Q_1}$ is called the inter quartile range. The difference ${Q_3 – Q_1}$ divided by 2 is called semi-inter quartile range or the quartile deviation. Formula ${Q.D. = frac{Q_3 – Q_1}{2}}$ Coefficient of Quartile Deviation A relative measure of dispersion based on the quartile deviation is known as the coefficient of quartile deviation. It is characterized as ${Coefficient of Quartile Deviation = frac{Q_3 – Q_1}{Q_3 + Q_1}}$ Example Problem Statement: Calculate the quartile deviation and coefficient of quartile deviation from the data given below: Maximum Load(short-tons) Number of Cables 9.3-9.7 22 9.8-10.2 55 10.3-10.7 12 10.8-11.2 17 11.3-11.7 14 11.8-12.2 66 12.3-12.7 33 12.8-13.2 11 Solution: Maximum Load(short-tons) Number of Cables(f) ClassBounderies CumulativeFrequencies 9.3-9.7 2 9.25-9.75 2 9.8-10.2 5 9.75-10.25 2 + 5 = 7 10.3-10.7 12 10.25-10.75 7 + 12 = 19 10.8-11.2 17 10.75-11.25 19 + 17 = 36 11.3-11.7 14 11.25-11.75 36 + 14 = 50 11.8-12.2 6 11.75-12.25 50 + 6 = 56 12.3-12.7 3 12.25-12.75 56 + 3 = 59 12.8-13.2 1 12.75-13.25 59 + 1 = 60 ${Q_1}$ Value of ${frac{n}{4}^{th}}$ item =Value of ${frac{60}{4}^{th}}$ thing = ${15^{th}}$ item. Thus ${Q_1}$ lies in class 10.25-10.75. $ {Q_1 = 1+ frac{h}{f}(frac{n}{4} – c) \[7pt] ,Where l=10.25, h=0.5, f=12, frac{n}{4}=15 and c=7 , \[7pt] , = 10.25+frac{0.5}{12} (15-7) , \[7pt] , = 10.25+0.33 , \[7pt] , = 10.58 }$ ${Q_3}$ Value of ${frac{3n}{4}^{th}}$ item =Value of ${frac{3 times 60}{4}^{th}}$ thing = ${45^{th}}$ item. Thus ${Q_3}$ lies in class 11.25-11.75. $ {Q_3 = 1+ frac{h}{f}(frac{3n}{4} – c) \[7pt] ,Where l=11.25, h=0.5, f=14, frac{3n}{4}=45 and c=36 , \[7pt] , = 11.25+frac{0.5}{14} (45-36) , \[7pt] , = 11.25+0.32 , \[7pt] , = 11.57 }$ Quartile Deviation $ {Q.D. = frac{Q_3 – Q_1}{2} \[7pt] , = frac{11.57 – 10.58}{2} , \[7pt] , = frac{0.99}{2} , \[7pt] , = 0.495 }$ Coefficient of Quartile Deviation ${Coefficient of Quartile Deviation = frac{Q_3 – Q_1}{Q_3 + Q_1} \[7pt] , = frac{11.57 – 10.58}{11.57 + 10.58} , \[7pt] , = frac{0.99}{22.15} , \[7pt] , = 0.045 }$ Print Page Previous Next Advertisements ”;

Statistics – Discrete Series Arithmetic Mode ”; Previous Next When data is given along with 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 In discrete series, Arithmetic Mode can be determined by inspection and finding the variable which has the highest frequency associated with it. However, when there is very less difference between the maximum frequency and the frequency preceding it or succeeding it, then grouping table method is used. Example Problem Statement − Calculate Arithmetic Mode for the following discrete data − Items 14 36 45 70 105 145 Frequency 2 5 1 3 12 0 Solution − The Arithmetic Mode of the given numbers is 105 as the highest frequency,12 is associated with 105. Calculator Print Page Previous Next Advertisements ”;

Statistics – Permutation with Replacement ”; Previous Next Each of several possible ways in which a set or number of things can be ordered or arranged is called permutation Combination with replacement in probability is selecting an object from an unordered list multiple times. Permutation with replacement is defined and given by the following probability function: Formula ${^nP_r = n^r }$ Where − ${n}$ = number of items which can be selected. ${r}$ = number of items which are selected. ${^nP_r}$ = Ordered list of items or permutions Example Problem Statement: Electronic device usually require a personal code to operate. This particular device uses 4-digits code. Calculate how many codes are possible. Solution: Each code is represented by r=4 permutation with replacement of set of 10 digits{0,1,2,3,4,5,6,7,8,9} ${^{10}P_4 = (10)^4 \[7pt] = 10000 }$ Print Page Previous Next Advertisements ”;

Statistics – Gamma Distribution ”; Previous Next The gamma distribution represents continuous probability distributions of two-parameter family. Gamma distributions are devised with generally three kind of parameter combinations. A shape parameter $ k $ and a scale parameter $ theta $. A shape parameter $ alpha = k $ and an inverse scale parameter $ beta = frac{1}{ theta} $, called as rate parameter. A shape parameter $ k $ and a mean parameter $ mu = frac{k}{beta} $. Each parameter is a positive real numbers. The gamma distribution is the maximum entropy probability distribution driven by following criteria. Formula ${E[X] = k theta = frac{alpha}{beta} gt 0 and is fixed. \[7pt] E[ln(X)] = psi (k) + ln( theta) = psi( alpha) – ln( beta) and is fixed. }$ Where − ${X}$ = Random variable. ${psi}$ = digamma function. Characterization using shape $ alpha $ and rate $ beta $ Probability density function Probability density function of Gamma distribution is given as: Formula ${ f(x; alpha, beta) = frac{beta^alpha x^{alpha – 1 } e^{-x beta}}{Gamma(alpha)} where x ge 0 and alpha, beta gt 0 }$ Where − ${alpha}$ = location parameter. ${beta}$ = scale parameter. ${x}$ = random variable. Cumulative distribution function Cumulative distribution function of Gamma distribution is given as: Formula ${ F(x; alpha, beta) = int_0^x f(u; alpha, beta) du = frac{gamma(alpha, beta x)}{Gamma(alpha)}}$ Where − ${alpha}$ = location parameter. ${beta}$ = scale parameter. ${x}$ = random variable. ${gamma(alpha, beta x)} $ = lower incomplete gamma function. Characterization using shape $ k $ and scale $ theta $ Probability density function Probability density function of Gamma distribution is given as: Formula ${ f(x; k, theta) = frac{x^{k – 1 } e^{-frac{x}{theta}}}{theta^k Gamma(k)} where x gt 0 and k, theta gt 0 }$ Where − ${k}$ = shape parameter. ${theta}$ = scale parameter. ${x}$ = random variable. ${Gamma(k)}$ = gamma function evaluated at k. Cumulative distribution function Cumulative distribution function of Gamma distribution is given as: Formula ${ F(x; k, theta) = int_0^x f(u; k, theta) du = frac{gamma(k, frac{x}{theta})}{Gamma(k)}}$ Where − ${k}$ = shape parameter. ${theta}$ = scale parameter. ${x}$ = random variable. ${gamma(k, frac{x}{theta})} $ = lower incomplete gamma function. Print Page Previous Next Advertisements ”;

Statistics – Discrete Series Arithmetic Median ”; Previous Next When data is given along with 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 In case of a group having even number of distribution, Arithmetic Median is found out by taking out the Arithmetic Mean of two middle values after arranging the numbers in ascending order. Formula Median = Value of ($frac{N+1}{2})^{th} item$. Where − ${N}$ = Number of observations Example Problem Statement − Let”s calculate Arithmetic Median for the following discrete data − Items, ${X}$ 14 36 45 70 105 145 Frequency, ${f}$ 2 5 2 3 12 4 Comulative Frequency, ${C_f}$ 2 7 9 12 24 28 Terms 1-2 3-7 8-9 10-12 13-24 25-28 Solution − Based on the above mentioned formula, Arithmetic Median M will be − $M = Value of (frac{N+1}{2})^{th} item. \[7pt] , = Value of (frac{28+1}{2})^{th} item. \[7pt] , = Value of 14.5^{th} item. \[7pt] , = Value of (frac{14^{th} item + 15^{th} item}{2})\[7pt] , = (frac{105 + 105}{2}) , = {105}$ The Arithmetic Median of the given numbers is 2. In case of a group having even number of distribution, Arithmetic Median is the middle number after arranging the numbers in ascending order. Example Let”s calculate Arithmetic Median for the following discrete data − Items, ${X}$ 14 36 45 70 105 Frequency, ${f}$ 2 5 1 4 13 Comulative Frequency, ${C_f}$ 2 7 8 12 25 Terms 1-2 3-7 8-8 9-12 13-25 Given numbers are 25, an odd number thus middle number, 12th term is the Arithmetic Median. ∴ The Arithmetic Median of the given numbers is 70. Calculator Print Page Previous Next Advertisements ”;

Statistics – Probability Additive Theorem ”; Previous Next For Mutually Exclusive Events The additive theorem of probability states if A and B are two mutually exclusive events then the probability of either A or B is given by ${P(A or B) = P(A) + P(B) \[7pt] P (A cup B) = P(A) + P(B)}$ The theorem can he extended to three mutually exclusive events also as ${P(A cup B cup C) = P(A) + P(B) + P(C) }$ Example Problem Statement: A card is drawn from a pack of 52, what is the probability that it is a king or a queen? Solution: Let Event (A) = Draw of a card of king Event (B) Draw of a card of queen P (card draw is king or queen) = P (card is king) + P (card is queen) ${P (A cup B) = P(A) + P(B) \[7pt] = frac{4}{52} + frac{4}{52} \[7pt] = frac{1}{13} + frac{1}{13} \[7pt] = frac{2}{13}}$ For Non-Mutually Exclusive Events In case there is a possibility of both events to occur then the additive theorem is written as: ${P(A or B) = P(A) + P(B) – P(A and B)\[7pt] P (A cup B) = P(A) + P(B) – P(AB)}$ Example Problem Statement: A shooter is known to hit a target 3 out of 7 shots; whet another shooter is known to hit the target 2 out of 5 shots. Find the probability of the target being hit at all when both of them try. Solution: Probability of first shooter hitting the target P (A) = ${frac{3}{7}}$ Probability of second shooter hitting the target P (B) = ${frac{2}{5}}$ Event A and B are not mutually exclusive as both the shooters may hit target. Hence the additive rule applicable is ${P (A cup B) = P (A) + P(B) – P (A cap B) \[7pt] = frac{3}{7}+frac{2}{5}-(frac{3}{7} times frac{2}{5}) \[7pt] = frac{29}{35}-frac{6}{35} \[7pt] = frac{23}{35}}$ Print Page Previous Next Advertisements ”;

Statistics – Geometric Mean ”; Previous Next Geometric mean of n numbers is defined as the nth root of the product of n numbers. Formula ${GM = sqrt[n]{x_1 times x_2 times x_3 … x_n}}$ Where − ${n}$ = Total numbers. ${x_i}$ = numbers. Example Problem Statement: Determine the geometric mean of following set of numbers. 1 3 9 27 81 Solution: Step 1: Here n = 5 $ {GM = sqrt[n]{x_1 times x_2 times x_3 … x_n} \[7pt] , = sqrt[5]{1 times 3 times 9 times 27 times 81} \[7pt] , = sqrt[5]{3^3 times 3^3 times 3^4} \[7pt] , = sqrt[5]{3^{10}} \[7pt] , = sqrt[5]{{3^2}^5} \[7pt] , = sqrt[5]{9^5} \[7pt] , = 9 }$ Thus geometric mean of given numbers is $ 9 $. Print Page Previous Next Advertisements ”;