Category: statistics

Statistics – F Test Table ”; Previous Next F-test is named after the more prominent analyst R.A. Fisher. F-test is utilized to test whether the two autonomous appraisals of populace change contrast altogether or whether the two examples may be viewed as drawn from the typical populace having the same difference. For doing the test, we calculate F-statistic is defined as: Formula ${F} = frac{Larger estimate of population variance}{smaller estimate of population variance} = frac{{S_1}^2}{{S_2}^2} where {{S_1}^2} gt {{S_2}^2}$ Procedure Its testing procedure is as follows: Set up null hypothesis that the two population variance are equal. i.e. ${H_0: {sigma_1}^2 = {sigma_2}^2}$ The variances of the random samples are calculated by using formula: ${S_1^2} = frac{sum(X_1- bar X_1)^2}{n_1-1}, \[7pt] {S_2^2} = frac{sum(X_2- bar X_2)^2}{n_2-1}$ The variance ratio F is computed as: ${F} = frac{{S_1}^2}{{S_2}^2} where {{S_1}^2} gt {{S_2}^2}$ The degrees of freedom are computed. The degrees of freedom of the larger estimate of the population variance are denoted by v1 and the smaller estimate by v2. That is, ${v_1}$ = degrees of freedom for sample having larger variance = ${n_1-1}$ ${v_2}$ = degrees of freedom for sample having smaller variance = ${n_2-1}$ Then from the F-table given at the end of the book, the value of ${F}$ is found for ${v_1}$ and ${v_2}$ with 5% level of significance. Then we compare the calculated value of ${F}$ with the table value of ${F_.05}$ for ${v_1}$ and ${v_2}$ degrees of freedom. If the calculated value of ${F}$ exceeds the table value of ${F}$, we reject the null hypothesis and conclude that the difference between the two variances is significant. On the other hand, if the calculated value of ${F}$ is less than the table value, the null hypothesis is accepted and concludes that both the samples illustrate the applications of F-test. Example Problem Statement: In a sample of 8 observations, the entirety of squared deviations of things from the mean was 94.5. In another specimen of 10 perceptions, the worth was observed to be 101.7 Test whether the distinction is huge at 5% level. (You are given that at 5% level of centrality, the basic estimation of ${F}$ for ${v_1}$ = 7 and ${v_2}$ = 9, ${F_.05}$ is 3.29). Solution: Let us take the hypothesis that the difference in the variances of the two samples is not significant i.e. ${H_0: {sigma_1}^2 = {sigma_2}^2}$ We are given the following: ${n_1} = 8 , {sum {(X_1 – bar X_1)}^2} = 94.5, {n_2} = 10, {sum {(X_2 – bar X_2)}^2} = 101.7, \[7pt] {S_1^2} = frac{sum(X_1- bar X_1)^2}{n_1-1} = frac {94.5}{8-1} = frac {94.5}{7} = {13.5}, \[7pt] {S_2^2} = frac{sum(X_2- bar X_2)^2}{n_2-1} = frac {101.7}{10-1} = frac {101.7}{9} = {11.3}$ Applying F-Test ${F} = frac{{S_1}^2}{{S_2}^2} = frac {13.5}{11.3} = {1.195}$ For ${v_1}$ = 8-1 = 7, ${v_2}$ = 10-1 = 9 and ${F_.05}$ = 3.29. The Calculated value of ${F}$ is less than the table value. Hence, we accept the null hypothesis and conclude that the difference in the variances of two samples is not significant at 5% level. Print Page Previous Next Advertisements ”;

Statistics – Chi Squared table ”; Previous Next The numbers in the table represent the values of the ${chi^2}$ statistics. Areas of the shaded region (A) are the column indexes. You can also use the Chi-Square Distribution to compute critical and p values exactly. df A=0.005 0.010 0.025 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.975 0.99 0.995 1 0.000039 0.00016 0.00098 0.0039 0.0158 0.102 0.455 1.32 2.71 3.84 5.02 6.63 7.88 2 0.0100 0.0201 0.0506 0.103 0.211 0.575 1.39 2.77 4.61 5.99 7.38 9.21 10.6 3 0.0717 0.115 0.216 0.352 0.584 1.21 2.37 4.11 6.25 7.81 9.35 11.3 12.8 4 0.207 0.297 0.484 0.711 1.06 1.92 3.36 5.39 7.78 9.49 11.1 13.3 14.9 5 0.412 0.554 0.831 1.15 1.61 2.67 4.35 6.63 9.24 11.1 12.8 15.1 16.7 6 0.676 0.872 1.24 1.64 2.20 3.45 5.35 7.84 10.6 12.6 14.4 16.8 18.5 7 0.989 1.24 1.69 2.17 2.83 4.25 6.35 9.04 12.0 14.1 16.0 18.5 20.3 8 1.34 1.65 2.18 2.73 3.49 5.07 7.34 10.2 13.4 15.5 17.5 20.1 22.0 9 1.73 2.09 2.70 3.33 4.17 5.9 8.34 11.4 14.7 16.9 19.0 21.7 23.6 10 2.16 2.56 3.25 3.94 4.87 6.74 9.34 12.5 16.0 18.3 20.5 23.2 25.2 11 2.60 3.05 3.82 4.57 5.58 7.58 10.3 13.7 17.3 19.7 21.9 24.7 26.8 12 3.07 3.57 4.40 5.23 6.30 8.44 11.3 14.8 18.5 21.0 23.3 26.2 28.3 13 3.57 4.11 5.01 5.89 7.04 9.3 12.3 16.0 19.8 22.4 24.7 27.7 29.8 14 4.07 4.66 5.63 6.57 7.79 10.2 13.3 17.1 21.1 23.7 26.1 29.1 31.3 15 4.60 5.23 6.26 7.26 8.55 11.0 14.3 18.2 22.3 25.0 27.5 30.6 32.8 16 5.14 5.81 6.91 7.96 9.31 11.9 15.3 19.4 23.5 26.3 28.8 32.0 34.3 17 5.70 6.41 7.56 8.67 10.1 12.8 16.3 20.5 24.8 27.6 30.2 33.4 35.7 18 6.26 7.01 8.23 9.39 10.9 13.7 17.3 21.6 26.0 28.9 31.5 34.8 37.2 19 6.84 7.63 8.91 10.1 11.7 14.6 18.3 22.7 27.2 30.1 32.9 36.2 38.6 20 7.43 8.26 9.59 10.9 12.4 15.5 19.3 23.8 28.4 31.4 34.2 37.6 40.0 21 8.03 8.90 10.3 11.6 13.2 16.3 20.3 24.9 29.6 32.7 35.5 38.9 41.4 22 8.64 9.54 11.0 12.3 14.0 17.2 21.3 26.0 30.8 33.9 36.8 40.3 42.8 23 9.26 10.2 11.7 13.1 14.8 18.1 22.3 27.1 32.0 35.2 38.1 41.6 44.2 24 9.89 10.9 12.4 13.8 15.7 19.0 23.3 28.2 33.2 36.4 39.4 43.0 45.6 25 10.5 11.5 13.1 14.6 16.5 19.9 24.3 29.3 34.4 37.7 40.6 44.3 46.9 26 11.2 12.2 13.8 15.4 17.3 20.8 25.3 30.4 35.6 38.9 41.9 45.6 48.3 27 11.8 12.9 14.6 16.2 18.1 21.7 26.3 31.5 36.7 40.1 43.2 47.0 49.6 28 12.5 13.6 15.3 16.9 18.9 22.7 27.3 32.6 37.9 41.3 44.5 48.3 51.0 29 13.1 14.3 16.0 17.7 19.8 23.6 28.3 33.7 39.1 42.6 45.7 49.6 52.3 30 13.8 15.0 16.8 18.5 20.6 24.5 29.3 34.8 40.3 43.8 47.0 50.9 53.7 31 14.5 15.7 17.5 19.3 21.4 25.4 30.3 35.9 41.4 45.0 48.2 52.2 55.0 32 15.1 16.4 18.3 20.1 22.3 26.3 31.3 37.0 42.6 46.2 49.5 53.5 56.3 33 15.8 17.1 19.0 20.9 23.1 27.2 32.3 38.1 43.7 47.4 50.7 54.8 57.6 34 16.5 17.8 19.8 21.7 24.0 28.1 33.3 39.1 44.9 48.6 52.0 56.1 59.0 35 17.2 18.5 20.6 22.5 24.8 29.1 34.3 40.2 46.1 49.8 53.2 57.3 60.3 36 17.9 19.2 21.3 23.3 25.6 30.0 35.3 41.3 47.2 51.0 54.4 58.6 61.6 37 18.6 20.0 22.1 24.1 26.5 30.9 36.3 42.4 48.4 52.2 55.7 59.9 62.9 38 19.3 20.7 22.9 24.9 27.3 31.8 37.3 43.5 49.5 53.4 56.9 61.2 64.2 39 20.0 21.4 23.7 25.7 28.2 32.7 38.3 44.5 50.7 54.6 58.1 62.4 65.5 40 20.7 22.2 24.4 26.5 29.1 33.7 39.3 45.6 51.8 55.8 59.3 63.7 66.8 41 21.4 22.9 25.2 27.3 29.9 34.6 40.3 46.7 52.9 56.9 60.6 65.0 68.1 42 22.1 23.7 26.0 28.1 30.8 35.5 41.3 47.8 54.1 58.1 61.8 66.2 69.3 43 22.9 24.4 26.8 29.0 31.6 36.4 42.3 48.8 55.2 59.3 63.0 67.5 70.6 44 23.6 25.1 27.6 29.8 32.5 37.4 43.3 49.9 56.4 60.5 64.2 68.7 71.9 45 24.3 25.9 28.4 30.6 33.4 38.3 44.3 51.0 57.5 61.7 65.4 70.0 73.2 df A=0.005 0.010 0.025 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.975 0.99 0.995 Print Page Previous Next Advertisements ”;

Statistics – Combination 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. Combination with replacement is defined and given by the following probability function − Formula ${^nC_r = frac{(n+r-1)!}{r!(n-1)!} }$ Where − ${n}$ = number of items which can be selected. ${r}$ = number of items which are selected. ${^nC_r}$ = Unordered list of items or combinations Example Problem Statement − There are five kinds of frozen yogurt: banana, chocolate, lemon, strawberry and vanilla. You can have three scoops. What number of varieties will there be? Solution − Here n = 5 and r = 3. Substitute the values in formula, ${^nC_r = frac{(n+r-1)!}{r!(n-1)!} \[7pt] = frac{(5+3+1)!}{3!(5-1)!} \[7pt] = frac{7!}{3!4!} \[7pt] = frac{5040}{6 times 24} \[7pt] = 35}$ Calculator Print Page Previous Next Advertisements ”;

Statistics – Central limit theorem ”; Previous Next If the population from which the sample has a been drawn is a normal population then the sample means would be equal to population mean and the sampling distribution would be normal. When the more population is skewed, as is the case illustrated in Figure, then the sampling distribution would tend to move closer to the normal distribution, provided the sample is large (i.e. greater then 30). According to Central Limit Theorem, for sufficiently large samples with size greater than 30, the shape of the sampling distribution will become more and more like a normal distribution, irrespective of the shape of the parent population. This theorem explains the relationship between the population distribution and sampling distribution. It highlights the fact that if there are large enough set of samples then the sampling distribution of mean approaches normal distribution. The importance of central limit theorem has been summed up by Richard. I. Levin in the following words: The significance of the central limit theorem lies in the fact that it permits us to use sample statistics to make inferences about population parameters without knowing anything about the shape of the frequency distribution of that population other than what we can get from the sample. Print Page Previous Next Advertisements ”;

Statistics – Deciles Statistics ”; Previous Next A system of dividing the given random distribution of the data or values in a series into ten groups of similar frequency is known as deciles. Formula ${D_i = l + frac{h}{f}(frac{iN}{10} – c); i = 1,2,3…,9}$ Where − ${l}$ = lower boundry of deciles group. ${h}$ = width of deciles group. ${f}$ = frequency of deciles group. ${N}$ = total number of observations. ${c}$ = comulative frequency preceding deciles group. Example Problem Statement: Calculate the deciles of the distribution for the following table:   fi Fi [50-60] 8 8 [60-60] 10 18 [70-60] 16 34 [80-60] 14 48 [90-60] 10 58 [100-60] 5 63 [110-60] 2 65   65   Solution: Calculation of First Decile $ {frac{65 times 1}{10} = 6.5 \[7pt] , D_1= 50 + frac{6.5 – 0}{8} times 10 , \[7pt] , = 58.12}$ Calculation of Second Decile $ {frac{65 times 2}{10} = 13 \[7pt] , D_2= 60 + frac{13 – 8}{10} times 10 , \[7pt] , = 65}$ Calculation of Third Decile $ {frac{65 times 3}{10} = 19.5 \[7pt] , D_3= 70 + frac{19.5 – 18}{16} times 10 , \[7pt] , = 70.94}$ Calculation of Fourth Decile $ {frac{65 times 4}{10} = 26 \[7pt] , D_4= 70 + frac{26 – 18}{16} times 10 , \[7pt] , = 75}$ Calculation of Fifth Decile $ {frac{65 times 5}{10} = 32.5 \[7pt] , D_5= 70 + frac{32.5 – 18}{16} times 10 , \[7pt] , = 79.06}$ Calculation of Sixth Decile $ {frac{65 times 6}{10} = 39 \[7pt] , D_6= 70 + frac{39 – 34}{14} times 10 , \[7pt] , = 83.57}$ Calculation of Seventh Decile $ {frac{65 times 7}{10} = 45.5 \[7pt] , D_7= 80 + frac{45.5 – 34}{14} times 10 , \[7pt] , = 88.21}$ Calculation of Eighth Decile $ {frac{65 times 8}{10} = 52 \[7pt] , D_8= 90 + frac{52 – 48}{10} times 10 , \[7pt] , = 94}$ Calculation of Nineth Decile $ {frac{65 times 9}{10} = 58.5 \[7pt] , D_9= 100 + frac{58.5 – 58}{5} times 10 , \[7pt] , = 101}$ Print Page Previous Next Advertisements ”;

Statistics – Harmonic Number ”; Previous Next Harmonic Number is the sum of the reciprocals of the first n natural numbers. It represents the phenomenon when the inductive reactance and the capacitive reactance of the power system becomes equal. Formula ${ H = frac{W_r}{W} \[7pt] , where W_r = sqrt{ frac{1}{LC}} } \[7pt] , and W = 2 pi f $ Where − ${f}$ = Harmonic resonance frequency. ${L}$ = inductance of the load. ${C}$ = capacitanc of the load. Example Calculate the harmonic number 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 number formula, let”s compute the number as: ${ H = frac{sqrt{ frac{1}{LC}}}{2 pi f} \[7pt] implies H = frac{sqrt{ frac{1}{6 times 5}} }{2 times 3.14 times 200} \[7pt] , = frac{0.18257}{1256} \[7pt] , = 0.0001 }$ Thus harmonic number is $ { 0.0001 }$. Print Page Previous Next Advertisements ”;

Statistics – Beta Distribution ”; Previous Next The beta distribution represents continuous probability distribution parametrized by two positive shape parameters, $ alpha $ and $ beta $, which appear as exponents of the random variable x and control the shape of the distribution. Probability density function Probability density function of Beta distribution is given as: Formula ${ f(x) = frac{(x-a)^{alpha-1}(b-x)^{beta-1}}{B(alpha,beta) (b-a)^{alpha+beta-1}} hspace{.3in} a le x le b; alpha, beta > 0 \[7pt] , where B(alpha,beta) = int_{0}^{1} {t^{alpha-1}(1-t)^{beta-1}dt} }$ Where − ${ alpha, beta }$ = shape parameters. ${a, b}$ = upper and lower bounds. ${B(alpha,beta)}$ = Beta function. Standard Beta Distribution In case of having upper and lower bounds as 1 and 0, beta distribution is called the standard beta distribution. It is driven by following formula: Formula ${ f(x) = frac{x^{alpha-1}(1-x)^{beta-1}}{B(alpha,beta)} hspace{.3in} le x le 1; alpha, beta > 0}$ Cumulative distribution function Cumulative distribution function of Beta distribution is given as: Formula ${ F(x) = I_{x}(alpha,beta) = frac{int_{0}^{x}{t^{alpha-1}(1-t)^{beta-1}dt}}{B(alpha,beta)} hspace{.2in} 0 le x le 1; p, beta > 0 }$ Where − ${ alpha, beta }$ = shape parameters. ${a, b}$ = upper and lower bounds. ${B(alpha,beta)}$ = Beta function. It is also called incomplete beta function ratio. Print Page Previous Next Advertisements ”;

Statistics – Cohen”s kappa coefficient ”; Previous Next Cohen”s kappa coefficient is a statistic which measures inter-rater agreement for qualitative (categorical) items. It is generally thought to be a more robust measure than simple percent agreement calculation, since k takes into account the agreement occurring by chance. Cohen”s kappa measures the agreement between two raters who each classify N items into C mutually exclusive categories. Cohen”s kappa coefficient is defined and given by the following function − Formula ${k = frac{p_0 – p_e}{1-p_e} = 1 – frac{1-p_o}{1-p_e}}$ Where − ${p_0}$ = relative observed agreement among raters. ${p_e}$ = the hypothetical probability of chance agreement. ${p_0}$ and ${p_e}$ are computed using the observed data to calculate the probabilities of each observer randomly saying each category. If the raters are in complete agreement then ${k}$ = 1. If there is no agreement among the raters other than what would be expected by chance (as given by ${p_e}$), ${k}$ ≤ 0. Example Problem Statement − Suppose that you were analyzing data related to a group of 50 people applying for a grant. Each grant proposal was read by two readers and each reader either said “Yes” or “No” to the proposal. Suppose the disagreement count data were as follows, where A and B are readers, data on the diagonal slanting left shows the count of agreements and the data on the diagonal slanting right, disagreements −   B Yes No A Yes 20 5 No 10 15 Calculate Cohen”s kappa coefficient. Solution − Note that there were 20 proposals that were granted by both reader A and reader B and 15 proposals that were rejected by both readers. Thus, the observed proportionate agreement is ${p_0 = frac{20+15}{50} = 0.70}$ To calculate ${p_e}$ (the probability of random agreement) we note that − Reader A said “Yes” to 25 applicants and “No” to 25 applicants. Thus reader A said “Yes” 50% of the time. Reader B said “Yes” to 30 applicants and “No” to 20 applicants. Thus reader B said “Yes” 60% of the time. Using formula P(A and B) = P(A) x P(B) where P is probability of event occuring. The probability that both of them would say “Yes” randomly is 0.50 x 0.60 = 0.30 and the probability that both of them would say “No” is 0.50 x 0.40 = 0.20. Thus the overall probability of random agreement is ${p_e}$ = 0.3 + 0.2 = 0.5. So now applying our formula for Cohen”s Kappa we get: ${k = frac{p_0 – p_e}{1-p_e} = frac{0.70 – 0.50}{1-0.50} = 0.40}$ Calculator Print Page Previous Next Advertisements ”;

Individual Series Arithmetic Mean ”; 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 For individual series, the Arithmetic Mean can be calculated using the following formula. Formula $bar{x} = sum_{i=1}^{n} X_{i}$ Alternatively, we can write same formula as follows − $bar{x} = frac{_{sum {x}}}{N}$ Where − $X_{1}, X_{2}, X_{3}, …. X_{n}$ = individual observation of variable. $sum {x}$ = sum of all observations of the variable ${N}$ = Number of observations Example Problem Statement − Calculate Arithmetic Mean for the following individual data − Items 14 36 45 70 105 Solution − Based on the above mentioned formula, Arithmetic Mean $bar{x}$ will be − $bar{x} = frac{14 + 36 + 45 + 70 + 105}{5} \[7pt] , = frac{270}{5} \[7pt] , = {54}$ The Arithmetic Mean of the given numbers is 54. Calculator Print Page Previous Next Advertisements ”;

Statistics – Adjusted R-Squared ”; Previous Next R-squared measures the proportion of the variation in your dependent variable (Y) explained by your independent variables (X) for a linear regression model. Adjusted R-squared adjusts the statistic based on the number of independent variables in the model.${R^2}$ shows how well terms (data points) fit a curve or line. Adjusted ${R^2}$ also indicates how well terms fit a curve or line, but adjusts for the number of terms in a model. If you add more and more useless variables to a model, adjusted r-squared will decrease. If you add more useful variables, adjusted r-squared will increase. Adjusted ${R_{adj}^2}$ will always be less than or equal to ${R^2}$. You only need ${R^2}$ when working with samples. In other words, ${R^2}$ isn”t necessary when you have data from an entire population. Formula ${R_{adj}^2 = 1 – [frac{(1-R^2)(n-1)}{n-k-1}]}$ Where − ${n}$ = the number of points in your data sample. ${k}$ = the number of independent regressors, i.e. the number of variables in your model, excluding the constant. Example Problem Statement − A fund has a sample R-squared value close to 0.5 and it is doubtlessly offering higher risk adjusted returns with the sample size of 50 for 5 predictors. Find Adjusted R square value. Solution − Sample size = 50 Number of predictor = 5 Sample R – square = 0.5.Substitute the qualities in the equation, $ {R_{adj}^2 = 1 – [frac{(1-0.5^2)(50-1)}{50-5-1}] \[7pt] , = 1 – (0.75) times frac{49}{44} , \[7pt] , = 1 – 0.8352 , \[7pt] , = 0.1648 }$ Calculator Print Page Previous Next Advertisements ”;