Category: statistics

Statistics – Pooled Variance (r) ”; Previous Next Pooled Variance/Change is the weighted normal for assessing the fluctuations of two autonomous variables where the mean can differ between tests however the genuine difference continues as before. Example Problem Statement: Compute the Pooled Variance of the numbers 1, 2, 3, 4 and 5. Solution: Step 1 Decide the normal (mean) of the given arrangement of information by including every one of the numbers then gap it by the aggregate include of numbers given the information set. ${Mean = frac{1 + 2 + 3 + 4 + 5}{5} = frac{15}{5} = 3 }$ Step 2 At that point, subtract the mean worth with the given numbers in the information set. ${Rightarrow (1 – 3), (2 – 3), (3 – 3), (4 – 3), (5 – 3) Rightarrow – 2, – 1, 0, 1, 2 }$ Step 3 Square every period”s deviation to dodge the negative numbers. ${Rightarrow (- 2)^2, (- 1)^2, (0)^2, (1)^2, (2)^2 Rightarrow 4, 1, 0, 1, 4 }$ Step 4 Now discover Standard Deviation utilizing the underneath equation ${S = sqrt{frac{sum{X-M}^2}{n-1}}}$ Standard Deviation = ${frac{sqrt 10}{sqrt 4} = 1.58113 }$ Step 5 ${Pooled Variance (r) = frac{((aggregate check of numbers – 1) times Var)}{(aggregate tally of numbers – 1)} , \[7pt] (r) = (5 – 1) times frac{2.5}{(5 – 1)}, \[7pt] = frac{(4 times 2.5)}{4} = 2.5}$ Hence, Pooled Variance (r) =2.5 Print Page Previous Next Advertisements ”;

Statistics – Laplace Distribution ”; Previous Next Laplace distribution represents the distribution of differences between two independent variables having identical exponential distributions. It is also called double exponential distribution. Probability density function Probability density function of Laplace distribution is given as: Formula ${ L(x | mu, b) = frac{1}{2b} e^{- frac{| x – mu |}{b}} }$ $ { = frac{1}{2b} } $ $ begin {cases} e^{- frac{x – mu}{b}}, & text{if $x lt mu $} \[7pt] e^{- frac{mu – x}{b}}, & text{if $x ge mu $} end{cases} $ Where − ${mu}$ = location parameter. ${b}$ = scale parameter and is > 0. ${x}$ = random variable. Cumulative distribution function Cumulative distribution function of Laplace distribution is given as: Formula ${ D(x) = int_{- infty}^x}$ $ = begin {cases} frac{1}{2}e^{frac{x – mu}{b}}, & text{if $x lt mu $} \[7pt] 1- frac{1}{2}e^{- frac{x – mu}{b}}, & text{if $x ge mu $} end{cases} $ $ { = frac{1}{2} + frac{1}{2}sgn(x – mu)(1 – e^{- frac{| x – mu |}{b}}) } $ Where − ${mu}$ = location parameter. ${b}$ = scale parameter and is > 0. ${x}$ = random variable. Print Page Previous Next Advertisements ”;

Statistics – One Proportion Z Test ”; Previous Next The test statistic is a z-score (z) defined by the following equation. ${z = frac{(p – P)}{sigma}}$ where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and ${sigma}$ is the standard deviation of the sampling distribution. Test Statistics is defined and given by the following function: Formula ${ z = frac {hat p -p_o}{sqrt{frac{p_o(1-p_o)}{n}}} }$ Where − ${z}$ = Test statistics ${n}$ = Sample size ${p_o}$ = Null hypothesized value ${hat p}$ = Observed proportion Example Problem Statement: A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim, a random sample of 100 doctors is obtained. Of these 100 doctors, 82 indicate that they recommend aspirin. Is this claim accurate? Use alpha = 0.05. Solution: Define Null and Alternative Hypotheses ${ H_0;p = .90 \[7pt] H_0;p ne .90 }$ Here Alpha = 0.05. Using an alpha of 0.05 with a two-tailed test, we would expect our distribution to look something like this: Here we have 0.025 in each tail. Looking up 1 – 0.025 in our z-table, we find a critical value of 1.96. Thus, our decision rule for this two-tailed test is: If Z is less than -1.96, or greater than 1.96, reject the null hypothesis.Calculate Test Statistic: ${ z = frac {hat p -p_o}{sqrt{frac{p_o(1-p_o)}{n}}} \[7pt] hat p = .82 \[7pt] p_o = .90 \[7pt] n = 100 \[7pt] z_o = frac {.82 – .90}{sqrt{frac{ .90 (1- .90)}{100}}} \[7pt] = frac{-.08}{0.03} \[7pt] = -2.667 }$ As z = -2.667 Thus as result we should reject the null hypothesis and as conclusion, The claim that 9 out of 10 doctors recommend aspirin for their patients is not accurate, z = -2.667, p < 0.05. Print Page Previous Next Advertisements ”;

Statistics – Multinomial Distribution ”; Previous Next A multinomial experiment is a statistical experiment and it consists of n repeated trials. Each trial has a discrete number of possible outcomes. On any given trial, the probability that a particular outcome will occur is constant. Formula ${P_r = frac{n!}{(n_1!)(n_2!)…(n_x!)} {P_1}^{n_1}{P_2}^{n_2}…{P_x}^{n_x}}$ Where − ${n}$ = number of events ${n_1}$ = number of outcomes, event 1 ${n_2}$ = number of outcomes, event 2 ${n_x}$ = number of outcomes, event x ${P_1}$ = probability that event 1 happens ${P_2}$ = probability that event 2 happens ${P_x}$ = probability that event x happens Example Problem Statement: Three card players play a series of matches. The probability that player A will win any game is 20%, the probability that player B will win is 30%, and the probability player C will win is 50%. If they play 6 games, what is the probability that player A will win 1 game, player B will win 2 games, and player C will win 3? Solution: Given: ${n}$ = 12 (6 games total) ${n_1}$ = 1 (Player A wins) ${n_2}$ = 2 (Player B wins) ${n_3}$ = 3 (Player C wins) ${P_1}$ = 0.20 (probability that Player A wins) ${P_1}$ = 0.30 (probability that Player B wins) ${P_1}$ = 0.50 (probability that Player C wins) Putting the values into the formula, we get: ${ P_r = frac{n!}{(n_1!)(n_2!)…(n_x!)} {P_1}^{n_1}{P_2}^{n_2}…{P_x}^{n_x} , \[7pt] P_r(A=1, B=2, C=3)= frac{6!}{1!2!3!}(0.2^1)(0.3^2)(0.5^3) , \[7pt] = 0.135 }$ Print Page Previous Next Advertisements ”;

Statistics – Poisson Distribution ”; Previous Next Poisson conveyance is discrete likelihood dispersion and it is broadly use in measurable work. This conveyance was produced by a French Mathematician Dr. Simon Denis Poisson in 1837 and the dissemination is named after him. The Poisson circulation is utilized as a part of those circumstances where the happening”s likelihood of an occasion is little, i.e., the occasion once in a while happens. For instance, the likelihood of faulty things in an assembling organization is little, the likelihood of happening tremor in a year is little, the mischance”s likelihood on a street is little, and so forth. All these are cases of such occasions where the likelihood of event is little. Poisson distribution is defined and given by the following probability function: Formula ${P(X-x)} = {e^{-m}}.frac{m^x}{x!}$ Where − ${m}$ = Probability of success. ${P(X-x)}$ = Probability of x successes. Example Problem Statement: A producer of pins realized that on a normal 5% of his item is faulty. He offers pins in a parcel of 100 and insurances that not more than 4 pins will be flawed. What is the likelihood that a bundle will meet the ensured quality? [Given: ${e^{-m}} = 0.0067$] Solution: Let p = probability of a defective pin = 5% = $frac{5}{100}$. We are given: ${n} = 100, {p} = frac{5}{100} , \[7pt] Rightarrow {np} = 100 times frac{5}{100} = {5}$ The Poisson distribution is given as: ${P(X-x)} = {e^{-m}}.frac{m^x}{x!}$ Required probability = P [packet will meet the guarantee] = P [packet contains up to 4 defectives] = P (0) +P (1) +P (2) +P (3) +P (4) $ = {e^{-5}}.frac{5^0}{0!} + {e^{-5}}.frac{5^1}{1!} + {e^{-5}}.frac{5^2}{2!} + {e^{-5}}.frac{5^3}{3!} +{e^{-5}}.frac{5^4}{4!}, \[7pt] = {e^{-5}}[1+frac{5}{1}+frac{25}{2}+frac{125}{6}+frac{625}{24}] , \[7pt] = 0.0067 times 65.374 = 0.438$ Print Page Previous Next Advertisements ”;

Statistics – Power Calculator ”; Previous Next Whenever a hypothesis test is conducted, we need to ascertain that test is of high qualitity. One way to check the power or sensitivity of a test is to compute the probability of test that it can reject the null hypothesis correctly when an alternate hypothesis is correct. In other words, power of a test is the probability of accepting the alternate hypothesis when it is true, where alternative hypothesis detects an effect in the statistical test. $ {Power = P( reject H_0 | H_1 is true) } $ Power of a test is also test by checking the probability of Type I error($ { alpha } $) and of Type II error($ { beta } $) where Type I error represents the incorrect rejection of a valid null hypothesis whereas Type II error represents the incorrect retention of an invalid null hypothesis. Lesser the chances of Type I or Type II error, more is the power of statistical test. Example A survey has been conducted on students to check their IQ level. Suppose a random sample of 16 students is tested. The surveyor tests the null hypothesis that the IQ of student is 100 against the alternative hypothesis that the IQ of student is not 100, using a 0.05 level of significance and standard deviation of 16. What is the power of the hypothesis test if the true population mean were 116? Solution: As distribution of the test statistic under the null hypothesis follows a Student t-distribution. Here n is large, we can approximate the t-distribution by a normal distribution. As probability of committing Type I error($ { alpha } $) is 0.05 , we can reject the null hypothesis ${H_0}$ when the test statistic $ { T ge 1.645 } $. Let”s compute the value of sample mean using test statistics by following formula. $ {T = frac{ bar X – mu}{ frac{sigma}{sqrt mu}} \[7pt] implies bar X = mu + T(frac{sigma}{sqrt mu}) \[7pt] , = 100 + 1.645(frac{16}{sqrt {16}})\[7pt] , = 106.58 } $ Let”s compute the power of statistical test by following formula. $ {Power = P(bar X ge 106.58 where mu = 116 ) \[7pt] , = P( T ge -2.36) \[7pt] , = 1- P( T lt -2.36 ) \[7pt] , = 1 – 0.0091 \[7pt] , = 0.9909 } $ So we have a 99.09% chance of rejecting the null hypothesis ${H_0: mu = 100 } $ in favor of the alternative hypothesis $ {H_1: mu gt 100 } $ where unknown population mean is $ {mu = 116 } $. Print Page Previous Next Advertisements ”;

Statistics – Kolmogorov Smirnov Test ”; Previous Next This test is used in situations where a comparison has to be made between an observed sample distribution and theoretical distribution. K-S One Sample Test This test is used as a test of goodness of fit and is ideal when the size of the sample is small. It compares the cumulative distribution function for a variable with a specified distribution. The null hypothesis assumes no difference between the observed and theoretical distribution and the value of test statistic ”D” is calculated as: Formula $D = Maximum |F_o(X)-F_r(X)|$ Where − ${F_o(X)}$ = Observed cumulative frequency distribution of a random sample of n observations. and ${F_o(X) = frac{k}{n}}$ = (No.of observations ≤ X)/(Total no.of observations). ${F_r(X)}$ = The theoretical frequency distribution. The critical value of ${D}$ is found from the K-S table values for one sample test. Acceptance Criteria: If calculated value is less than critical value accept null hypothesis. Rejection Criteria: If calculated value is greater than table value reject null hypothesis. Example Problem Statement: In a study done from various streams of a college 60 students, with equal number of students drawn from each stream, are we interviewed and their intention to join the Drama Club of college was noted.   B.Sc. B.A. B.Com M.A. M.Com No. in each class 5 9 11 16 19 It was expected that 12 students from each class would join the Drama Club. Using the K-S test to find if there is any difference among student classes with regard to their intention of joining the Drama Club. Solution: ${H_o}$: There is no difference among students of different streams with respect to their intention of joining the drama club. We develop the cumulative frequencies for observed and theoretical distributions. Streams No. of students interested in joining ${F_O(X)}$ ${F_T(X)}$ ${|F_O(X)-F_T(X)|}$   Observed(O) Theoretical(T)       B.Sc. 5 12 5/60 12/60 7/60 B.A. 9 12 14/60 24/60 10/60 B.COM. 11 12 25/60 36/60 11/60 M.A. 16 12 41/60 48/60 7/60 M.COM. 19 12 60/40 60/60 60/60 Total n=60         Test statistic ${|D|}$ is calculated as: $D = Maximum {|F_0 (X)-F_T (X)|} \[7pt] , = frac{11}{60} \[7pt] , = 0.183$ The table value of D at 5% significance level is given by ${D_0.05 = frac{1.36}{sqrt{n}}} \[7pt] , = frac{1.36}{sqrt{60}} \[7pt] , = 0.175$ Since the calculated value is greater than the critical value, hence we reject the null hypothesis and conclude that there is a difference among students of different streams in their intention of joining the Club. K-S Two Sample Test When instead of one, there are two independent samples then K-S two sample test can be used to test the agreement between two cumulative distributions. The null hypothesis states that there is no difference between the two distributions. The D-statistic is calculated in the same manner as the K-S One Sample Test. Formula ${D = Maximum |{F_n}_1(X)-{F_n}_2(X)|}$ Where − ${n_1}$ = Observations from first sample. ${n_2}$ = Observations from second sample. It has been seen that when the cumulative distributions show large maximum deviation ${|D|}$ it is indicating towards a difference between the two sample distributions. The critical value of D for samples where ${n_1 = n_2}$ and is ≤ 40, the K-S table for two sample case is used. When ${n_1}$ and/or ${n_2}$ > 40 then the K-S table for large samples of two sample test should be used. The null hypothesis is accepted if the calculated value is less than the table value and vice-versa. Thus use of any of these nonparametric tests helps a researcher to test the significance of his results when the characteristics of the target population are unknown or no assumptions had been made about them. Print Page Previous Next Advertisements ”;

Statistics – Pie Chart ”; Previous Next A pie chart (or a pie graph) is a circular statistical graphical chart, which is divided into slices in order to explain or illustrate numerical proportions. In a pie chart, centeral angle, area and an arc length of each slice is proportional to the quantity or percentages it represents. Total percentages should be 100 and total of the arc measures should be 360° Following illustration of pie graph depicts the cost of construction of a house. From this graph, one can compare the sum spent on cement, steel and so on. One can also compute the actual sum spent on each individual expense. Consider an example, where we want to know how much more is the labour cost when compared to cost of steel. $ { Amount spent on labor = frac{90}{60} times 600000 = $ 150000 \[7pt] Sum spent on steel = frac{54}{360} times 600000 = $ 90000 \[7pt] Excess = 150000 – 90000 = $ 60000 \[7pt] Let 60000=x% of 600000. \[7pt] implies frac{x}{100} times 600000 = $ 60000. \[7pt] implies x = 10% of total expense. } $ 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 ”;

Statistics – Negative Binomial Distribution ”; Previous Next Negative binomial distribution is a probability distribution of number of occurences of successes and failures in a sequence of independent trails before a specific number of success occurs. Following are the key points to be noted about a negative binomial experiment. The experiment should be of x repeated trials. Each trail have two possible outcome, one for success, another for failure. Probability of success is same on every trial. Output of one trial is independent of output of another trail. Experiment should be carried out until r successes are observed, where r is mentioned beforehand. Negative binomial distribution probability can be computed using following: Formula ${ f(x; r, P) = ^{x-1}C_{r-1} times P^r times (1-P)^{x-r} }$ Where − ${x}$ = Total number of trials. ${r}$ = Number of occurences of success. ${P}$ = Probability of success on each occurence. ${1-P}$ = Probability of failure on each occurence. ${f(x; r, P)}$ = Negative binomial probability, the probability that an x-trial negative binomial experiment results in the rth success on the xth trial, when the probability of success on each trial is P. ${^{n}C_{r}}$ = Combination of n items taken r at a time. Example Robert is a football player. His success rate of goal hitting is 70%. What is the probability that Robert hits his third goal on his fifth attempt? Solution: Here probability of success, P is 0.70. Number of trials, x is 5 and number of successes, r is 3. Using negative binomial distribution formula, let”s compute the probability of hitting third goal in fifth attempt. ${ f(x; r, P) = ^{x-1}C_{r-1} times P^r times (1-P)^{x-r} \[7pt] implies f(5; 3, 0.7) = ^4C_2 times 0.7^3 times 0.3^2 \[7pt] , = 6 times 0.343 times 0.09 \[7pt] , = 0.18522 }$ Thus probability of hitting third goal in fifth attempt is $ { 0.18522 }$. Print Page Previous Next Advertisements ”;