Intro Statistics

Question 1

central limit theorem

Answer 1

If x_bar is the mean of a random sample X1, X2, … , Xn of size n from a distribution with a finite mean mu and a finite positive variance sigma ², then the distribution of W = (x_bar -mu)/ (sigma/sqrt(n)) is N(0,1) in the limit as n approaches infinity.

This means that the variable is distributed N(mu,sigma/sqrt(n)).

Question 2

binomial distribution

Answer 2

with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question

P(x=k) = (n,k) * p^k * (1 - p)^(n-k)

(n,k) = n! / (k! (n - k)!)

Mu = n*p
Sigma = n*p*(1-p)

Question 3

Accuracy

Answer 3

the proportion of true results (both true positives and true negatives) among the total number of cases examined.[

accuracy = tp + tn / (tp + tn + fp + fn)

Question 4

Precision

Answer 4

precision (also called positive predictive value) is the fraction of relevant instances among the retrieved instances

precision = tp / (tp + fp)

Question 5

Recall

Answer 5

recall (also known as sensitivity) is the fraction of relevant instances that have been retrieved over the total amount of relevant instances

recall = tp / (tp + fn)

Question 6

type I error

Answer 6

a type I error is the rejection of a true null hypothesis (also known as a “false positive” finding)

a type I error is to falsely infer the existence of something that is not there

Question 7

type II error

Answer 7

type II error is retaining a false null hypothesis (also known as a “false negative” finding)

a type II error is to falsely infer the absence of something that is

Question 8

kullback liebler divergence

Answer 8

a measure of how one probability distribution diverges from a second, expected probability distribution

Question 9

kolmogrov smirnoff test

Answer 9

is a nonparametric test of the equality of continuous, one-dimensional probability distributions that can be used to compare a sample with a reference probability distribution (one-sample K–S test), or to compare two samples (two-sample K–S test)

Question 10

Bootstrap

Answer 10

statistical method for estimating the sampling distribution of an estimator by sampling with replacement from the original sample, most often with the purpose of deriving robust estimates of standard errors and confidence intervals of a population parameter like a mean, median, proportion, odds ratio, correlation coefficient or regression coefficient.

Question 11

Jackknife

Answer 11

The jackknife estimator of a parameter is found by systematically leaving out each observation from a dataset and calculating the estimate and then finding the average of these calculations.

Question 12

Permutation test

Answer 12

the distribution of the test statistic under the null hypothesis is obtained by calculating all possible values of the test statistic under rearrangements of the labels on the observed data points. In other words, the method by which treatme

Question 13

Two tailed test

Answer 13

appropriate if the estimated value may be more than or less than the reference value, for example, whether a test taker may score above or below the historical average

Question 14

One tailed test

Answer 14

appropriate if the estimated value may depart from the reference value in only one direction, for example, whether a machine produces more than one-percent defective products

Question 15

Assessing normality

Answer 15

Subtract mean divide by variance, compare to standard normal values — nscore

Question 16

Box plot

Answer 16

Categorical variables, shows shape of distribution, central value and variability

Median black center line
Box top bottom are first and third quartiles
Vertical lines 1.5 times IQR
Outside lines points shown

Question 17

IQR

Answer 17

Inter quartile range

Distance between first and third quartiles

Question 18

Two way table

Answer 18

two-way table presents categorical data by counting the number of observations that fall into

Question 19

Correlation coefficient

Answer 19

R = 1/(n-1) * Sum( ((x-x_mean)/std_x ) * ((y-y_mean)/std_y)) )

Question 20

ANOVA

Answer 20

Analysis of variance is a statistical method used to test differences between two or more means of variance

Question 21

Parameter

Answer 21

parameter is a number describing a population, such as a percentage or proportion.

true proportion of defective items in the entire population

Question 22

Statistic

Answer 22

is a number which may be computed from the data observed in a random sample without requiring the use of any unknown parameters, such as a sample mean.

takes a sample 300 items and observes that 15 of these are defective- computes the statistic , p_hat = 15/300 = 0.05 an estimate of the parameter p

Question 23

Biased estimator

Answer 23

statistic is systematically skewed away from the true parameter p, it is considered to be a biased estimator of the parameter

Question 24

Unbiased estimator

Answer 24

unbiased estimator will have a sampling distribution whose mean is equal to the true value of the parameter.

Question 25

Variability of statistic

Answer 25

Determined by the spread of its sampling distribution. In general, larger samples will have smaller variability

Question 26

Probability model

Answer 26

mathematical representation of a random phenomenon. It is defined by its sample space, events within the sample space, and probabilities associated with each event.

Question 27

Sample space

Answer 27

set of all possible outcomes

Question 28

Probability

Answer 28

numerical value assigned to a given event A. The probability of an event is written P(A), and describes the long-run relative frequency of the event.

Rule 1: Any probability P(A) is a number between 0 and 1 (0 < P(A) < 1).

Rule 2: The probability of the sample space S is equal to 1 (P(S) = 1).

Question 29

Probability disjoint

Answer 29

If two events have no outcomes in common

Rule 3: If two events A and B are disjoint, then the probability of either event is the sum of the probabilities of the two events:
P(A or B) = P(A) + P(B). I’ll

Question 30

Probability union

Answer 30

chance of any (one or more) of two or more events occurring is called the union of the events. The probability of the union of disjoint events is the sum of their individual probabilities.

If two events A and B are not disjoint, then the probability of their union (the event that A or B occurs) is equal to the sum of their probabilities minus the sum of their intersection.

Question 31

Probability complement

Answer 31

Rule 4: The probability that any event A does not occur is P(Ac) = 1 - P(A).

Question 32

Probability independence

Answer 32

If the outcome of the first event has no effect on the probability of the second event,

Rule 5: If two events A and B are independent, then the probability of both events is the product of the probabilities for each event:
P(A and B) = P(A)P(B)

Question 33

Probability intersection

Answer 33

chance of all of two or more events occurring

For independent events, the probability of the intersection of two or more events is the product of the probabilities.

Question 34

Conditional probability

Answer 34

event B is the probability that the event will occur given the knowledge that an event A has already occurred

If events A and B are not independent, then the probability of the intersection of A and B (the probability that both events occur) is defined by
P(A and B) = P(A)P(B|A).

From this definition, the conditional probability P(B|A) is easily obtained by dividing by P(A):

P(B|A)= P(A and B)/P(A)

Question 35

Random variable

Answer 35

is a variable whose possible values are numerical outcomes of a random phenomenon

Question 36

Law of large numbers

Answer 36

law of large numbers states that the observed random mean from an increasingly large number of observations of a random variable will always approach the distribution mean

Question 37

Properties of random variate means

Answer 37

Mu_a+by = a+b*mu_x

Mu_x+y = mu_x + mu_y

Question 38

Properties of random variate variance

Answer 38

Sigma^2_a+by = b^2*sigma^2

Sigma^2_x+y = sigma_x^2 + sigma_y^2

Question 39

Sample mean and variance

Answer 39

Mu_x = mu
Sigma_x = sigma/sqrt(n)

Sample mean sigma gets smaller as n goes up

If distribution of population is normal then distribution of sample mean is normal with mean mu and stdev sigma/sqrt(n)

Question 40

Tests of Significance for Two Unknown Means and Known Standard Deviations

Answer 40

two-sample z statistic

from two normal populations of size n1 and n2 with unknown means and and known standard deviations and , the test statistic comparing the means is known as the two-sample z statistic

z = ((x1 - x2) - (mu1 - mu2))/ sqrt((sigma1^2/n1) + (sigma2^2/n2))

Question 41

Tests of Significance for Two Unknown Means and Unknown Standard Deviations

Answer 41

two-sample t-statistic

t = ((x1 - x2) - (mu1 - mu2))/ sqrt((s1^2/n1) + (s2^2/n2))

confidence interval

(x1 - x2) +/- t*(sqrt(s1^2/n1 + s2^2/n2))

conservative P-values may be obtained using the t(k) distribution where k represents the smaller of n1-1 and n2-1

Question 42

Pooled t-statistic

Answer 42

same variance for both

s_p^2 = (n1 - 1)s1^2 + (n2 -1)s2^2/ (n1 + n2 - 2)

t = ((x1 - x2) - (mu1 - mu2))/ s_p * sqrt((1/n1) + (1/n2))

Question 43

sample proportion (categorical)

Answer 43

given a simple random sample of size n from a population, the number of “successes” X divided by the sample size n gives us p_hat the sample proportion

This proportion follows a binomial distribution with mean p and variance (p(1-p))/n

An approximate level C confidence interval for p is p_hat +/- z* (sqrt((p(1-p))/n) where z is the upper (1-C)/2 critical value from the standard normal distribution.

Question 44

Confidence Intervals for Unknown Mean and Known Standard Deviation

Answer 44

For a population with unknown mean mu and known standard deviation sigma, a confidence interval for the population mean, based on a simple random sample (SRS) of size n, is x_bar +/- z(sigma/sqrt(n)), where z is the upper (1-C)/2 critical value for the standard normal distribution.

Question 45

Level C

Answer 45

gives the probability that the interval produced by the method employed includes the true value of the parameter theta

Question 46

Confidence Intervals for Unknown Mean and Unknown Standard Deviation

Answer 46

For a population with unknown mean mu and unknown standard deviation, a confidence interval for the population mean, based on a simple random sample (SRS) of size n, is x_bar +/- t* (s/sqrt(n)), where t* is the upper (1-C)/2 critical value for the t distribution with n-1 degrees of freedom, t(n-1).

s = standard error (estimated stddev)

Question 47

Significance Tests for Unknown Mean and Known Standard Deviation

Answer 47

For claims about a population mean from a population with a normal distribution or for any sample with large sample size n (for which the sample mean will follow a normal distribution by the Central Limit Theorem), if the standard deviation sigma is known, the appropriate significance test is known as the z-test, where the test statistic is defined as

z = (z_bar - mu_theta)/(sigma/sqrt(n))

Question 48

Power of a test

Answer 48

the probability that a fixed level significance test will reject the null hypothesis H0 when a particular alternative value of the parameter is true.

Question 49

Significance Tests for Unknown Mean and Unknown Standard Deviation

Answer 49

claims about a population mean from a population with a normal distribution or for any sample with large sample size n (for which the sample mean will follow a normal distribution by the Central Limit Theorem) with unknown standard deviation, the appropriate significance test is known as the t-test, where the test statistic is defined as

t = (x_bar - mu_theta)/(s/sqrt(n))

Question 50

Sign test

Answer 50

o perform a sign test on matched pairs data, take the difference between the two measurements in each pair and count the number of non-zero differences n. Of these, count the number of positive differences X. Determine the probability of observing X positive differences for a B(n,1/2) distribution, and use this probability as a P-value for the null hypothesis.

Question 51

categorical test single proportion

Answer 51

To test the null hypothesis H0: p = p0 against a one- or two-sided alternative hypothesis Ha, replace p with p0 in the test statistic

z = (p - p0)/(sqrt((p0*(1-p0))/n)

Question 52

categorical sample size single proportion

Answer 52

n = (z/m)²p(1-p*).

The margin of error is maximized when p* = 0.5, in which case

n = (z*/2m)².

Question 53

Comparison of Two Proportions

Answer 53

An approximate level C confidence interval for p1 - p2 is p_hat1 - p_hat2 + zsD where z is the upper (1-C)/2 critical value from the standard normal distribution.

sD = sqrt( (p1_hat(1-p1_hat)/n1) + (p2_hat(1-p2_hat)/n2)

Question 54

Test two proportions

Answer 54

To test the null hypothesis H0: p1 = p2 against a one- or two-sided alternative hypothesis Ha, first compute a pooled estimate for the parameter =
(X1 + X2)/(n1 + n2), where X1 and X2 represent the number of “successes” in each population sample

sP = sqrt(p_hat(1-p_hat)(1/n1 + 1/n2))

z = (1 - 2)/sp follows the standard normal distribution (with mean = 0 and standard deviation = 1). The test statistic z is used to compute the P-value

Question 55

chi-squared statistic

Answer 55

chi^2 = Sum( (observed - expected)^2/expected )

Question 56

chi-squared distribution

Answer 56

random variable is said to have a chi-square distribution with m degrees of freedom if it is the sum of the squares of m independent standard normal random variables

he distribution of the chi-square test statistic based on k counts is approximately the chi-square distribution with m = k-1 degrees of freedom, denoted chi^2(k-1).

Question 57

chi-squared fitting

Answer 57

In general, if we estimate d parameters under the null hypothesis with k possible counts the degrees of freedom for the associated chi-square distribution will be k - 1 - d.

Question 58

chi-squared hypothesis test

Answer 58

use the chi-square test to test the validity of a distribution assumed for a random phenomenon. The test evaluates the null hypotheses H0 (that the data are governed by the assumed distribution) against the alternative (that the data are not drawn from the assumed distribution).

Let p1, p2, …, pk denote the probabilities hypothesized for k possible outcomes. In n independent trials, we let Y1, Y2, …, Yk denote the observed counts of each outcome which are to be

The chi-square test statistic is qk-1 =

= (Y1 - np1)² + (Y2 - np2)² + … + (Yk - npk)²
———- ———- ——–
np1 np2 npk

Reject H0 if this value exceeds the upper critical value of the (k-1) distribution, where is the desired level of significance.

Question 59

Permutations and combinations

Answer 59

Permutations are for lists (order matters)

Combinations are for groups (order doesn’t matter)

Question 60

Permutation formula

Answer 60

P(n,k)=n!/(n-k)!

You have n items and want to find the number of ways k of those items can be ordered

N pick k

Question 61

Combination formula

Answer 61

C(n,k) = n!/(k!(n-k!))

Question 62

Multinomial coefficient

Answer 62

n!/(k1!k2!k3!…*km!)

as the number of ways of depositing n distinct objects into m distinct bins, with k1 objects in the first bin, k2 objects in the second bin, and so on.

the number of distinct ways to permute a multiset of n elements, and ki are the multiplicities of each of the distinct elements