Quiz
Non-parametric and chi squared tests
This quiz has question types that open in the usual player.
Questions in this quiz (10)
1data might not be normally distributed if the data is not evenly distributed around the [blank_start]mean[blank_end], might have very long tails, might be skewed in either direction
- data might not be [blank_start]continuous[blank_end], might be categorical data > which cannot follow a normal distribution...
NON-parametric methods helps with this: where t tests are [blank_start]not[blank_end] appropriate
- do not assume any underlying distributions
- when we cannot [blank_start]transform[blank_end] the data to fit normality
The appropriate non-parametric test for comparing 2 independent samples is called the [blank_start]Mann[blank_end]-Whitney test
- the mann-whitney tests for differences in [blank_start]medians[blank_end]
- conducted via minitab- Amean
- Bcontinuous
- Cnot
- Dtransform
- EMann
- Fmedians
2Minitab- Mann whitney U
stat > non-parametric > [blank_start]Mann[blank_end]-Whitney >
- the [blank_start]null[blank_end] hypothesis = the 2 medians are the same
- alternative = the 2 medians are [blank_start]different[blank_end]
gives us 95% confident intervals; difference in medians is somewhere in this interval, if 0 is not in this range
- if the medians were the same they range would be close to [blank_start]0[blank_end]- AMann
- Bnull
- Cdifferent
- D0
3The mann-whitney test is [blank_start]less[blank_end] powerful that a t-test if the data is normally distributed or can be [blank_start]transformed[blank_end] to normality > trade-off for not assuming distribution
- will obtain a [blank_start]bigger[blank_end] p value
generally tho, similar conclusions are made
- it should be used when a t test is not possible
- it assume that the distributions in the two samples are similar except for a possible shift of location e.g. similar [blank_start]standard[blank_end] deviations but shift of location
- thus it should not be used if the shapes of the distributions are very [blank_start]different[blank_end] > in particular if the standard [blank_start]deviations[blank_end] are very different- Aless
- Btransformed
- Cbigger
- Dstandard
- Edifferent
- Fdeviations
4When the data are in counts: the [blank_start]poisson[blank_end] distribution
- frequencies of something
the poisson distribution can be used to model counts of [blank_start]random[blank_end] events in a certain area or time interval when there is [blank_start]no[blank_end] fixed upper limit
e.g...
> counts of radioactive emissions per second at a location
> numbers of individuals of a particular species in areas of a fixed size- Apoisson
- Brandom
- Cno
5the [blank_start]Poisson[blank_end] distribution
- rate parameter (lamba) is equal to both the [blank_start]mean[blank_end] and the variance
- if data follows the poisson distribution then the mean and the [blank_start]variance[blank_end] are equal
10 x 10 grid example
- split up into equal sized squares = in each count number of a particular species of molluscs and count how many in each square
- make a table of frequencies: 7 occurrences with 0 in, with cases with 1 and 2, 14 cases with 14
- no cases where 1 of the boxes have more than 12
- can calculate the sample mean and variance
- remember variance = the [blank_start]square[blank_end] of our standard deviation
if the sample mean and variance are [blank_start]not[blank_end] the same, it suggests the data does not conform to the Poisson distribution > to assess this more exactly (more solid conclusion) we need to carry out a [blank_start]goodness[blank_end] of fit test...
- APoisson
- Bmean
- Cvariance
- Dsquare
- Enot
- Fgoodness
6does this data conform to a [blank_start]poisson[blank_end] distribution? check using a goodness of fit test
- here we address the question of whether or not the observed data appears to fit to a proposed probability distribution
E.g. are the counts in the example before consistent with data from a poisson distribution?
- method to check = first form table of observed frequencies e.g. species counts
- compare what we have actually observed to what we would [blank_start]expect[blank_end] to observe under the proposed probability distribution
with this goodness-of-fit test the [blank_start]null[blank_end] hypothesis = that the proposed probability distribution is the correct model for the data
- if the observed frequencies are very [blank_start]different[blank_end] from the expected frequencies then this will be evidence against the null hypothesis- Apoisson
- Bexpect
- Cnull
- Ddifferent
7Minitab will calculate all the expected frequencies for us
- if our probability distribution is correct, then we would expect observed and expected frequencies to be close
- in order to properly assess whether or not the poisson distribution is a good fit, we must carry out a hypothesis test
- the appropriate test for comparing observed and expected frequencies is called the [blank_start]chi[blank_end]-squared goodness of fit test > this is because it uses the chi-square distribution
stat > basic statistics > goodness of fit test for poisson
- variable is the counts
- frequency variable = the frequencies of those
makes a table with with [blank_start]contribution[blank_end] to chi squared statistic
big values relative to the degrees of freedom imply the theory does not hold, the p value = <0.001- Achi
- Bcontribution
8a [blank_start]small[blank_end] p value following the chi-square goodness of fit test, is evidence against the null hypothesis
- this means that our counts are not consistent with data from a [blank_start]Poisson[blank_end] distribution
- we have too many big and small values, implying the species is clustered
- minitab gives a helpful picture which highlights where the departure from the theory are
- Asmall
- BPoisson
9what about if you assume equal frequencies in each group
- in the previous, we are assessing the fit to a particular distribution (Poisson)
- a common distribution to want to compared to is uniform distribution i.e. where the expected numbers in each group are the [blank_start]same[blank_end]
- the same procedure is used but now the degrees of freedom, v = (no. of categories) - [blank_start]1[blank_end]; not estimating anything from data e.g. mean
stat > tables > chi-square goodness of fit test (one variable)- Asame
- B1
10Tests of Independence
- categorical data, but now we want to test for any association between any categorical variables
- IQ tests of students > drinkers and [blank_start]non[blank_end]-drinkers
- looks for association between drinking and IQ
gonna use chi-squared goodness of fit test
- when looking at association between 2 variables, the degrees of freedom = (number of rows -1) [blank_start]x[blank_end] (number of columns -1)
- assess the null hypothesis of independence: small p values are usual evidence against the null hypothesis, i.e. [blank_start]small[blank_end] p values are in favour of an association between the variables- Anon
- Bx
- Csmall
Information
- Description
- Stats Quiz on Non-parametric and chi squared tests, created by sophie waddington on 02/01/2019.
- Author
- sophie waddington
- Language
- English
- Views
- 0
- Tags
- Rating
- No ratings yet Rate this quiz