WEEK 6 Flashcards
It uses mostly
Categorical variables
Steps to do it
1st set up null and alternative hypothesis
2nd decide on the significance level normally 5%
3rd calculate test statistics
4th decide whether to accept or reject the null hypothesis
Since the TS is greater than CV
It’s in rejections region and reject the null hypothesis and accept the alternative hypothesis
expected 40% and 60%
we take the observation value and x it by 0.40 and 0.60 to get the expected value then we subtract from observed and the data is added to find the critical value
2 variable we do DF
(r-1) x (c-1) gives degree of freddom and the r is the male and female and c is the options like agree, disagree or neither
iF THE TS>CV
so there is significant difference we reject the null and we conclude
also the right side is reject and left accept (higher)
the null hypothesis and alternative
that there are no differences between this and the other
Ho = O = E
and alternative is H1 = O = (with line) E
the critical value depends on the
degrees of freedom which is n - 1
this is not symmetric it is
right skewed distribution tail right it doesn’t take a value
2 types of chi-squared tests
test of association ad goodness of fit test
Test of association
tests to see if there is any association between categories in a two-way table
often testing to discover an association between two categorical variable
Goodness of fit test
this is a test to see if the data fits some distribution often investigating the distribution of one categorical (nominal or ordinal) variable
its written as formula
X^2 = Observation minus expectation squared / by expected then sum of all is added
Chi-squared test is considered
non parametric test it doesnt have a parameter such as mean or proportion
if the p value is > than the significance level 0.05 we
fail to reject the Ho hypothesis
if CV>TS WE
fail to reject the Ho hypothesis
test of association 1st step we find the df by doing the
R-1 (row) times by c-1 (column)
2nd step of test association
give the hypothesis and find the umber from the table using the 5% and 2 DF
we then do calculations for each row and column
expected value = row total x column total / grand total
then we do TS
observed number minus the expected we found then square it and divide by expected then add all and we find the TS
we then compare the CV AND TS
if the TS is greater than CV we reject the null hypothesis and give a conclusion that there is an association between the 2 variables
