statistical tests Flashcards
(23 cards)
test for associations between measurements
spearman’s rank correlation
test for frequencies (number of individuals in categories)
chi-squared
test for differences between measurements
student’s t-test
what is the null hypothesis for spearman’s rank correlation
no correlation between … and …
how to calculate spearman’s rank correlation
1 assign ranks from lowest to highest
2 tied values will share ranks + be given average
3 calculate difference in rankings
4 square each difference
5 calculate rs
6 use numbers of pairs of measurements to find critical value
7 compare rs to critical value
formula for spearman’s rank
(rs value)
rs smaller than critical value
accept null hypothesis
rs greater than critical value
reject null hypothesis
less than 5% probability that correlation is due to chance
what is important to remember about critical value for spearman’s rank
could be negative or positive depending on correlation of data
what is the null hypothesis for chi-squared test
no difference between observed and expected frequencies
how to calculate expected frequency
total of observed frequencies / number of categories
formula for chi-squared
(X2 value)
how to calculate chi-squared
categories
observed frequencies
expected frequencies
O-E for each category
(O-E)2 for each category
calculate X2
degrees of freedom
find critical value
compare X2 to critical value
how to calculate degrees of freedom for chi-squared
number of categories - 1
x2 smaller than critical value
accept null hypothesis
x2 greater than critical value
reject null hypothesis
less than a 5% probability that difference us due to chance
what is the null hypothesis for student’s t test
no significant difference between the …. of …. and ….
how to calculate degrees of freedom for student’s t test
(number of measurements in first sample + number of measurements in second sample) - 2
formula for student’s t test
formula for standard deviation
n = number of measurements
x = mean
xi = data value
how to calculate students’s t test
1 calculate t value
2 degrees of freedom
3 find critical value
4 compare t value to critical value
t value smaller than critical value
accept null hypothesis
t value greater than critical value
reject null hypothesis
less than a 5% probability that difference is due to chance
