STATISTICAL TESTS Flashcards
What are the 2 types of statistical tests?
Descriptive statistics
Inferential statistics
What are descriptive statistics?
Provide an understanding of the data
What is inferential statistics ?
Enable differences about a population based on the sample of data that has been collected
What are prescriptive statistics ?
Range, +2 standard deviation Al from the mean, median, mode
Inferential statistics examples?
Chi squared, spearman’s rank, students t test
What is a null hypothesis ?
A statement which is usually that there is no difference between the samples being studied.
So as a result of a statistical test either disproves or fails to disprove that null hypothesis; it can never prove a hypothesis to be true
What is a chi squared test?
Observed outcomes vs expected
-frequencies and categoric data
- the measurements relate to the number of individuals in particular categories
- the observed number can be compared with an expected number which is calculated from a theory
What is spearman’s rank?
Correlation : strength of relationships
Used when you have two sets of measurement variables and you want to see whether as one variable increases (correlation), the other variable tends to increase or decrease
- between 7 +30 pairs of measurements
What is a students t test?
Comparing means
Use this test when you are looking for the difference between two means and you want to know if the difference is significant or not
What is SD?
The spread around the mean
What is a critical value?
Helps to decide regions where the left statistics is unlikely to lie
What is a p value, and the number for it ?
A measure of how confident you can be that your results are not due to chance.
P value of 0.05. We can be confident results are significant if there is a less than 5% probability they were due to chance.
If the calculated chi -squared value is higher than the critical value at the p=0.05 level, then we REJECT the null hypothesis
What are the degrees of freedom?
N-1, where n is the number of categories
What is the null hypothesis for a chi squared test?
“There s no significant difference in the frequency of ___________ in category ‘a’ and category ‘b’
How do you calculate expected frequency ?
Expected frequency = total number / number of categories
What are steps for a statistical test?
- Form a null hypothesis
- Choose correct statistical test + justify your answer
- Calculate value of t using formula
- Determine degrees of freedom and find critical value to compare our statistical test to.
- Interpret value of statistics by making a conclusion (is there significant difference or not, reject or accept null hypothesis)
Explain the steps of chi-squared (detailed explanation):
- Formulate a null hypothesis : “there is no significant difference in the frequency of ____ in category ‘a’ and category ‘b’.
- Find the expected frequency : E= total number/number of categories.
Minus this answer from your observed frequency (O-E)
Then square your answers, then add it all together for the sum - Then divide this number of expected frequency.
- Equation for this : X^2 = sum of (O-E)^2/ E
- Find the degrees of freedom : n-1, where n is the number of categories
- Always use 5%/0.05.
- If chi-squared value > critical value, REJECT null hypothesis. There is significant difference between …
Less than 5% is due to chance.
Explain the steps of student’s t test (detailed explanation).
- Form a null hypothesis
- Calculate the value of t, using the formula:
t=__ __
x1 - X2 / square root (S1^2/n1) + (S2^2/n2)
S1 and S2 ~> SD of 1st and 2nd sample. __
Work out SD using s= square root :sum of (X- x )^2 / n-1
- If unpaired t test (n1,n2), then df = (n1+n2) - 2
If paired t test (2 sets of data from same individual . Results are paired, both groups have same sample size), then df=n-1 - If t > critical value, REJECT null hypothesis , there is significant difference between …
Less than 5% due to chance
Explain the steps of Spearman’s rank (detailed example):
- Formulate a null hypothesis - “there will be no significant correlation between “
- Convert the data into ranks
E.g.
Temp | Rank | Rate of reaction | Rank | D |D^2
10 | 1 | 5 | 1 | 0 |
20 | 2 | 13 | 2 | 0 |
30 | 3 | 28 | 3.5 | 0.5 |
40 | 4 | 35 | 6 | 2 |
50 | 5 | 40 | 7 | 2 |
60 | 6 | 30 | 8 | 1 |
70 | 7 | 28 | 3.5 | 3.5 |
If the same numbers have the same rank, find the average of the 2 ranks.
E.g. look at example above, rank 3 and rank 4. 3+4= 7 7/2 = 3.5 - Find difference between ranked values ~> D
- Then square the differences [D^2], then add it all up ~> sum of D^2
- Input this into formula : rs = 1 - [6 X sum of D^2/ n^3 - n]
- No degrees of freedom, just number of paired measurements.
- If our calculated value> critical value, we REJECT null hypothesis. There is a significant correlation between ….
Less than 5% due to chance.
