STATISTICAL TESTS

Question 1

What are the 2 types of statistical tests?

Answer 1

Descriptive statistics
Inferential statistics

Question 2

What are descriptive statistics?

Answer 2

Provide an understanding of the data

Question 3

What is inferential statistics ?

Answer 3

Enable differences about a population based on the sample of data that has been collected

Question 4

What are prescriptive statistics ?

Answer 4

Range, +2 standard deviation Al from the mean, median, mode

Question 5

Inferential statistics examples?

Answer 5

Chi squared, spearman’s rank, students t test

Question 6

What is a null hypothesis ?

Answer 6

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

Question 7

What is a chi squared test?

Answer 7

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

Question 8

What is spearman’s rank?

Answer 8

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

Question 9

What is a students t test?

Answer 9

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

Question 10

What is SD?

Answer 10

The spread around the mean

Question 11

What is a critical value?

Answer 11

Helps to decide regions where the left statistics is unlikely to lie

Question 12

What is a p value, and the number for it ?

Answer 12

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

Question 13

What are the degrees of freedom?

Answer 13

N-1, where n is the number of categories

Question 14

What is the null hypothesis for a chi squared test?

Answer 14

“There s no significant difference in the frequency of ___________ in category ‘a’ and category ‘b’

Question 15

How do you calculate expected frequency ?

Answer 15

Expected frequency = total number / number of categories

Question 16

What are steps for a statistical test?

Answer 16

1. Form a null hypothesis
2. Choose correct statistical test + justify your answer
3. Calculate value of t using formula
4. Determine degrees of freedom and find critical value to compare our statistical test to.
5. Interpret value of statistics by making a conclusion (is there significant difference or not, reject or accept null hypothesis)

Question 17

Explain the steps of chi-squared (detailed explanation):

Answer 17

1. Formulate a null hypothesis : “there is no significant difference in the frequency of ____ in category ‘a’ and category ‘b’.
2. 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
3. Then divide this number of expected frequency.
4. Equation for this : X^2 = sum of (O-E)^2/ E
5. Find the degrees of freedom : n-1, where n is the number of categories
6. Always use 5%/0.05.
7. If chi-squared value > critical value, REJECT null hypothesis. There is significant difference between …
   Less than 5% is due to chance.

Question 18

Explain the steps of student’s t test (detailed explanation).

Answer 18

1. Form a null hypothesis
2. 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

3. 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
4. If t > critical value, REJECT null hypothesis , there is significant difference between …
   Less than 5% due to chance

Question 19

Explain the steps of Spearman’s rank (detailed example):

Answer 19

1. Formulate a null hypothesis - “there will be no significant correlation between “
2. 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
3. Find difference between ranked values ~> D
4. Then square the differences [D^2], then add it all up ~> sum of D^2
5. Input this into formula : rs = 1 - [6 X sum of D^2/ n^3 - n]
6. No degrees of freedom, just number of paired measurements.
7. If our calculated value> critical value, we REJECT null hypothesis. There is a significant correlation between ….
   Less than 5% due to chance.