Stats tests

Question 1

What are the uses for a Spearman’s rank coefficient test

Answer 1

To determine if there is a significant correlation between 2 continuous variables.

Question 2

What is the null hypothesis of a spearman’s rank coefficient test

Answer 2

There is no significant correlation between x and y.

Question 3

What are the degrees of freedom in a spearman’s rank coefficient test

Answer 3

The number of pairs of data

Question 4

How do you interpret the calculated value against the critical value in a spearman’s rank coefficient test

Answer 4

• If the calculated value < critical value at p=0.05 (5%) then there is NO significant correlation between x and y, so ACCEPT null hypothesis
• If the calculated value is > or equal to the critical value at p=0.05 (5%) then there IS a significant correlation between x and y, so REJECT null hypothesis

Question 5

What are the reasons for using a Chi squared test

Answer 5

To determine if there is a significant difference between the observed and expected numbers in categories x and y.

Question 6

What is the null hypothesis for a Chi squared test

Answer 6

There is no significant difference between the observed and expected numbers for x and y.

Question 7

How do you find the degrees of freedom for a Chi squared test

Answer 7

Number of categories -1 (e.g. if there were 7 categories, the degrees of freedom would be 6).

Question 8

How do you interpret the calculated value against the critical value in a Chi squared test

Answer 8

• If the calculated value is < critical value at p=0.05 (5%), there is NO significant difference so ACCEPT null hypothesis
• If calculated value is > critical value at p=0.05 (5%), there IS a significant difference so REJECT null hypothesis

Question 9

What are the reasons for using a student t test

Answer 9

To determine of there is a significant difference between the means of two or more datasets.

Question 10

What is the null hypothesis of a student t test

Answer 10

There is no significant difference between the mean (number of)… and …

Question 11

How do you find the degrees of freedom for a student t test

Answer 11

number of values in dataset one + number of values in dataset two - number of means

Question 12

How do you interpret the calculated value against the critical value in a student t test

Answer 12

• If calculated value is < critical value at p=0.05 (5%), there is NO significant difference between means so ACCEPT null hypothesis
• If calculated value is > critical value at p=0.05 (5%), there IS a significant difference between the means so REJECT the null hypothesis

Question 13

Why do you reject the null hypothesis if the calculated value is less than the critical value

Answer 13

• if the calculated value is less than the critical value, it shows that there is a greater than 5% probability that the differences/correlation in data is due to random chance
• This implies that there is some correlation/difference
• This means you can reject the null hypothesis as the null hypothesis states there is no relationship/difference

Question 14

Why do you reject the null hypothesis if the calculated value is greater than the critical value

Answer 14

• If the calculated value is greater than the critical value, it shows that there is a less than 5% probability that the differences/correlation is due to random chance
• This implies that the difference/correlation in the dataset is most likely completely random
• This means you can accept the null hypothesis as the null hypothesis states that there is no relationship/difference