Lecture 8: Non-parametric tests

Question 1

What does the chi-square tests?

Answer 1

whether there is a relationship between two categorical variables.

Question 2

What leads to chi-squared test?

Answer 2

Q: What sort of measurement? A: Categorical (in this case counts or frequencies)
Q:How many predictor variables? A: One
Q: What type of predictor variable? A: Categorical
Q: How many levels of the categorical predictor? A: Not relevant
Q: Same or Different participants for each predictor level? A: Different
This leads us to and Chi-square test for independence of groups

Question 3

IV and DV in chi-squared tests - (2)

Answer 3

One categorical DV (because of frequencies)

with one categorical IV with different participants at each predictor level

Question 4

In chi-square since we are using categorical variables we can not use

Answer 4

mean or any similar statistic hence cannot use any parametric tests

Question 5

In chi-square test when measuring categorical variables we are interested in

Answer 5

frequencies (number of items that fall into combination of categories)

Question 6

What does chi-square compare?

Answer 6

observed frequencies from the data with frequencies which would be expected if there was no relationship between the two variables.

Question 7

In chi-square test, participants is allocated to one and only one category such as - (3)

Answer 7

pass or fail,

pregnant or not pregnant,

win, draw or lose

Question 8

What is assumptions of chi-square test? - (3)

Answer 8

Data values that are a simple random sample from the population of interest.

Two categorical or nominal variables. Don’t use the independence test with continuous variables that define the category combinations. However, the counts for the combinations of the two categorical variables will be continuous.

For each combination of the levels of the two variables, we need at least five expected values. When we have fewer than five for any one combination, the test results are not reliable

Question 9

Since each participant is allocated to one category in chi-squared test each individual therefore

Answer 9

contributes to the frequency or count with which a category occurs

Question 10

In chi-squared categorical outcomes, the null hypothesis is set

Answer 10

up on the basis of expected frequencies, four all four variable combinations, based on the idea that the variable of interest has no effect on frequencies

Question 11

Example of scenario using chi-square

Answer 11

We have a list of movie genres; this is our first variable. Our second variable is whether or not the patrons of those genres bought snacks at the theatre. Our idea (or, in statistical terms, our null hypothesis) is that the type of movie and whether or not people bought snacks are unrelated. The owner of the movie theatre wants to estimate how many snacks to buy. If movie type and snack purchases are unrelated, estimating will be simpler than if the movie types impact snack sales.

Question 12

We have a list of movie genres; this is our first variable. Our second variable is whether or not the patrons of those genres bought snacks at the theatre. Our idea (or, in statistical terms, our null hypothesis) is that the type of movie and whether or not people bought snacks are unrelated. The owner of the movie theatre wants to estimate how many snacks to buy. If movie type and snack purchases are unrelated, estimating will be simpler than if the movie types impact snack sales.

Is the Chi-square test of independence an appropriate method to evaluate the relationship between movie type and snack purchases? - (3)

Answer 12

We have a simple random sample of 600 people who saw a movie at our theatre. We meet this requirement.

Our variables are the movie type and whether or not snacks were purchased. Both variables are categorical.

But last requirement is for more than five expected values for each combination of the two variables. To confirm this, we need to know the total counts for each type of movie and the total counts for whether snacks were bought or not. = check later

Question 13

We have a list of movie genres; this is our first variable. Our second variable is whether or not the patrons of those genres bought snacks at the theatre. Our idea (or, in statistical terms, our null hypothesis) is that the type of movie and whether or not people bought snacks are unrelated. The owner of the movie theatre wants to estimate how many snacks to buy. If movie type and snack purchases are unrelated, estimating will be simpler than if the movie types impact snack sales.

Diagram of contigency table in Chi-square and calculating row totals and colum and grand total - (7)

Answer 13

50 + 125 + 90 +45 = 310

75 + 175 + 30 + 10 = 290

50 + 75 = 125

125 + 175 = 300

90 + 30 = 120

45 + 10 = 55

310 + 290 = 600

Question 14

We have a list of movie genres; this is our first variable. Our second variable is whether or not the patrons of those genres bought snacks at the theatre. Our idea (or, in statistical terms, our null hypothesis) is that the type of movie and whether or not people bought snacks are unrelated. The owner of the movie theatre wants to estimate how many snacks to buy. If movie type and snack purchases are unrelated, estimating will be simpler than if the movie types impact snack sales.

Diagram of contigency table in Chi-square of calculating eexpected counts

Answer 14

e.g., for action and snacks it would be column total (310) * row total (125) divided by grand total of 600 = 65

Question 15

How to calculate chi-square test statistic? - (4)

Answer 15

1. Calculate the difference from actual and expected for each Movie-Snacks combination.
2. square that difference.
3. Divide by the expected value for the combination.
4. We add up these values for each Movie-Snacks combination. This gives us our test statistic.

Question 16

Example of calculating chi-square from table

Answer 16

For this it would be 65.03

Question 17

How to understand your test statistic from chi-squared? - (5) if you have test statistic of 65.03

Answer 17

1. Set your significance level = .05
2. Calculate the test statistic -> 65.03
3. Find your critical value from chi-squared distribution table based on df & significance level
4. Degrees of freedom: df (r – 1) x (C-1)
   For the movie example this is; Df = (4-1) x (2-1) = 3 -> 7.815
5. compare test statistic with critical level
   65.03 > 7.82 so reject the idea that movie type and snack purchases are independent

Question 18

Example of research question and hypothesis and sig level of chi-square test of independence- (4)

Answer 18

Research question:
Does the area of psychology that a person prefers depend on whether they would select a cat or a dog as a pet?

Hypotheses:
H0: The area of interest in psychology and type of pet preferred are independent of each other.
H1: The area of interest in psychology and type of pet preferred are not independent of each other. That is the primary area of interest in psychology depends on whether you prefer a cat or a dog.
Significance level: α = .05

Question 19

Does the area of psychology that a person prefers depend on whether they would select a cat or a dog as a pet? - chi-square test of independence

Chi-square example we need to check the assumptions below - (2)

Answer 19

Independence
Each item or entity contributes to only one cell of the contingency table.

The expected frequencies should be greater than 5.
In larger contingency tables up to 20% of expected frequencies can be below 5, but there is a loss of statistical power.
Even in larger contingency tables no expected frequencies should be below 1.

Question 20

Does the area of psychology that a person prefers depend on whether they would select a cat or a dog as a pet? - chi-square test of independence

Chi-square example we need to check the assumptions of The expected frequencies should be greater than 5.

What does it show? - (4)

Answer 20

Here we see that all the expected counts in the cat group and one expected count in the dog group are below 5.

We also have one in the cat group that is below 1.

So, SPSS has flagged that we have 60% of the expected counts falling below 5.

So assmption of expected frequencies greater than 5 is not assumed

Question 21

If chi-square assumption that The expected frequencies should be greater than 5 is not satisfied then do - chi-square test of independence

Answer 21

We should use Fisher’s Exact Test which can correct for this.

Question 22

Does the area of psychology that a person prefers depend on whether they would select a cat or a dog as a pet? - - chi-square test of independence

If assumptions were met (expected frequencies greater than 5) then.. report - (2)

Answer 22

A chi-square independence test was performed to examine whether there was a relationship between their area of studies in psychology and their preference for cats or dogs.

The relationship between these variables was not significant, χ²(4, N = 46) = 1.46, p = .834, so we fail to reject H0.

Question 23

Are directional hypotheses possible with chi-square?

A.Yes, but only when you have a 2 × 2 design.
B.Yes, but only when there are 12 or more degrees of freedom.
C.Directional hypotheses are never possible with the chi-squared test.
D.Yes, but only when your sample is greater than 200.

Answer 23

A = only when you have 2 variables to compare and can’t do non-directional in chi-square have to use loglinear or goodness of fit tests

Question 24

Example situations you can do chi-square directional and not possible - (5)

Answer 24

If we are just comparing pet preferences between males and females, we can make a directional hypothesis (2 x 2 – male/female, cats/dogs).

Males prefer cats or females prefer dogs.

However, when we start adding variables to the design it gets complicated.

If we wanted to compare drink preferences at different times of the day for students/lecturers, we couldn’t form a directional hypothesis.

This is because we have 3 main effects and several interactions to consider. We need to use loglinear analyses to do this.

Question 25

Table scenario in which cats can be trained to dance more effectively with food or affection at reward - chi-squared test

Answer 25

Question 26

Table scenario in which cats can be trained to dance more effectively with food or affection at reward - chi-squared test what are the four categories? - (4)

Answer 26

* could they dance - yes * could they dance - no * food as reward * affection as reward

Question 27

Table scenario in which cats can be trained to dance more effectively with food or affection at reward - chi-squared test highlight the frequencies for four categories

Answer 27

Question 28

Table scenario in which cats can be trained to dance more effectively with food or affection at reward - chi-squared test what do the rows give?

Answer 28

Row totals give frequencies of dancing and non-dancing cats

Question 29

Table scenario in which cats can be trained to dance more effectively with food or affection at reward - chi-squared test what do the columns give? - (2)

Answer 29

The column totals give frequencies of food and affection as reward These are the numbers in each group

Question 30

Loglinear analysis is a .... of chi-square

Answer 30

extension

Question 31

Chi-square only analyses two variables at a time, whilst log-linear models

Answer 31

can determine complex interactions in multidimensional contingency tables with more than two categorical variables.

Question 32

Loglinear is appropriate when

Answer 32

there’s no clear distinction between response and explanatory variables

Question 33

think of

Answer 33

Think of chi-square like t-tests (2 groups) and log-linear like ANOVA (more than 2 groups).

Question 34

Example of RQ, hypothesis and sig level of loglinear - (3)

Answer 34

Research question: Is the new treatment associated with improvements in health in cats and dogs? Hypotheses: H0: Treatment, type of animal and improvements are independent of each other. H1: Treatment, type of animal and improvements are associated with each other. Significance level: α = .05

Question 35

Assumptions of log linear - (2)

Answer 35

Independence Expected counts > 5

Question 36

Research question: Is the new treatment associated with improvements in health in cats and dogs? Checking assumption of expected counts - (3):

Answer 36

Here we have 3 things we are comparing: animal (cat/dog), treatment (yes/no) and improvement (yes/no) all of which are categorical. We look and see that all of the expected counts are above 5. So met assumption of independence and expected counts

Question 37

Research question: Is the new treatment associated with improvements in health in cats and dogs? Here we have 3 things we are comparing: animal (cat/dog), treatment (yes/no) and improvement (yes/no) all of which are categorical. In loglinear model selection it begins with - (2)

Answer 37

all terms present (all main effects and all possible interactions main effects: Animal, Treatment and Improvement interactions: Animal * Treatment, Animal * Improvement, Treatment * Improvement and Treatment* Animal* Improvement

Question 38

Research question: Is the new treatment associated with improvements in health in cats and dogs? Here we have 3 things we are comparing: animal (cat/dog), treatment (yes/no) and improvement (yes/no) all of which are categorical. In loglinear model selection after including all main effects and interactions then it - (4)

Answer 38

Remove a term and compares the new model with the one in which the term was present. Starts with the highest-order interaction (including max number of variables/categories) Uses the likelihood ratio to ‘compare’ models below: If the new model is no worse than the old, then the term is removed and the next highest-order interactions are examined, and so on.

Question 39

Model selection of loglinear - what does it show? Research question: Is the new treatment associated with improvements in health in cats and dogs? Here we have 3 things we are comparing: animal (cat/dog), treatment (yes/no) and improvement (yes/no) all of which are categorical. - (3)

Answer 39

We can see that the model selection worked in a way that it first tried to remove the 3-way interaction. However, we can see here that it * affected the fit of the model, so it was left in. Since removing the highest-order interaction made a * difference to the fit of the model, we get a final model that is the saturated model (it contains all main effects and interactions).

Question 40

Loglinear SPSS K way and Higher order effects what does it show? Research question: Is the new treatment associated with improvements in health in cats and dogs? Here we have 3 things we are comparing: animal (cat/dog), treatment (yes/no) and improvement (yes/no) all of which are categorical. - (2)

Answer 40

we are using the likelihood ratio here because that’s how we compare the models to find the best fit . We see that all main effects and interactions are significantly contributing to explaining the variance in the data

Question 41

loglinear what does K represent and what does K = 1,2 and 3 represent? - (4)

Answer 41

K represents the level of the terms. For example, K=1 would be the main effects, K=2 would be our 2-way interactions and K=3 is our 3-way interaction.

Question 42

Loglinear SPSS - what does parameter estimates show? Research question: Is the new treatment associated with improvements in health in cats and dogs? Here we have 3 things we are comparing: animal (cat/dog), treatment (yes/no) and improvement (yes/no) all of which are categorical. - (3)

Answer 42

There is a significant three-way interaction between animal, treatment and improvement, as well as two significant two-way interactions between animal and improvement and treatment and improvement (p < .001) a * 3-way interaction between animal, treatment and improvement as well as two * 2-way interaction between animal/improvement and treatment/improvement. Like our post-hoc tests, this is telling us where the * differences are.

Question 43

Loglinear after seeing statistical tests we go to raw data showing that...

Answer 43

Based on the raw data, there seems to be indication that the cats responded better to treatment than dogs, this should be followed up by chi-square tests separately for cats and dogs to determine whether the association between treatment and improvement is present in both cats and dogs

Question 44

When conducting a loglinear analysis, if our model is a good fit of the data then the goodness-of-fit statistic for the final model should be: A. Significant (p should be smaller than .05) B. Non-significant (p should be bigger than .05) C. Less than 5 but greater than 1 D. Greater than 5

Answer 44

B

Question 45

The goodness of fit tests in log linear tests

Answer 45

hypothesis that frequencies predicted by model (expected frequencies) are sig different from actual frequencies in data (obsevered)

Question 46

A significant goodness of fit result mean

Answer 46

our model was significantly different from our data (i.e., the model is a bad fit to the data).

Question 47

A recent story in the media has claimed that women who eat breakfast every day are more likely to have boy babies than girl babies. Imagine you conducted a study to investigate this in women from two different age groups (18–30 and 31–43 years). Looking at the output tables below, which of the following sentences best describes the results? = chi-square A. Women who ate breakfast were significantly more likely to give birth to baby boys than girls. B. There was a significant two-way interaction between eating breakfast and age group of the mother. C. Whether or not a woman eats breakfast significantly affects the gender of her baby at any age. D. The model is a poor fit of the data.

Answer 47

C

Question 48

Chi square and log linear are both

Answer 48

non-parametric methods

Question 49

Non-parametric tests used when

Answer 49

When data violate the assumptions of parametric tests we can sometimes find a nonparametric equivalent eg. normality of distribution

Question 50

Non-parametric tests work on the principle of

Answer 50

randomization or ranking the data for each group

Question 51

Ranking data gets rid of in non parametric

Answer 51

outliers and skew

Question 52

How does ranking work in non-parametric? - (2)

Answer 52

Add up the ranks for the two groups and take the lowest of these sums to be our test statistic The analysis is carried out on the ranks rather than the actual data.

Question 53

Non-parametric equivalent of independent/unrelated t-tests

Answer 53

Mann-Whitney or Wilcoxon rank-sum test

Question 54

Non-parametric equivalent of repeated t-test

Answer 54

Wilcoxon signed-rank test

Question 55

Non-parametric equivalent of : One-way independent (between-subjects) ANOVA

Answer 55

Kruskall-Wallis or (for trends) Jonckheere-Terpstra

Question 56

Non-parametric equivalent of one-way repeated ANOVA

Answer 56

Friedmanʼs ANOVA

Question 57

Non-parametric equivalent of Multi-way between or within-subjects ANOVA

Answer 57

Loglinear analysis (categorical outcome, with participants as a factor)

Question 58

Non-parametric equivalent of correlation

Answer 58

Spearman’s Rho or Kendall’s Tau

Question 59

Mann-Whitney/Wilcoxon rank-sum Test - Compares

Answer 59

two independent groups of scores

Question 60

Wilcoxon signed rank Test - Compare

Answer 60

two dependent groups of scores

Question 61

Kruskal-Wallis Test - Compares

Answer 61

> 2 independent groups of scores

Question 62

Friedman’s Test - Compares

Answer 62

> 2 dependent groups of scores

Question 63

Spearman’s Rho & Kendall’s Tau - Measures the extent to which

Answer 63

two continuous variables are related (pattern of responses across variables)

Question 64

Logic behind Wilcoxon's rank sum test, what does SPSS do? - (3)

Answer 64

Step 1: Get some not normally distributed data Step 2: Rank it (regardless of group) Step 3: Significance testing Does one of the groups have more of the higher ranking scores than the other?

Question 65

What is DF of chi-square?

Answer 65

(r-1)(c-1)

Question 66

The likelhood ratio in loglinear model preferred

Answer 66

small sample sizes

Question 67

DF of likelhood ratio in loglinear

Answer 67

df = (r-1)(c-1)

Question 68

Decision tree of Mann Whitney - (4)

Answer 68

1 DV = Ordinal (e.g., high school, bachelors, order is meaningful) or continous 1 IV = Categorical and 2 levels Different partiicpants Does not meet assumption of parametric

Question 69

Wilcoxon rank sum and Man Whitney U is

Answer 69

same procedure and used to compare two independent groups and assess whether samples come from same distribution

Question 70

For Mann-Whitney U/Wilcoxon Rank Sum they comparing 2 independent conditions the two steps - (2)

Answer 70

Rank all the data on the the basis of the scores irrespective of the group compute the sum of ranks of each group

Question 71

For wilcoxon rank sum, the statistic Ws is

Answer 71

the lower of the two sums of ranks

Question 72

For Mann-Whitney, the statistic U use the

Answer 72

sum of ranks for group 1, R1, as follows

Question 73

Example of table where comparing 2 independent conditions of Wilcoxon rank sum or Mann Whitney U test

Answer 73

Here we have data for two groups; one taking alcohol, the other ecstasy. The scores for a measure of depression. Scores were obtained on two days; Sunday and Wednesday. The drugs were administered on Saturday.

Question 74

Example of table where comparing 2 independent conditions of Wilcoxon rank sum or Mann Whitney U test Here we have data for two groups; one taking alcohol, the other ecstasy. The scores for a measure of depression. Scores were obtained on two days; Sunday and Wednesday. The drugs were administered on Saturday. Two steps for both statistics: Rank all the data on the the basis of the scores irrespective of the group compute the sum of ranks of each group - (5)

Answer 74

The graphic here shows how we can list the scores in order and as a result assign each score a rank. When scores tie, we give them the average of the ranks. If we ensure we keep track of the group the scores came from we can relatively easily add the ranks up for each group. Note, that if there was little difference between the groups the sums of their ranks would be similar, as they are for the data shown her for Sunday. However, the sum of ranks differ considerably for the data obtained on Wednesday.

Question 75

For Wilcoxon sum of ranks = comparing 2 independent groups the W s statistic, the group sizes of n1 and n2 the mean of W2 is given:

Answer 75

Question 76

For Wilcoxon sum of ranks = comparing 2 independent groups the W s statistic, the standard error of Ws is given

Answer 76

Question 77

For Wilcoxon sum of ranks = comparing 2 independent groups the z score of Ws can be calculated

Answer 77

Question 78

For Mann Whitney , the statistic U use the the sum of ranks for group 1, R1, as follows

Answer 78

Question 79

For Mann Whitney , the statistic U use the the sum of ranks for group 1, R1, as follows Specificy equation - (2)

Answer 79

The first terms involving n1 and n2 actually compute the maximum possible sum of ranks for group 1. U is zero when all those in group one have scores that exceed the scores of those in group 2.

Question 80

In Mann Whitney U there is a standardised test statistic which is z score that can allow you to compute

Answer 80

effect size so r = z / square root of N (number of pps

Question 81

What is decision tree of Wilcoxon signed rank test? - (4)

Answer 81

1 IV categorical with 2 levels Same participants in each predictor level 1 DV - Ordinal or continous Does not meet assumption of parametric tests

Question 82

Steps of Wilcoxon signed rank test - (4)

Answer 82

1. Compute the difference between scores for the two conditions 2. Note the sign of the difference (positive or negative) 3. Rank the differences ignoring the sign and also exclude any zero differences from the ranking 4. Sum the ranks for positive and negative ranks

Question 83

Example of Wilcoxon signed rank test carrying out steps - (9)

Answer 83

The table shown here has the Depression Scores taken on Sunday and Wednesday for those taking ecstasy on Saturday. Data for Sunday are in the first column and Wednesday in the second column. The third column shows the difference between scores obtained on Sunday and Wednesday. NOte some could be negative, some positive. In this example however the difference is always positive apart from two values when the difference is zero. The fourth column notes the sign of the difference or notes it is going to be excluded because the difference was zero. The fifth column ranks the differences in terms of their size, but not sign. The sixth and seventh column list the ranks that were for positive and negative differences, respectively. It is these two columns that are summed to get the relevant statistics, called T+ and T-. Because T+ and T- are not independent, we take only the T+ value.

Question 84

For Wilcoxon signed rank test the group size n the mean T is given

Answer 84

Question 85

For Wilcoxon signed rank test thestandard error fo T is given:

Answer 85

Question 86

For Wilcoxon signed rank test compute z score of T by

Answer 86

Question 87

Kruskal Wallis decision tree like one-way independent ANOVA - (4)

Answer 87

1 DV of ordinal 1 IV categorical predictor of more than 2 levels Diff participants in each predictor level Does not meet assumption of parametric

Question 88

Kruskal Wallis steps - (2)

Answer 88

Rank all the data on the the basis of the scores irrespective of the group Compute the sum of ranks of each group, Ri , where i is the group number

Question 89

For Kruskal-Wallis, the statistic H is as follows

Answer 89

Question 90

What is decision tree of Friedman test? - (4)

Answer 90

1 DV continous or ordinal 1 IV predictor categorical with more than 2 levels Same participants in each predictor level Doesnot meet assumption of parametric tests

Question 91

What is steps of Friedman test? - (2)

Answer 91

Rank the scores or each individual - that means you will have ranks varying from 1 to the number of conditions the participants took part in Compute the sum of ranks, Ri , for each condition

Question 92

For Friedman, the statistic F is as follows

Answer 92

K = conditions N = number of pps

Question 93

Example when using chi-square test - (3)

Answer 93

- In this example, they wanted to look at whether attendance at lectures had an impact on their exam performance on whether they passed or failed - Attendence was coded as 1 if participants generally attended lectures , barirng illness, and 2 if they did not attend - Exam was scored as 1 = Pass and 2 = Fail

Question 94

- In this example, they wanted to look at whether attendance at lectures had an impact on their exam performance on whether they passed or failed - chi square What does it show? - (4)

Answer 94

- Attendence, Attended Lectures , Count = this is people who attended lecture and number of people who passed was 84 and people who failed was 29 - % Within attendence give same info so 74.3% passed and 25.7% failed when attended lectures -Going to didn’t attend lectures, 22 people passed and 35 failed and below is in percentages: - Easier using percentages writing up

Question 95

- In this example, they wanted to look at whether attendance at lectures had an impact on their exam performance on whether they passed or failed - chi square What does it show? - (2)

Answer 95

- At top row, pearson chi-squared which chi-square statistic which was 20.617, DF which is 1 and p-value was 0.000 - 0 cells have a count less than 5  met assumption of chi-square test that expected counts greater than 5

Question 96

- DF is always ... in two-by-two chi-square

Answer 96

1

Question 97

If SPSS output shows below in chi-square that

Answer 97

0 cells have a count less than 5  met assumption of chi-square test

Question 98

- In this example, they wanted to look at whether attendance at lectures had an impact on their exam performance on whether they passed or failed - chi square What does it this effect size show? - (2)

Answer 98

- x^2 (1) = 20.62, p < 0.001 - Cramer's V = 0.35 , indicating a medium effect size

Question 99

Effect size guideline of r correlation coefficient - (3)

Answer 99

Small effect = 0.1 Medium effect = 0.3 Large = 0.5 and above

Question 100

Cramer's V can be interpreted similar to

Answer 100

correlation coefficient:

Question 101

In chi-square we can calculate odds

Answer 101

ratio

Question 102

Example of calculating odds ratio for chi-square - (3)

Answer 102

odds of passing/failing for students who attended lecture = no. of students who attended and passed (84) / no. of students who attedned and failed (29) = 2.897 odds of passing/failing for students who did not attend = no. of students who did not attend lectures and passed (22) / no of students who did not attended and fail (35) Odds ratio = odds of P/F of attended/ odds of P/F of not attended = 2.897/ 0.629 = 4.606 saying for an individual who attended lectures lead them to be more likely to pass exam

Question 103

Example research scenario of Mann Whitney - (4)

Answer 103

- Independent sample design - One IV, two conditions = existing vs new medication - One DV (symptoms) but this time on ordinal (scale from 1 to 5) and got combination of non-normally distributed data and small sample size (very problematic for t-tests) - Mann Whitney U Test

Question 104

Example of using Mann Whitney U = skew

Answer 104

Question 105

What does this Mann Whitney U show? - Independent sample design - One IV, two conditions = existing vs new medication - One DV (symptoms) but this time on ordinal (scale from 1 to 5) and got combination of non-normally distributed data and small sample size (very problematic for t-tests) - Mann Whitney U Test

Answer 105

- This box summarises the p-value ( p = 0.026) and tells you whether to accept or reject the null hypothesis.

Question 106

What does this output show of Mann Whitney U? - (3)

Answer 106

- Mann Whitney U test statistic is 166.000 to report and also people report standardised test statistic 2.292 which is z score so handy to report as know if its above +/-1.96 then p-value we get out of test is significant - P-value of exact significant is p = 0.026 - This is significant difference between the 2 groups

Question 107

- Next we would want to look at the median scores to see which group is scoring highest and lowest after sig Mann Whitney U test What does this output show? - (3) - Independent sample design - One IV, two conditions = existing vs new medication - One DV (symptoms) but this time on ordinal (scale from 1 to 5) and got combination of non-normally distributed data and small sample size (very problematic for t-tests) - Mann Whitney U Test

Answer 107

- For existing treatment, median score was 3. And new treatment the median score was 4. It suggests new treatment was more effective in reducing symptons than the existing treatment

Question 108

Example research scenario of Friedman ANOVA - (5)

Answer 108

- Again we got ordinal data for DV not sure distances between levels is going to be the same - Related design - One IV, 3 conditions - One DV (level reached to video game) - Friedman’s ANOVA = more than 2 groups in related design

Question 109

What does this Friedman ANOVA output show?

Answer 109

- We got total sample size which is 30 and test statistic which is 21.788, DF = 2 and p value is 0.000 so significant difference between the 3 groups

Question 110

For Friedman's ANOVA we do

Answer 110

post hoc tests for pairwise comparisons to look where the differences are

Question 111

What do this Friedman ANOVA test post hoc tests show? - (7)

Answer 111

- First one is Joy stick vs Vyper Max - Second one is Joystick vs Evo Pro etc… - Notice it gives two p-values of sig and adjusted sig - Adjusted sig control for multiple comparison and make correcitons to p-value (use this - Difference between joystick vs Vyper Max was sig at p = 0.005 - Difference between Joystick vs Evo Pro was sig at p = 0.00 - Difference between VyperMax vs Evopro is non-significant as p = 0.660

Question 112

- The problem with non-parametric tests is that they have less power

Answer 112

to detect sig effects compared to parametric effects so maybe issue of dealing with power so may have median scores higher in one then another but not sig

Question 113

Non-parametric tests are used when A. The assumptions of parametric tests have not been met. B. You want to increase the power of your experiment. C. You have more than the maximum number of tied scores in your data set. D. All of these.

Answer 113

A = non parametric have fewer assumptions than parametric

Question 114

With 2  2 contingency tables (i.e., two categorical variables both with two categories) no expected values should be below ____. A. 5 B. 1 C. 0.8 D. 10

Answer 114

A

Question 115

Which of the following statements about the chi-square test is false? A. Which of the following statements about the chi-square test is false? B. The chi-square test can be used to check how well a model fits the data. C. The chi-square test is used to quantify the relationship between two categorical variables. D. The chi-square test is based on the idea of comparing the frequencies you observe in certain categories to the frequencies you might expect to get in those categories by chance.

Answer 115

A = correct, because it is false. Chi-square can be used on categorical variables only

Question 116

When conducting a loglinear analysis, if our model is a good fit of the data then the goodness-of-fit statistic for the final model should be: Hint: The goodness-of-fit test tests the hypothesis that the frequencies predicted by the model (the expected frequencies) are significantly different from the actual frequencies in our data (the observed frequencies).) A. Non-significant (p should be bigger than .05) B. Significant (p should be smaller than .05) C.Greater than 5 D. Less than 5 but greater than 1

Answer 116

A = If our model is a good fit of the data then the observed and expected frequencies should be very similar (i.e., not significantly different

Question 117

What is the parametric equivalent of the Wilcoxon signed-rank test? A. The paired samples t-test B. The independent t-test C. Independent ANOVA D. Pearson’s r correlation

Answer 117

A

Question 118

Are directional hypotheses possible with chi-square? A. Yes, but only when you have a 2 × 2 design. B. Yes, but only when there are 12 or more degrees of freedom. C. Directional hypotheses are never possible with the chi-squared test. D. Yes, but only when your sample is greater than 200.

Answer 118

A =

Question 119

A psychologist was interested in whether there was a gender difference in the use of email. She hypothesized that because women are generally better communicators than men, they would spend longer using email than their male counterparts. To test this hypothesis, the researcher sat by the computers in her research methods laboratory and when someone started using email, she noted whether they were male or female and then timed how long they spent using email (in minutes). How should she analyse the differences in males and females (use the output below to help you decide)? A. Mann–Whitney test B. Paired t-test C.Wilcoxon signed-rank test D. Independent t-test

Answer 119

A.Mann Whitney Test

Question 120

What is the Jonckheere–Terpstra test used for? A. To test for an ordered pattern to the medians of the groups you’re comparing. B. To test whether the variances in your data set are approximately equal. C. To test for an ordered pattern to the means of the groups you’re comparing. D.To control for the familywise error rate.

Answer 120

A