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[mystery-two]

Topic: The Impact of Different Teaching Methods on High School Student Performance and Engagement Scenario: You want to evaluate how different teaching methods (e.g., traditional lecture-based, project-based learning, and flipped classroom) affect high school students' academic performance and engagement in class. You also want to examine whether there’s a relationship between student engagement and performance.

Data Collected:

Teaching Method (traditional, project-based, flipped classroom) Student Performance (measured by grades or test scores) Student Engagement (measured by participation, attendance, and classroom behavior) Test 1: ANOVA (Analysis of Variance) Question: Does the type of teaching method (traditional, project-based, flipped classroom) significantly affect student performance (grades or test scores)? Purpose: Use ANOVA to compare the mean test scores or grades of students across the three teaching methods and determine if there are significant differences in academic performance. Test 2: Chi-Square Test Question: Is there an association between the teaching method used (traditional, project-based, flipped classroom) and the level of student engagement (high, medium, low)? Purpose: Use Chi-Square to test if the distribution of student engagement levels (high, medium, low) differs significantly across the three teaching methods. Test 3: Correlation Question: Is there a correlation between student engagement (participation, attendance) and student performance (grades or test scores)? Purpose: Use correlation to examine whether higher levels of student engagement (e.g., more active participation or better attendance) are associated with higher performance on tests or assignments. Summary: This topic allows you to explore the relationship between teaching methods, student engagement, and academic performance. It provides a comprehensive way to test different statistical approaches:

ANOVA to compare the effect of different teaching methods on performance. Chi-Square to analyze categorical data like engagement levels. Correlation to investigate how engagement relates to performance. It’s a straightforward yet effective topic for assessing a potential moderator’s ability to analyze educational data.

Testing moderation with three variables in the context of ANOVA involves examining the interaction between two independent variables (IVs) on a continuous dependent variable (DV). Here's an

Independent Variable 1 (IV1): instruction_method (Method of instruction: traditional, online, or blended)

Independent Variable 2 (IV2): learning_style (Learning style: visual, auditory, or kinesthetic)

Dependent Variable (DV): exam_score (Exam score)

We'll use a 2-way ANOVA to test the moderation effect.

Research Question

Is the effect of instruction_method on exam_score moderated by learning_style?

Hypotheses

Main effect of instruction_method on exam_score

Main effect of learning_style on exam_score

Interaction effect between instruction_method and learning_style on exam_score

ANOVA Table Source DF SS MS F p-value instruction_method 2 100 50 5.00 0.01 learning_style 2 50 25 2.50 0.05 instruction_method x learning_style 4 150 37.5 3.75 0.01 Error 90 900 10 Total 98 1200

Interpretation

The significant main effect of instruction_method (p = 0.01) indicates that exam_score differs across instruction_method groups.

The significant main effect of learning_style (p = 0.05) indicates that exam_score differs across learning_style groups.

The significant interaction effect (p = 0.01) indicates moderation: the relationship between instruction_method and exam_score varies depending on learning_style.

Post-hoc Tests

Perform post-hoc tests (e.g., Tukey's HSD) to examine the nature of the interaction:

Compare exam_score means across instruction_method groups within each learning_style group.

Compare exam_score means across learning_style groups within each instruction_method group.

Visualizing Moderation

Plot the interaction using a 3D bar chart or an interaction plot:

import matplotlib.pyplot as plt import pandas as pd

Sample data

data = { 'instruction_method': ['traditional', 'online', 'blended', 'traditional', 'online', 'blended', 'traditional', 'online', 'blended'], 'learning_style': ['visual', 'visual', 'visual', 'auditory', 'auditory', 'auditory', 'kinesthetic', 'kinesthetic', 'kinesthetic'], 'exam_score': [80, 90, 85, 70, 95, 88, 75, 92, 89] }

df = pd.DataFrame(data)

Plot interaction

plt.figure(figsize=(10, 6)) plt.bar(df['instruction_method'], df['exam_score'], color=['blue' if x == 'visual' else 'red' if x == 'auditory' else 'green' for x in df['learning_style']]) plt.xlabel('Instruction Method') plt.ylabel('Exam Score') plt.title('Moderation Effect') plt.legend(['Visual', 'Auditory', 'Kinesthetic']) plt.show()

[mystery-two]

Generating a Correlation Coefficient

Topic:  Alcohol intake among young adults in the morning, afternoon, and evening on a weekly basis.

X = {categorical: morning, afternoon, evening}

Y= {1, 2, 3, 4, 5, 6… 30} number of participants.

Sample size = 30

Numerical values were assigned to the categories

Morning = 1

Afternoon= 2

Evening = 3

PARTICIPANTS

TIME OF DAY

1

1

2

2

3

3

4

1

5

3

6

3

7

2

8

2

9

1

10

2

11

1

12

3

13

1

14

1

15

2

16

1

17

2

18

2

19

3

20

2

21

2

22

3

23

3

24

1

25

1

26

3

27

2

28

2

29

3

30

1

Mean Calculation: 

X = first term+ last term/2 =1+30/2= 15.5

Y=  1+2 +3+1+3+3+2+2+1+2+1+3+1+1+2+1+2+2+3+2+2+3+3+1+1+3+2+2+3+1/30=65/30=2.17

Deviation Score for X and Y

X={ 1, 2, 3, ……30}

Formula 1 - mean = deviation score

X [−14.5,−13.5,−12.5,−11.5,−10.5,−9.5,−8.5,−7.5,−6.5,−5.5,−4.5,−3.5,−2.5,−1.5,−0.5,0.5,1.5,2.5,3.5,4.5,5.5,6.5,7.5,8.5,9.5,10.5,11.5,12.5,13.5,14.5]

 Y= [-1.17,−0.17,0.83,−1.17,0.83,0.83,−0.17,−0.17,−1.17,−0.17,−1.17,0.83,−1.17,−1.17,−0.17,−1.17,−0.17,−0.17,0.83,−0.17,−0.17,0.83,0.83,−1.17,−1.17,0.83,−0.17,−0.17,0.83,−1.17]

Product of Deviation Scores

Formula:- deviation score X x deviation score Y.

Example -14.5 ×-1.17=16.965

[16.965, 2.295, −10.375, 13.455, −8.715, −7.885, 1.445, 1.275, 7.605, 0.935, 5.265, -2.905, 2.925, 1.755, 0.085, −0.585, −0.255, −0.425 , 2.905 ,−0.765 ,−0.935, 5.395, 6.225 ,−9.945, −11.115, 8.715,

− 1.955,  −2 .125, 11.205, −16.965]

Positive numbers sum 

16.965 + 2.295 + 13.455 + 1.444 + 1.275 + 7.605 + 0.935 + 5.265 + 2.925 + 1.755 + 0.085 + 2.905 + 5.395 + 6.225 + 8.715 + 11.205 =88.450

Negative numbers sum

−10.375 + -8.715+ -7.885 + −2.905 + -0.585 + -0.255 + -0.425 + -0.765 + - 0.935 +- 9.945 + -11.115 + -1.955 +-2.125 + -16.965 = -74.948

Sum of the positive numbers+ negative numbers= 

88.450+(−74.948)=13.502

Standard Deviation

Square all the X values

X = 210.25 + 182.25 + 156.25 + 132.25 +110.25 + 90.25 + 72.25 + 56.25 + 42.25 + 30.25 + 20.25+ 12.25 + 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 + 12.25 + 20.25 + 30.25 + 42.25 + 56.25 + 72.25 + 90.25 + 110.25 + 132.25 + 156.25 + 182.25 + 210.25  

Sum all the values of X

2247.5/30 = 74.916

Then find the square 74.916 = 8.6554

Square all the values of Y

1.3689 + 0.0289 + 0.6889 + 1.3689 + 0.6889 + 0.6889 + 0.0289 + 0.0289 + 1.3689 + 0.0289 + 1.3689 + 0.6889 + 1.3689 + 1.3689 + 0.0289 + 1.3689 + 0..0289 + 0.0289 + 0.6889 + 0.0289 + 0.0289 + 0.6889 + 0.6889 + 1.3689 + 1.3689 + 0.6889 + 0.0289 + 0.0289 + 0.6889 + 1.3689 = 20.207

Sum of Y deviation scores/30

20.207/30= 0.6735

Then the square root= 0.8206

Find the square root of 0.6735 = 0.8206

Final calculation

The sum of the deviation = 13.502

Standard Deviation X= 8.6554

Standard Deviation Y= 0.8206

N=30

13.502/ 30-1 x  8.6554 × 0.8206

13.502/205.9760 = 0.06556

r = 0.06556  a very weak correlation

[mystery-two]

Calculating expected frequencies

Effectiveness

Facebook

Instagram

Twitter

TikTok

YouTube

Total

High

10

5

15

5

10

45

Moderate

30

15

10

10

20

85

Low

20

20

5

15

10

70

Total

60

40

30

30

40

200

Calculate expected frequencies:

E=row total X column total / overall sample size

Effectiveness

Facebook

Instagram

Twitter

TikTok

YouTube

Total

High

(45*60)/200=

13.5

(45*40)/200=

9.0

(45*30)/200=

6.75

(45*30)/200=

6.75

(45*40)/200=

9.0

45

Moderate

(85*60)/200=

25.5

(85*40)/200=

17.0

(85*30)/200=

12.75

(85*30)/200=

12.75

(85*40)/200=

17.0

85

Low

(70*60)/200=

21.0

(70*40)/200=

14.0

(70*30)/200=

10.5

(70*30)/200=

10.5

(70*40)/200=

14.0

70

Total

60

40

30

30

40

200

Chi Square value X2= (0-E) 2/E

Effectiveness

Facebook

(O,E)

Instagram

(O,E)

Twitter

(O,E)

Tik Tok

YouTube

High

(10, 13.5)

(5, 9.0)

(15, 6.75)

(5, 6.75)

(10, 9.0)

Moderate

(30, 25.5)

(15, 17.0)

(10, 12.75)

(10, 12.75)

(20, 17.0)

Low

(20, 21.0)

(20, 14.0)

(5, 10.5)

(15, 10.5)

(10, 14.0

Calculating Degrees of Freedom

df= (number 0f rows-1) x (number of columns-1)

df= (3-1)x (5-1)= 2x4=8

choose a significance level of 0.05

The critical value = X2 = 0.05, 8 = 15.51 using python

Conclusion

The observation made is that effectiveness ratings and social media platforms is statistically significant. It is unlikely to be due to random chance

[mystery-two]

study of Diabetes in young children

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Sunsets are proof that endings can often be beautiful,too.

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"Nature is not a place to visit. It is home."