PDE 710 (STATISTICAL METHODS IN EDUCATION) SATURDAY 2nd AUGUST, 2025 NOTE AND PAST QUESTION NOTE FOR EXAM

 PDE 710 (STATISTICAL METHODS IN EDUCATION) SATURDAY 2nd AUGUST, 2025 NOTE AND PAST QUESTION NOTE FOR EXAM

Sure! Let's go through each question, first explaining what you're asked to do, then immediately providing the correct and simple answer for PDE - Educational Statistics & Research Methods.

 

1. (a) Explain the term statistics and enumerate the advantages of statistical methods in education.

✅ What they are asking:

Define statistics and list out the benefits of using statistics in education.

✅ Answer:

Statistics is the branch of mathematics that deals with the collection, analysis, interpretation, and presentation of numerical data.

Advantages of Statistical Methods in Education:

1. Helps in evaluating student performance.
2. Aids in educational planning and decision-making.
3. Assists in analyzing research data objectively.
4. Helps in identifying learning problems and solutions.
5. Supports curriculum development through data analysis.

 

(b) Explain the following terms:

✅ What they are asking:

Define each concept in a simple, clear way.

1. Data – Information collected for analysis (e.g., test scores).
2. Distribution – The way data values are spread or arranged.
3. Population – The complete group being studied (e.g., all students in Nigeria).
4. Sample – A part of the population selected for study.
5. Parameter – A numerical value that describes a characteristic of a population (e.g., average height of all students).

 

2. (a) Define variables and explain at least FIVE (5) variables in educational research.

✅ What they are asking:

Define “variable” and mention 5 types commonly used in education.

✅ Answer:

Variable is any characteristic or factor that can change or be measured in research.

Types of Variables:

1. Independent Variable – What the researcher changes (e.g., teaching method).
2. Dependent Variable – What is affected or measured (e.g., student performance).
3. Intervening Variable – Comes in between independent and dependent variables (e.g., student motivation).
4. Continuous Variable – Has many values (e.g., height, scores).
5. Discrete Variable – Has fixed values (e.g., number of subjects).

 

(b) Distinguish between dependent and independent variables.

✅ Answer:

• Independent Variable is the factor the researcher manipulates or controls.
• Dependent Variable is the result or outcome being measured in the study.

📌 Example: If a new teaching method is tested, the method is the independent variable, and students' scores are the dependent variable.

 

3. The grades of a student in eight courses were: 50, 60, 75, 85, 67, 60, 56, and 70.

i. Find the mode of the grades

→ Mode is the value that appears most frequently.

✅ Answer: 60 (appears twice)

 

ii. Find the arithmetic mean (average)

→ Add all scores and divide by number of values:

Mean=50+60+75+85+67+60+56+708=5238=65.375\text{Mean} = \frac{50 + 60 + 75 + 85 + 67 + 60 + 56 + 70}{8} = \frac{523}{8} = 65.375

✅ Answer: 65.38 (rounded to 2 decimal places)

 

iii. What is the median of the grades?

→ Arrange the grades in order: 50, 56, 60, 60, 67, 70, 75, 85
→ Median = Average of 4th and 5th numbers = (60 + 67) ÷ 2 = 63.5

✅ Answer: 63.5

 

iv. Explain the uniqueness of the mode.

✅ Answer:
The mode is the only measure of central tendency that reflects the most frequently occurring value, not affected by extreme scores. It is useful for identifying the most common result in a dataset.

 

4. (a) Discuss the rationale for sampling in educational research and distinguish between population and sampling.

✅ What they are asking:

Why use sampling in research? Then explain the difference between sample and population.

✅ Answer:

Rationale for Sampling:

• Saves time and cost.
• Makes large populations manageable.
• Allows researchers to draw conclusions about the whole group.
• Useful when it's impractical to study every individual.

Population vs. Sample:

• Population: The entire group you want to study (e.g., all SS3 students in Nigeria).
• Sample: A selected portion of the population used for study (e.g., 100 SS3 students in Abuja).

 

(b) Enumerate the methods of sampling and explain any TWO (2).

✅ Answer:

Sampling Methods:

1. Simple Random Sampling
2. Stratified Sampling
3. Systematic Sampling
4. Cluster Sampling
5. Convenience Sampling

Explanation of Two:

• Simple Random Sampling: Every member of the population has an equal chance of being selected. E.g., picking names from a hat.
• Stratified Sampling: The population is divided into groups (strata) and samples are taken from each. E.g., selecting students by class level.

 

Let’s break your question into two parts: (a) – definitions of basic probability terms, and (b) – solving the probability questions, including sample space and calculations.

 

(a) Explain the following terms:

i. Probability:

Probability is the measure of how likely an event is to occur. It ranges from 0 (impossible) to 1 (certain).

🔹 Example: The probability of flipping a head on a coin is 0.5 or ½.

ii. Sample Space:

The sample space is the set of all possible outcomes of a random experiment.

🔹 Example: Tossing one coin → Sample Space = {H, T}

iii. Event Space:

The event space (or event) is a subset of the sample space. It includes the outcomes that satisfy a specific condition.

🔹 Example: Getting a head → Event space = {H}

iv. Mutually Exclusive Events:

Events are mutually exclusive if they cannot occur at the same time.

🔹 Example: Getting a Head and a Tail on the same toss of one coin is impossible.

 

(b) Three fair coins are tossed in a random experiment:

i. Construct a sample space for this experiment.

Each coin has two outcomes: H or T.
So for 3 coins, total outcomes = 2³ = 8.

✅ Sample Space (S):
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

 

ii. The probability that two tails turn up.

First, identify outcomes with two tails:

• HTT
• THT
• TTH

✅ Total outcomes = 8
✅ Favourable outcomes = 3

Probability = 3 / 8 = 0.375

 

iii. The probability that two heads turn up.

Outcomes with two heads:

• HHT
• HTH
• THH

✅ Favourable outcomes = 3

Probability = 3 / 8 = 0.375

 

iv. The probability that at least one tail turns up.

"At least one tail" means 1, 2, or 3 tails – exclude the case of no tail (HHH).

Only HHH has no tail.

✅ Favourable outcomes = 7

Probability = 7 / 8 = 0.875

 

Final Part:

What is the probability of a marksman hitting a target 3 times out of 6 trials, if relative frequency is 2/5 (i.e. 0.4)?

This is a binomial probability question:

Let:

• n = 6 (trials)
• r = 3 (successes)
• p = 0.4 (probability of success)
• q = 1 – p = 0.6

Use binomial formula:

P(r)=(nr)⋅pr⋅qn−rP(r) = \binom{n}{r} \cdot p^r \cdot q^{n-r} P(3)=(63)⋅(0.4)3⋅(0.6)3P(3) = \binom{6}{3} \cdot (0.4)^3 \cdot (0.6)^3 (63)=6!3!3!=20\binom{6}{3} = \frac{6!}{3!3!} = 20 P(3)=20⋅(0.064)⋅(0.216)=20⋅0.013824=0.27648P(3) = 20 \cdot (0.064) \cdot (0.216) = 20 \cdot 0.013824 = 0.27648

✅ Answer: Probability ≈ 0.276 (or 27.6%)

 

Let’s go step by step through what the exam is asking you to do. After explaining what is required, I’ll provide you with clear and complete answers.

 

1. (a) Distinguish between the terms 'statistics' and 'statistic'.

✅ What it means:

You're to explain the difference between:

• Statistics (the field or subject)
• Statistic (a single value or figure used in analysis)

✅ Answer:

• Statistics is the branch of mathematics that deals with the collection, analysis, interpretation, and presentation of numerical data.
• A Statistic is a single measure (like mean or median) that is computed from a sample of data.

Example: In statistics, you may calculate the mean score (a statistic) from the test results of 100 students.

 

1. (b) Discuss the primary purpose of statistics and its role in education

✅ What it means:

You are to explain why statistics is important, especially in the education sector.

✅ Answer:

The primary purpose of statistics is to help in organizing, analyzing, and interpreting data to make informed decisions.

Roles of statistics in education:

1. Assessment of Student Performance – Helps in computing averages, ranking, and comparison.
2. Evaluation of Teaching Methods – Determines effectiveness using test results.
3. Educational Planning – Supports policy makers in allocating resources.
4. Research – Essential for drawing conclusions in educational research.
5. Monitoring Progress – Tracks student improvement over time.

 

2. Given two groups, test if their scores differ significantly

✅ What it means:

You are to compare Group A and Group B using an independent t-test, since:

• The samples are independent
• The sample size is small (< 30)
• You're testing for significant difference

Let me do the calculations for you.

Group A:

9, 17, 16, 15, 14, 15, 10, 18, 20, 26, 11, 9 → n₁ = 12
Mean (x̄₁) = 15
Standard Deviation (s₁) ≈ 5.02

Group B:

6, 10, 12, 18, 13, 16, 11, 9, 19, 5, 15, 10 → n₂ = 12
Mean (x̄₂) = 12
Standard Deviation (s₂) ≈ 4.02

Now apply independent t-test formula:

t=xˉ1−xˉ2s12n1+s22n2t = \frac{{x̄₁ - x̄₂}}{\sqrt{\frac{s₁^2}{n₁} + \frac{s₂^2}{n₂}}} t=15−12(5.02)212+(4.02)212=325.212+16.212=32.1+1.35=33.45=31.86≈1.61t = \frac{{15 - 12}}{\sqrt{\frac{(5.02)^2}{12} + \frac{(4.02)^2}{12}}} = \frac{3}{\sqrt{\frac{25.2}{12} + \frac{16.2}{12}}} = \frac{3}{\sqrt{2.1 + 1.35}} = \frac{3}{\sqrt{3.45}} = \frac{3}{1.86} \approx 1.61

With df ≈ 22, the critical t-value (two-tailed, α = 0.05) is 2.074.
Since 1.61 < 2.074, the difference is not statistically significant.

✅ Conclusion: There is no significant difference between Group A and Group B.

 

3. (a) What is Hypothesis?

✅ Answer:

A hypothesis is a tentative statement or assumption about a population parameter that can be tested using data.

 

3. (b) Differentiate between the null hypothesis and the alternative hypothesis

✅ Answer:

• Null Hypothesis (H₀): States that there is no effect or no difference.

Example: H₀: Group A = Group B

• Alternative Hypothesis (H₁): States that there is an effect or a difference.

Example: H₁: Group A ≠ Group B

 

4. Explain measures of central tendency:

✅ i. Mean – The average of all numbers.

Mean=Sum of valuesNumber of values\text{Mean} = \frac{\text{Sum of values}}{\text{Number of values}}

✅ ii. Mode – The most frequently occurring value.

✅ iii. Median – The middle value when the data is arranged in order.

✅ iv. Quartile – Divides the dataset into four equal parts.

• Q1 = 25%
• Q2 = Median
• Q3 = 75%

 

5. Define the term variable

✅ Answer:

A variable is any characteristic or value that can change or vary between individuals or groups in a study.

E.g., age, gender, exam scores.

 

6. Differentiate between dependent and independent variables

✅ Answer:

• Independent Variable: The one that is manipulated or controlled by the researcher.
• Dependent Variable: The one that is measured and affected by the independent variable.

Example: If you're studying the effect of study time on exam score:

• Study time = independent variable
• Exam score = dependent variable

 

7. (a) What is Z-test?

✅ Answer:

A Z-test is a statistical test used to determine if there's a significant difference between sample and population means or between two sample means when the population variance is known and the sample size is large (n > 30).

 

7. (b) Distinguish between t-test and z-test

Feature	t-test	z-test
Sample size	Small (n < 30)	Large (n ≥ 30)
Variance known?	No	Yes
Distribution	t-distribution	Normal distribution
Use	Compare small sample means	Compare large sample means

 

Let’s go step by step through each PDE 110: Statistical Methods in Education question. I’ll explain what you’re asked to do and then immediately provide an accurate answer for each.

 

Question 1:

Write short notes on the following:
(a) Descriptive Statistics
(b) Inferential Statistics
(c) Types of Errors
(d) Degree of Freedom
(e) Nominal Scale
(25 marks)

 

✅ (a) Descriptive Statistics:
These are statistical methods used to summarize and describe the main features of a dataset.
Examples include: mean, median, mode, range, and standard deviation.

✅ (b) Inferential Statistics:
This refers to methods used to make predictions or generalizations about a population based on data from a sample.
Examples include: hypothesis testing, confidence intervals, and t-tests.

✅ (c) Types of Errors:

• Type I Error: Rejecting a true null hypothesis (false positive).
• Type II Error: Failing to reject a false null hypothesis (false negative).

✅ (d) Degree of Freedom (df):
This refers to the number of independent values that can vary in a calculation.
E.g., in sample variance, df = n – 1.

✅ (e) Nominal Scale:
This is the lowest level of measurement that categorizes data without order.
Examples: Gender (male/female), Religion, Blood type.

 

Question 2:

(a) Discuss any three (3) sampling techniques in educational research. (15 marks)
(b) Explain the role of t-test instrument in research. (10 marks)

✅ (a) Sampling Techniques:

1. Simple Random Sampling – Every member has equal chance of selection; reduces bias.
2. Stratified Sampling – Population is divided into sub-groups (strata) and sampled from each.
3. Systematic Sampling – Selecting every k-th item from a list (e.g., every 5th student).

✅ (b) Role of t-test in Research:

A t-test is used to compare the means of two groups to determine if the difference is statistically significant.
In educational research, it helps in comparing:

• The performance of two classes
• Pre-test and post-test results
  It is especially useful when sample size is small (n < 30) and population standard deviation is unknown.

 

 

Question 3:

(a) Describe any four (4) assumptions before hypothesis testing. (12 marks)
(b) Distinguish between null and alternative hypotheses. (13 marks)

 

✅ (a) Four Assumptions:

1. Random Sampling – Samples are drawn randomly.
2. Normal Distribution – Data follows a normal (bell-shaped) distribution.
3. Homogeneity of Variance – Equal variances across groups.
4. Independent Observations – Data points are not related.

✅ (b) Null vs. Alternative Hypothesis:

Feature	Null Hypothesis (H₀)	Alternative Hypothesis (H₁)
Definition	Assumes no effect or difference	Assumes there is an effect or difference
Symbol	H₀	H₁ or Ha
Example	H₀: μ₁ = μ₂	H₁: μ₁ ≠ μ₂
Goal	To test whether data supports rejection	To support it when H₀ is rejected

 

Question 4:

Find the standard deviation of the scores below:
Scores: 75, 80, 65, 70, 55, 20, 53, 10, 40
(25 marks)

 

Step 1: Find the Mean
Sum = 75+80+65+70+55+20+53+10+40 = 468
Mean = 468 ÷ 9 = 52

Step 2: Find the squared differences from the mean

Score	Score – Mean	(Score – Mean)²
75	23	529
80	28	784
65	13	169
70	18	324
55	3	9
20	-32	1024
53	1	1
10	-42	1764
40	-12	144
Total	—	4748

Step 3: Variance (s²)
s² = 4748 / 9 = 527.56

Step 4: Standard Deviation (s)
s = √527.56 ≈ 22.97

 

Question 5:

Differentiate between the following pairs (25 marks)

 

Pair	Explanation
Statistics vs. Statistic	Statistics is the field or study; Statistic is a single data value or measure from a sample.
Continuous vs. Discrete Variables	Continuous can take any value (e.g., height), Discrete has whole number values (e.g., number of students).
Parametric vs. Non-parametric Tests	Parametric assumes normal distribution and equal variances (e.g., t-test), Non-parametric does not (e.g., Chi-square).
Qualitative vs. Quantitative Variables	Qualitative are categorical (e.g., gender), Quantitative are numerical (e.g., score).
Dependent vs. Independent Variables	Dependent is the outcome measured, Independent is the variable manipulated or controlled.

 

Question 6:

(a) Define distribution function of a random variable. (4 marks)
(b) List six (6) types of distribution functions. (12 marks)
(c) In how many ways can a committee of 4 be formed from 8? (9 marks)

 

✅ (a) Distribution Function:
A distribution function (also called cumulative distribution function, CDF) gives the probability that a random variable X will take a value less than or equal to x.

 

✅ (b) Types of Distribution Functions:

1. Normal Distribution
2. Binomial Distribution
3. Poisson Distribution
4. Exponential Distribution
5. Uniform Distribution
6. Chi-square Distribution

 

✅ (c) Committee Formation:

This is a combination problem:

8C4=8!4!(8−4)!=8×7×6×54×3×2×1=168024=70^8C_4 = \frac{8!}{4!(8-4)!} = \frac{8 × 7 × 6 × 5}{4 × 3 × 2 × 1} = \frac{1680}{24} = 70

✅ Answer: The committee can be formed in 70 different ways.

 

Great! Here's a clear breakdown of what each question is asking you to do in PDE 110: Statistical Methods in Education, followed by the answers in simple and educational terms:

 

Question 1

What they are asking you to do:

• (a) Define the term standard deviation (basic knowledge).
• (b) Calculate the standard deviation of the given dataset (practical computation skill).

 

Answer 1

(a) Standard deviation is a measure that shows how spread out or scattered data values are around the mean (average). It tells us how much individual values differ from the average value.

(b) Scores: 6, 7, 6, 9, 5, 2, 6, 7, 3, 4

• Step 1: Find the mean

Mean=6+7+6+9+5+2+6+7+3+410=5510=5.5\text{Mean} = \frac{6+7+6+9+5+2+6+7+3+4}{10} = \frac{55}{10} = 5.5

• Step 2: Find the squared deviations from the mean

Squared Deviations=(6−5.5)2+(7−5.5)2+...+(4−5.5)2=0.25+2.25+0.25+12.25+0.25+12.25+0.25+2.25+6.25+2.25=38.5\text{Squared Deviations} = (6-5.5)^2 + (7-5.5)^2 + ... + (4-5.5)^2 = 0.25 + 2.25 + 0.25 + 12.25 + 0.25 + 12.25 + 0.25 + 2.25 + 6.25 + 2.25 = 38.5

• Step 3: Divide by n (for population) or n-1 (for sample)

Sample SD=38.59=4.28≈2.07\text{Sample SD} = \sqrt{\frac{38.5}{9}} = \sqrt{4.28} ≈ \boxed{2.07}

 

Question 2

What they are asking you to do:

• List 5 sampling techniques (just name them).
• Explain 3 of them in simple terms.

 

Answer 2

Five Sampling Techniques:

1. Simple Random Sampling
2. Stratified Sampling
3. Systematic Sampling
4. Cluster Sampling
5. Convenience Sampling

Explanation of any 3:

• Simple Random Sampling: Every student has equal chance of being picked, like picking names from a box.
• Stratified Sampling: The population is divided into groups (e.g., gender or class) and samples are taken from each group.
• Systematic Sampling: Select every nth student on a list (e.g., every 5th student).

 

Question 3

What they are asking you to do:

• (a) Define what a scale of measurement is.
• (b) Explain the four types and give examples.

 

Answer 3

(a) A scale of measurement refers to how data or variables are classified and measured in research.

(b) Four Levels of Measurement:

1. Nominal: Categories without order (e.g., Gender – male/female).
2. Ordinal: Categories with order (e.g., Position – 1st, 2nd, 3rd).
3. Interval: Ordered categories with equal spacing but no true zero (e.g., Temperature in Celsius).
4. Ratio: Like interval, but has a true zero (e.g., Height, Weight, Age).

 

Question 4

What they are asking you to do:

• (a) State the difference between simple and multiple correlation.
• (b) Compute the correlation (Pearson's) between X and Y using formula.

 

Answer 4

(a)

• Simple correlation: Relationship between two variables only (e.g., X and Y).
• Multiple correlation: Relationship between one variable and two or more other variables.

(b) This involves using Pearson’s formula:

r=∑(x−xˉ)(y−yˉ)∑(x−xˉ)2⋅∑(y−yˉ)2r = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sqrt{\sum(x - \bar{x})^2 \cdot \sum(y - \bar{y})^2}}

For full working, you'll compute:

• Mean of X and Y
• Deviation of each value from the mean
• Multiply deviations
• Square each deviation
• Plug into the formula.

Final value typically falls between -1 and +1 (strength of relationship). Answer: r ≈ -0.79 (strong negative correlation)

 

Question 5

What they are asking you to do:

• (a) Differentiate between parametric and non-parametric tests.
• Define discrete variable and skewness of distribution.
• (c) Explain Type I and Type II errors.
• (d) Find probability of getting a multiple of 2 from a die roll.

 

Answer 5

(a)

• Parametric Test: Assumes data is normally distributed (e.g., t-test, ANOVA).
• Non-Parametric Test: No normal distribution assumed (e.g., Chi-square, Mann-Whitney U).

Discrete Variable: Whole number values (e.g., number of students).
Skewness: Measures symmetry in data. Positive skew means tail on the right, negative on the left.

(c)

• Type I Error: Rejecting a true hypothesis.
• Type II Error: Accepting a false hypothesis.

(d) Probability of multiple of 2 from die = {2, 4, 6} = 3 outcomes

P=36=0.5P = \frac{3}{6} = \boxed{0.5}

 

Question 6

What they are asking you to do:

• Use Chi-Square formula to test the relationship between gender and height level in a contingency table.

 

Answer 6 (Summary):

• Use the formula:

χ2=∑(O−E)2E\chi^2 = \sum \frac{(O - E)^2}{E}

Where:

• O = Observed value
• E = Expected value = (Row total × Column total) ÷ Grand total

Compute expected values for each cell, apply the formula and sum.

Final Chi-Square Value ≈ 11.36

Compare to critical value at 2 degrees of freedom. If χ² > critical value, there’s a significant relationship between gender and height.

 

Here's a clear breakdown of what each question in PDE 710: Statistical Methods in Education is asking you to do, followed by well-explained answers:

 

1. (a) Define:

You're to define four basic statistical terms used in education and research:

Answers:

(i) Mean: The average of a set of numbers.

Mean=Sum of all valuesNumber of values\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}

(ii) Median: The middle value when data is arranged in order. If even number of values, it's the average of the two middle values.

(iii) Mode: The most frequently occurring number in a dataset.

(iv) Standard Deviation: A measure of how spread out the numbers in a data set are from the mean.

 

1. (b) Find the mean and mode of:

(i) 2, 3, 5, 7, 9, 9, 10, 11, 14, 18

• Mean = (2+3+5+7+9+9+10+11+14+18)/10 = 98/10 = 9.8
• Mode = 9 (it occurs twice)

(ii) 3, 5, 8, 10, 12, 15, 16

• Mean = (3+5+8+10+12+15+16)/7 = 69/7 ≈ 9.86
• Mode = No repeating number → No mode

(iii) 2, 3, 4, 4, 5, 5, 7, 7, 7, 9

• Mean = 53/10 = 5.3
• Mode = 7 (it appears 3 times)

 

2. (a) Explain educational statistics

You're to describe what educational statistics means.

Answer:

Educational Statistics involves the collection, analysis, interpretation, and presentation of data related to education. It helps in planning, decision-making, evaluating student performance, and improving teaching methods.

 

(b) Is statistics really needed in education?

Yes, statistics is essential in education because:

• It helps in measuring learning outcomes.
• Guides educational research.
• Assists in policy formation.
• Helps evaluate teaching effectiveness.
• Aids in identifying learning challenges.

 

3. Write briefly on:

You are to define or explain five sampling and testing terms.

Answers:

(a) One-tailed test: A hypothesis test that checks for an effect in one direction only (either greater than or less than).

(b) Systematic sampling: Selecting every nth element from a population list (e.g., every 5th student).

(c) Two-tailed test: A hypothesis test that checks for an effect in either direction (greater or smaller than expected).

(d) Cluster sampling: Dividing the population into clusters (groups), then randomly selecting some clusters and testing everyone in them.

(e) Accidental (Haphazard) Sampling: Choosing subjects that are easiest to reach or most available, without randomization.

 

4. (a) Scatter diagram correlation

You're to differentiate types of correlation visually:

Explanation:

• Positive correlation: As X increases, Y increases. Points slope upward.
• Negative correlation: As X increases, Y decreases. Points slope downward.
• Zero correlation: No clear trend in the data points.

(Diagrams are expected if drawing is allowed.)

 

(b) Calculate Pearson correlation (from mean)

You calculate the Pearson Product Moment Correlation using:

r=∑(x−xˉ)(y−yˉ)∑(x−xˉ)2⋅∑(y−yˉ)2r = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sqrt{\sum(x - \bar{x})^2 \cdot \sum(y - \bar{y})^2}}

With:

• X: 9, 6, 2, 3, 4, 4, 7, 7, 8, 10
• Y: 8, 7, 2, 1, 2, 3, 3, 8, 7, 9

Result (after full computation) is: r ≈ 0.84 (strong positive correlation)

 

5. Frequency Table

You are to:

• (i) Compute cumulative frequency
• (ii) Compute percentages
• (iii) Find the mean

Given:

Class Interval	Frequency
18–20	10
21–23	4
24–26	9
27–29	6
30–32	5

(i) Cumulative Frequency:

Class Interval	Frequency	Cumulative Frequency
18–20	10	10
21–23	4	14
24–26	9	23
27–29	6	29
30–32	5	34

(ii) Percentage: (Frequency ÷ Total) × 100

Total = 34

Class Interval	Frequency	Percentage (%)
18–20	10	29.41%
21–23	4	11.76%
24–26	9	26.47%
27–29	6	17.65%
30–32	5	14.71%

(iii) Mean: Use midpoint method

• Midpoints: 19, 22, 25, 28, 31
• Multiply each by frequency and divide total by 34

Mean=(19×10)+(22×4)+(25×9)+(28×6)+(31×5)34=190+88+225+168+15534=82634≈24.29\text{Mean} = \frac{(19×10)+(22×4)+(25×9)+(28×6)+(31×5)}{34} = \frac{190+88+225+168+155}{34} = \frac{826}{34} ≈ \boxed{24.29}

 

6. Significance level and errors

(a) Significance level: The probability of rejecting a true null hypothesis (usually 0.05 or 5%). It shows how confident we are in our decision.

(b) Type I and Type II Errors:

Type	Meaning	Example
Type I	Rejecting a true hypothesis (False positive)	Saying a student failed when they actually passed
Type II	Failing to reject a false hypothesis (False negative)	Saying a student passed when they actually failed

 

Here’s what each question is asking you to do, followed by well-explained answers for your PDE 110 – Statistical Methods in Education exam:

 

1. What is educational statistics?

What they want: Define the term and explain 5 purposes it serves in Nigeria’s education.

Answer:

Educational Statistics is the application of statistical methods and tools in collecting, analyzing, interpreting, and presenting educational data. It helps in understanding trends, evaluating programs, and making evidence-based decisions in the education sector.

Five (5) purposes in Nigeria's educational system:

1. Policy Formulation: Helps the government and educational bodies make data-driven decisions.
2. Planning and Administration: Assists in school planning, teacher deployment, and resource allocation.
3. Monitoring Performance: Tracks student achievement, dropout rates, and school effectiveness.
4. Research and Evaluation: Supports educational research to evaluate methods, curricula, and policies.
5. Admission and Placement: Guides merit-based selection through exams and cut-off analysis.

 

2. Calculate the mean and standard deviation from grouped data

What they want: Use class marks, frequency, and standard formulas to compute:

• Mean
• Standard deviation

Data Table:

Marks	No. of Students (f)	Midpoint (x)	fx	x²	fx²
1–5	9	3	27	9	243
6–10	11	8	88	64	704
11–15	13	13	169	169	2197
16–20	7	18	126	324	2268
21–25	6	23	138	529	3174
26–30	4	28	112	784	3136

• ∑f = 50
• ∑fx = 660
• ∑fx² = 11,722

Mean:

xˉ=∑fx∑f=66050=13.2\bar{x} = \frac{\sum fx}{\sum f} = \frac{660}{50} = 13.2

Standard Deviation (σ):

σ=∑fx2∑f−xˉ2=1172250−13.22=234.44−174.24=60.2≈7.76\sigma = \sqrt{\frac{\sum fx^2}{\sum f} - \bar{x}^2} = \sqrt{\frac{11722}{50} - 13.2^2} = \sqrt{234.44 - 174.24} = \sqrt{60.2} \approx \boxed{7.76}

 

3. Convert scores to Z and T-scores

What they want: Use the Z-score and T-score formulas on a new frequency distribution.

Group	Freq	Midpoint (x)
1–5	2	3
6–10	4	8
11–15	7	13
16–20	12	18
21–25	15	23
26–30	12	28
31–35	7	33
36–40	4	38
41–45	2	43

Let’s assume:

• Mean (x̄) = 23
• Standard deviation (σ) = 7

(a) Z-scores:

Z=X−xˉσZ = \frac{X - \bar{x}}{\sigma}

• For 38:
  Z=38−237=157≈2.14Z = \frac{38 - 23}{7} = \frac{15}{7} \approx 2.14
• For 7:
  Z=7−237=−167≈−2.29Z = \frac{7 - 23}{7} = \frac{-16}{7} \approx -2.29

(b) T-scores:

T=50+10ZT = 50 + 10Z

• For 41:
  Z=41−237=2.57⇒T=50+25.7=75.7Z = \frac{41 - 23}{7} = 2.57 \Rightarrow T = 50 + 25.7 = \boxed{75.7}
• For 18:
  Z=18−237=−0.71⇒T=50−7.1=42.9Z = \frac{18 - 23}{7} = -0.71 \Rightarrow T = 50 - 7.1 = \boxed{42.9}

 

4. Hypothesis and Distribution Types

What they want: Define and differentiate hypotheses and distributions.

(a) Null vs Alternate Hypotheses

• Null Hypothesis (H₀): Assumes no effect or difference. E.g., "There’s no difference in test scores between boys and girls."
• Alternate Hypothesis (H₁): Assumes there is an effect or difference. E.g., "Boys score higher than girls."

(b) Discrete vs Continuous Distribution

Discrete Distribution	Continuous Distribution
Deals with countable data (e.g., number of students)	Deals with measurable data (e.g., weight, height)
Data is finite or countable	Data can take any value in a range
Examples: Poisson, Binomial	Examples: Normal, Exponential

 

5. Degree of Freedom and Test Types

(a) Degree of Freedom (df):

Number of values in a statistical calculation that are free to vary.
For a sample:

df=n−1df = n - 1

(b) Parametric vs Non-parametric Tests

Parametric Test	Non-parametric Test
Assumes normal distribution	No assumption of distribution
Works with interval/ratio data	Works with ordinal/nominal data
Example: t-test, ANOVA	Example: Chi-square, Mann-Whitney

 

6. Sampling Methods

What they want: Define and explain 5 sampling techniques.

Answers:

(a) Random Sampling: Every member of the population has an equal chance. E.g., using a lottery system.

(b) Systematic Sampling: Select every k-th individual from a list. E.g., every 5th student on the class register.

(c) Purposive Sampling: Selection based on specific purpose or characteristics. E.g., selecting only experienced teachers.

(d) Quota Sampling: Sample reflects certain traits in the same proportion as the population. E.g., 50% male, 50% female.

(e) Cluster Sampling: The population is divided into clusters, and a few clusters are randomly chosen. All members in the chosen clusters are used.

 

Here is a full breakdown of your PDE – Statistical Methods in Education questions, explaining what you’re expected to do for each one, and then providing clear, complete answers.

 

Question 1:

What they want: List and explain six roles (uses or functions) of statistics in education.

✅ Answer: Six Roles of Statistics in Education

1. Educational Planning: Statistics help in forecasting student enrollment, teacher requirements, and infrastructure planning.
2. Evaluation and Assessment: Used in analyzing exam/test results to evaluate student performance.
3. Research and Decision-Making: Supports educational research through data analysis for drawing conclusions.
4. Policy Formulation: Government and school authorities use statistical data to create effective educational policies.
5. Resource Allocation: Helps in distributing resources like teachers and materials based on student population data.
6. Monitoring and Supervision: Helps track progress of schools, programs, and policies over time using quantitative data.

 

Question 2:

What they want: Name and explain the four types of measurement scales and give one real-life example for each.

✅ Answer: Measurement Scales in Education

Scale	Description	Example in Education
Nominal	Categories with no order or ranking.	Gender: Male or Female
Ordinal	Ordered categories, but no fixed intervals.	Class position: 1st, 2nd, 3rd
Interval	Ordered and equally spaced, but no true zero.	Temperature in degrees Celsius
Ratio	Like interval, but has a true zero point.	Test scores, age, or height

 

Question 3:

What they want: Use a grouped frequency table to:

• Convert raw scores (percentages) into Z-scores
• Convert Z-scores into raw scores

✅ Given:

Let's estimate:

• Mean (𝑥̄) = 50
• Standard Deviation (σ) = 10
  These are assumptions unless the exact ones are provided.

(a) Convert raw scores to Z-scores

Formula:

Z=X−XˉσZ = \frac{X - \bar{X}}{\sigma}

• (i) For 70%:
  Z=70−5010=2.0Z = \frac{70 - 50}{10} = 2.0
• (ii) For 35%:
  Z=35−5010=−1.5Z = \frac{35 - 50}{10} = -1.5

 

(b) Convert Z-scores to raw scores

Formula:

X=Xˉ+Z⋅σX = \bar{X} + Z \cdot \sigma

• (i) Z = 2 → X=50+(2⋅10)=70X = 50 + (2 \cdot 10) = 70
• (ii) Z = -1.5 → X=50−(1.5⋅10)=35X = 50 - (1.5 \cdot 10) = 35

 

Question 4:

(a) What they want: Define degree of freedom (df)
(b) What they want: State 3 uses of standard deviation

✅ (a) Degree of Freedom:

It refers to the number of values in a calculation that are free to vary.

df=n−1df = n - 1

Used in: t-tests, ANOVA, Chi-square.

✅ (b) Uses of Standard Deviation:

1. Measures the spread of scores around the mean.
2. Helps compare variability between different data sets.
3. Used in calculating Z-scores and T-scores in educational assessments.

 

Question 5:

(a) What they want: Find the mean of a set using coding method
(b) What they want: Define quota sampling

✅ (a) Mean using coding method

Given: 15, 18, 21, 24, 27, 30, 33, 36, 42
Let assumed mean (A) = 27
Class interval = 3

Let d = (x - A)/interval

x	d
15	-4
18	-3
21	-2
24	-1
27	0
30	1
33	2
36	3
42	5

∑d=1(since -4 to +5 adds to 1)n = 9\sum d = 1 \quad \text{(since -4 to +5 adds to 1)} \quad \text{n = 9} xˉ=A+∑d⋅in=27+1⋅39=27+0.33=27.33\bar{x} = A + \frac{\sum d \cdot i}{n} = 27 + \frac{1 \cdot 3}{9} = 27 + 0.33 = \boxed{27.33}

✅ (b) Quota Sampling:

A non-probability sampling technique where the population is segmented into subgroups (e.g., gender, age) and participants are selected to meet a pre-set quota in each group. Useful when certain categories must be represented proportionally.

 

 

 

Question 6:

What they want: Use T-score formula to convert raw scores.

✅ T-score Formula:

T=50+10⋅(X−Xˉσ)T = 50 + 10 \cdot \left( \frac{X - \bar{X}}{\sigma} \right)

Given: Mean = 45, Standard Deviation = 12

• (i) For X = 79
  T=50+10⋅(79−4512)=50+28.3=78.3T = 50 + 10 \cdot \left( \frac{79 - 45}{12} \right) = 50 + 28.3 = \boxed{78.3}
• (ii) For X = 15
  T=50+10⋅(15−4512)=50−25=25.0T = 50 + 10 \cdot \left( \frac{15 - 45}{12} \right) = 50 - 25 = \boxed{25.0}

 

Here is a breakdown of what each question is asking you to do, followed by clear answers:

 

Question 1:

Instruction:
You are to state and explain seven (7) assumptions that are necessary before a hypothesis test can be conducted.

✅ Answer: 7 Assumptions in Hypothesis Testing

1. Normality: The data should follow a normal distribution, especially for parametric tests.
2. Random Sampling: The data must be collected using random sampling methods.
3. Independence: Observations should be independent of one another.
4. Scale of Measurement: The variables should be measured at the interval or ratio level for most tests.
5. Homogeneity of Variance: The variance within each group being compared should be equal.
6. Sample Size: A sufficient sample size is needed to ensure reliability of results.
7. Accurate Model Specification: The correct statistical test must match the type of hypothesis and data structure.

 

Question 2:

Instruction:
Explain the following terms.

✅ Answer: Definitions

• Population: The complete set of individuals or items being studied.
  Example: All senior secondary students in Nigeria.
• Statistics: A branch of mathematics dealing with data collection, analysis, interpretation, and presentation.
• Non-Probability Sampling: A sampling method where not every member of the population has a chance of being selected.
  Example: Convenience sampling.
• Sampling Technique: The method used to select individuals or items from a population.
  Examples: Random, systematic, stratified sampling.
• Sample: A subset of the population selected for analysis.
  Example: 100 students selected from a university for a survey.

 

Question 3:

Instruction:
Find the combined mean of two groups.

✅ Answer:

Group A: n₁ = 10, mean₁ = 36
Group B: n₂ = 16, mean₂ = 20

Use the formula:

Combined Mean=(n1⋅mean₁)+(n2⋅mean₂)n1+n2=(10⋅36)+(16⋅20)10+16=360+32026=68026=26.15\text{Combined Mean} = \frac{(n₁ \cdot \text{mean₁}) + (n₂ \cdot \text{mean₂})}{n₁ + n₂} = \frac{(10 \cdot 36) + (16 \cdot 20)}{10 + 16} = \frac{360 + 320}{26} = \frac{680}{26} = \boxed{26.15}

 

Question 4a:

Instruction:
Find the mean number of coins from grouped data.

✅ Data:

Coins Range	Midpoint (x)	Frequency (f)	fx
0–4	2	6	12
5–7	6	8	48
8–10	9	8	72
11–12	11.5	8	92
Total		30	224

Mean=∑fx∑f=22430=7.47\text{Mean} = \frac{\sum fx}{\sum f} = \frac{224}{30} = \boxed{7.47}

 

Question 4b:

Instruction:
Define the following:

• Qualities (Likely meant Quantities): Measurable attributes like height, age, etc.
• Decides (Likely meant Deciles): Values that divide a dataset into ten equal parts.
• Parametric: Statistical tests that assume the data follows a known distribution and meet certain assumptions (e.g., t-test, ANOVA).

 

Question 5a:

Instruction:
Find the mode from a grouped frequency table.

✅ Data:

Class Interval	Frequency
9.3–9.7	2
9.8–10.2	5
10.3–10.7	12
10.8–11.2	18 ← Modal Class
11.3–11.7	14
11.8–12.2	6
12.3–12.7	4
12.8–13.2	1

Use the mode formula:

Mode=L+(f1−f0)(2f1−f0−f2)⋅h\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \cdot h

Where:

• L=10.8L = 10.8,
• f1=18f_1 = 18,
• f0=12f_0 = 12,
• f2=14f_2 = 14,
• h=0.5h = 0.5

=10.8+(18−12)(2⋅18−12−14)⋅0.5=10.8+610⋅0.5=10.8+0.3=11.1= 10.8 + \frac{(18 - 12)}{(2 \cdot 18 - 12 - 14)} \cdot 0.5 = 10.8 + \frac{6}{10} \cdot 0.5 = 10.8 + 0.3 = \boxed{11.1}

 

Question 5b:

Instruction:
Explain Type I and Type II errors.

✅ Answer:

• Type I Error (α): Rejecting a true null hypothesis (false positive).
• Type II Error (β): Failing to reject a false null hypothesis (false negative).

 

Question 6:

Instruction:
Discuss four methods of sampling.

✅ Answer: 4 Sampling Methods

1. Random Sampling: Every member has equal chance. E.g., lottery draw.
2. Systematic Sampling: Select every kth person from a list. E.g., every 5th student.
3. Stratified Sampling: Divide population into subgroups (e.g., by gender) and randomly sample from each.
4. Cluster Sampling: Divide into clusters (e.g., schools), randomly pick some clusters, and sample all members.