| | Decision: Reject H₀ | Decision: Fail to Reject H₀ |
|-----------------|-------------------------------------|-------------------------------------|
| **H₀ is TRUE** | Type 1 Error (False Positive) | ✅ Correct Decision (True Negative) |
| **H₀ is FALSE** | ✅ Correct Decision (True Positive) | Type 2 Error (False Negative) |
✅ Definition: Rejecting the null hypothesis (H₀) when it is actually TRUE.
✅ Probability of Occurrence: Denoted by α (Significance Level), typically set to 0.05 (5%).
✅ Consequence: You detect an effect that DOES NOT exist.
✅ Definition: Failing to reject the null hypothesis (H₀) when it is actually FALSE.
✅ Probability of Occurrence: Denoted by β, related to statistical power (1 - β).
✅ Consequence: You MISS detecting a real effect.
| Error Type | What Happens? | Example | Probability |
|--------------|-----------------------------------------------------------------|--------------------------------------|------------------------|
| Type 1 Error | Rejecting H₀ when it is actually TRUE (False Positive) | Finding an effect that doesn’t exist | α (Significance Level) |
| Type 2 Error | Failing to reject H₀ when it is actually FALSE (False Negative) | Missing a real effect | β (Related to Power) |
🔹 Lower Type 1 Error (α):
- Decrease significance level (α = 0.01 instead of 0.05).
- Use larger sample sizes to improve accuracy.
🔹 Lower Type 2 Error (β):
- Increase sample size for more statistical power.
- Increase effect size (look for bigger differences).
- **Type 1 Error** → False Positive (Detecting something that isn’t there).
- **Type 2 Error** → False Negative (Missing something that actually exists).
Colleges in Karnataka have an average placement rate of **85%**.
A new college has a sample of **150 students** with a placement rate of **88%** and a standard deviation of **4%**.
Does this college have a different placement rate? We will conduct both **One-Tailed and Two-Tailed Tests**.
🔹 **Null Hypothesis (H₀)**: The new college has the same placement rate as other Karnataka colleges.
- H₀: p = 0.85
🔹 **Alternative Hypothesis (H₁) - Two-Tailed Test**: The new college has a different placement rate.
- H₁: p ≠ 0.85 (We check if placement rate is significantly different in either direction)
🔹 **Alternative Hypothesis (H₁) - One-Tailed Test**: The new college has a higher placement rate.
- H₁: p > 0.85 (We check if placement rate is significantly higher)
Formula:
Z = (p̂ - p) / (σ / √n)
Where:
- p̂ = 0.88 (Sample Placement Rate)
- p = 0.85 (Population Placement Rate)
- σ = 0.04 (Standard Deviation)
- n = 150 (Sample Size)
Plugging in values:
Z = (0.88 - 0.85) / (0.04 / √150)
Z = 0.03 / (0.04 / 12.247)
Z = 0.03 / 0.003267
Z ≈ 9.18
For a **Two-Tailed Test** at **α = 0.05**:
- Critical Z-values: ±1.96
- Decision Rule: If |Z| > 1.96, we reject H₀.
For a **One-Tailed Test** (Right-Tailed) at **α = 0.05**:
- Critical Z-value: 1.645
- Decision Rule: If Z > 1.645, we reject H₀.
Our calculated Z-score: **9.18**
🔹 **For the Two-Tailed Test**:
- |9.18| > 1.96 → Reject H₀.
- Conclusion: The new college has a **significantly different** placement rate.
🔹 **For the One-Tailed Test (Right-Tail)**:
- 9.18 > 1.645 → Reject H₀.
- Conclusion: The new college has a **significantly higher** placement rate.
✅ **Final Answer**: The new college has a placement rate that is significantly higher than 85%.
On a quantitative test of the CAT, the standard deviation is known to be 100. A sample of 25 test takers has a mean score of 520.
Construct a 95% confidence interval about the population mean.
Step 1: Identify the given values
- Sample Mean (X̄): 520
- Population Standard Deviation (σ): 100
- Sample Size (n): 25
- Confidence Level: 95%
Step 2: Find the Z-Score for 95% Confidence Level
- For a 95% confidence level, the Z-value ≈ 1.96.
Step 3: Calculate the Standard Error (SE)
SE = σ / √n
SE = 100 / √25
SE = 100 / 5
SE = 20
Step 4: Calculate the Confidence Interval
CI = X̄ ± Z * SE
CI = 520 ± 1.96 * 20
CI = 520 ± 39.2
So, the 95% Confidence Interval is: **(480.8, 559.2)**
Conclusion: We are 95% confident that the true population mean lies between 480.8 and 559.2.
In the population, the average IQ is 100 with a standard deviation of 15. Researchers want to test a new medication to see if
it has a positive, negative, or no effect on intelligence. A sample of 30 people who have taken the medication has a mean IQ of 140.
Does the medication affect intelligence?
Null Hypothesis (H₀): μ = 100 (No effect)
Alternative Hypothesis (Hₐ): μ ≠ 100 (Has an effect)
Given:
- Sample Mean (X̄) = 140
- Population Mean (μ) = 100
- Standard Deviation (σ) = 15
- Sample Size (n) = 30
Step 1: Compute Standard Error (SE)
SE = σ / √n
SE = 15 / √30
SE = 15 / 5.477
SE ≈ 2.737
Step 2: Compute Z-Score
Z = (X̄ - μ) / SE
Z = (140 - 100) / 2.737
Z ≈ 14.62
Step 3: Compare with Critical Value
- Critical Z-Value at 95% Confidence (α = 0.05, two-tailed) = ±1.96
- Since |14.62| > 1.96, we reject the null hypothesis.
Conclusion:
- The test is statistically significant.
- The medication has a significant effect on intelligence.
In the population, the average IQ is 100 with a standard deviation of 15. Researchers want to test a new medication to see if it has a positive, negative, or no effect on intelligence. A sample of 30 people who have taken the medication has a mean IQ of 140. Does the medication affect intelligence?
Null Hypothesis (H₀): μ = 100 (No effect)
Alternative Hypothesis (Hₐ): μ ≠ 100 (Has an effect)
Given:
- Sample Mean (X̄) = 140
- Population Mean (μ) = 100
- Sample Standard Deviation (s) = 15
- Sample Size (n) = 30
Step 1: Compute Standard Error (SE)
SE = s / √n
SE = 15 / √30
SE = 15 / 5.477
SE ≈ 2.737
Step 2: Compute t-Score
t = (X̄ - μ) / SE
t = (140 - 100) / 2.737
t ≈ 14.62
Step 3: Determine Degrees of Freedom
df = n - 1 = 30 - 1 = 29
Step 4: Compare with Critical Value
- Critical t-Value at 95% Confidence (α = 0.05, two-tailed) = ±2.045
- Since |14.62| > 2.045, we reject the null hypothesis.
Conclusion:
- The test is statistically significant.
- The medication has a significant effect on intelligence.
A pharmaceutical company develops a new drug to lower blood pressure. The average blood pressure in the population is 120 mmHg. A sample of 50 patients who took the drug had an average blood pressure of 112 mmHg, with a standard deviation of 10 mmHg.
H₀: μ = 120 (No effect)
Hₐ: μ < 120 (The drug lowers blood pressure)
A one-sample t-test can determine if the drug significantly lowers blood pressure.
A coffee shop chain wants to check if its new marketing strategy increases average daily sales. Historically, the average daily sales were $1,000. After the new strategy, a sample of 20 stores showed an average daily sale of $1,150, with a standard deviation of $200.
H₀: μ = 1000 (No effect)
Hₐ: μ > 1000 (Marketing strategy increased sales)
A t-test can determine whether the increase in sales is significant.
A school wants to know if a new online learning program improves student math scores. The average math test score before the program was 75. After using the program, a sample of 40 students had an average score of 80, with a standard deviation of 12.
H₀: μ = 75 (No improvement)
Hₐ: μ > 75 (The program improves scores)
A t-test will determine if the program significantly improved scores.
A company producing light bulbs claims that their bulbs last an average of 1,000 hours. A sample of 30 bulbs from a new batch shows an average lifespan of 950 hours, with a standard deviation of 80 hours.
H₀: μ = 1000 (No change)
Hₐ: μ ≠ 1000 (Quality has changed)
A t-test will determine if the observed lifespan is significantly different from 1,000 hours.
A psychologist wants to know if meditation reduces stress. The average stress level in the population (on a scale of 1-10) is 6.5. After a 30-day meditation program, a sample of 25 participants had an average stress level of 5.8, with a standard deviation of 1.2.
H₀: μ = 6.5 (No reduction)
Hₐ: μ < 6.5 (Meditation reduces stress)
A one-sample t-test will determine if the reduction in stress levels is significant.