Learning Objectives
By the end of this chapter, you will be able to:
- Distinguish between point and interval estimates
- Construct confidence intervals for population mean
- Construct confidence intervals for population proportion
- Interpret confidence intervals correctly
- Choose appropriate confidence levels
Point Estimation vs Interval Estimation
flowchart TD
A[Estimation]
A --> B[Point Estimate]
A --> C[Interval Estimate]
B --> B1["Single value<br/>μ ≈ 52"]
C --> C1["Range with confidence<br/>μ ∈ (48, 56) at 95%"]
Point Estimate
A point estimate is a single value used to estimate a parameter.
Example: “The average processing time is 45 minutes.”
Interval Estimate (Confidence Interval)
An interval estimate (or confidence interval) provides a range of plausible values with an associated confidence level.
Example: “The average processing time is between 42 and 48 minutes with 95% confidence.”
Confidence Interval Structure
\[\text{Point Estimate} \pm \text{Margin of Error}\]Or:
\[\text{CI} = (\text{Lower Limit}, \text{Upper Limit})\]Components
flowchart LR
A["Point Estimate<br/>(x̄ or p̂)"] --> B["+/-"]
B --> C["Margin of Error<br/>(z × SE)"]
D["Confidence Level<br/>(90%, 95%, 99%)"] --> E["z-value"]
F["Sample Size & SD"] --> G["Standard Error"]
E --> C
G --> C
Confidence Level
The confidence level represents how confident we are that the interval contains the true parameter.
| Confidence Level | z-value (z*) | α | α/2 |
|---|---|---|---|
| 90% | 1.645 | 0.10 | 0.05 |
| 95% | 1.96 | 0.05 | 0.025 |
| 99% | 2.576 | 0.01 | 0.005 |
Interpretation
“95% confident” means: If we took 100 different samples and built 100 confidence intervals, approximately 95 of them would contain the true population parameter.
⚠️ Common Mistake: It does NOT mean there’s a 95% probability the parameter is in this interval.
Confidence Interval for Population Mean (σ Known)
When population standard deviation (σ) is known:
\[\bar{x} \pm z^* \cdot \frac{\sigma}{\sqrt{n}}\]Or:
\[\left(\bar{x} - z^* \cdot \frac{\sigma}{\sqrt{n}}, \quad \bar{x} + z^* \cdot \frac{\sigma}{\sqrt{n}}\right)\]Where:
- $\bar{x}$ = sample mean
- $z^*$ = z-value for desired confidence level
- $\sigma$ = population standard deviation
- $n$ = sample size
Step-by-Step Example 1: CI for Mean (σ Known)
Problem: A sample of 64 government employees has a mean salary of NPR 48,000. The population standard deviation is known to be NPR 8,000. Construct a 95% confidence interval for the population mean salary.
Solution:
Step 1: Identify given values
- $\bar{x}$ = 48,000
- σ = 8,000
- n = 64
- Confidence level = 95% → z* = 1.96
Step 2: Calculate standard error \(SE = \frac{\sigma}{\sqrt{n}} = \frac{8,000}{\sqrt{64}} = \frac{8,000}{8} = 1,000\)
Step 3: Calculate margin of error \(ME = z^* \times SE = 1.96 \times 1,000 = 1,960\)
Step 4: Construct interval \(CI = \bar{x} \pm ME = 48,000 \pm 1,960\) \(CI = (46,040, \quad 49,960)\)
Answer: We are 95% confident that the true mean salary is between NPR 46,040 and NPR 49,960.
Confidence Interval for Population Mean (σ Unknown)
When population standard deviation is unknown (more common in practice), use the t-distribution:
\[\bar{x} \pm t^* \cdot \frac{s}{\sqrt{n}}\]Where:
- $s$ = sample standard deviation
- $t^*$ = t-value with df = n - 1
t-Distribution
- Similar to normal but with heavier tails
- Depends on degrees of freedom (df)
- As df increases, t approaches z
Step-by-Step Example 2: CI for Mean (σ Unknown)
Problem: A sample of 25 employees took a mean of 42 minutes to complete a task with a sample standard deviation of 10 minutes. Construct a 95% confidence interval for the population mean.
Solution:
Step 1: Identify given values
- $\bar{x}$ = 42
- s = 10
- n = 25
- df = 25 - 1 = 24
- Confidence level = 95%
Step 2: Find t-value From t-table with df = 24 and α/2 = 0.025: t* = 2.064
Step 3: Calculate standard error \(SE = \frac{s}{\sqrt{n}} = \frac{10}{\sqrt{25}} = \frac{10}{5} = 2\)
Step 4: Calculate margin of error \(ME = t^* \times SE = 2.064 \times 2 = 4.128\)
Step 5: Construct interval \(CI = 42 \pm 4.128 = (37.87, \quad 46.13)\)
Answer: We are 95% confident that the true mean completion time is between 37.87 and 46.13 minutes.
t-Table (Selected Values)
| df | 90% (t₀.₀₅) | 95% (t₀.₀₂₅) | 99% (t₀.₀₀₅) |
|---|---|---|---|
| 10 | 1.812 | 2.228 | 3.169 |
| 15 | 1.753 | 2.131 | 2.947 |
| 20 | 1.725 | 2.086 | 2.845 |
| 25 | 1.708 | 2.060 | 2.787 |
| 30 | 1.697 | 2.042 | 2.750 |
| 60 | 1.671 | 2.000 | 2.660 |
| ∞ | 1.645 | 1.960 | 2.576 |
Confidence Interval for Population Proportion
For proportions (yes/no data):
\[\hat{p} \pm z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Where:
- $\hat{p} = \frac{x}{n}$ = sample proportion
- x = number of successes
- n = sample size
Conditions
- $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$
Step-by-Step Example 3: CI for Proportion
Problem: In a survey of 400 citizens, 280 support a new policy. Construct a 95% confidence interval for the true proportion of supporters.
Solution:
Step 1: Calculate sample proportion \(\hat{p} = \frac{280}{400} = 0.70\)
Step 2: Check conditions
- $n\hat{p} = 400 \times 0.70 = 280 \geq 10$ ✓
- $n(1-\hat{p}) = 400 \times 0.30 = 120 \geq 10$ ✓
Step 3: Calculate standard error \(SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.70 \times 0.30}{400}} = \sqrt{\frac{0.21}{400}} = \sqrt{0.000525} = 0.0229\)
Step 4: Find z-value For 95%: z* = 1.96
Step 5: Calculate margin of error \(ME = 1.96 \times 0.0229 = 0.0449\)
Step 6: Construct interval \(CI = 0.70 \pm 0.0449 = (0.6551, \quad 0.7449)\)
Answer: We are 95% confident that the true proportion of supporters is between 65.51% and 74.49%.
Step-by-Step Example 4: Exam-Style Problem
Problem: A random sample of 36 households shows average monthly electricity consumption of 280 kWh with standard deviation 45 kWh.
(a) Construct a 90% confidence interval for population mean. (b) Construct a 99% confidence interval for population mean. (c) Compare the widths and explain the difference.
Solution:
Given: $\bar{x}$ = 280, s = 45, n = 36, df = 35
(a) 90% Confidence Interval:
t* for df=35, 90% ≈ 1.690
\[SE = \frac{45}{\sqrt{36}} = \frac{45}{6} = 7.5\] \[ME = 1.690 \times 7.5 = 12.675\] \[CI_{90\%} = 280 \pm 12.675 = (267.33, \quad 292.67)\](b) 99% Confidence Interval:
t* for df=35, 99% ≈ 2.724
\[ME = 2.724 \times 7.5 = 20.43\] \[CI_{99\%} = 280 \pm 20.43 = (259.57, \quad 300.43)\](c) Comparison:
| Confidence Level | Width |
|---|---|
| 90% | 25.34 |
| 99% | 40.86 |
Explanation: Higher confidence requires a wider interval. The 99% CI is wider because we need more “room” to be more certain the true parameter is captured.
Factors Affecting Interval Width
flowchart TD
A[Interval Width]
A --> B[Confidence Level ↑]
A --> C[Sample Size ↓]
A --> D[Variability ↑]
B --> B1["Wider interval"]
C --> C1["Wider interval"]
D --> D1["Wider interval"]
| Factor | Effect on Width |
|---|---|
| ↑ Confidence level | ↑ Width |
| ↑ Sample size | ↓ Width |
| ↑ Standard deviation | ↑ Width |
Margin of Error Summary
| Parameter | Margin of Error |
|---|---|
| Mean (σ known) | $z^* \cdot \frac{\sigma}{\sqrt{n}}$ |
| Mean (σ unknown) | $t^* \cdot \frac{s}{\sqrt{n}}$ |
| Proportion | $z^* \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ |
Practice Problems
Problem 1
A sample of 49 items has mean 85 and population σ = 14. Construct: (a) 90% confidence interval (b) 95% confidence interval (c) 99% confidence interval
Problem 2
A sample of 20 employees has mean productivity of 120 units with s = 18. Construct a 95% CI for population mean.
Problem 3
In a sample of 500 voters, 225 favor a candidate. Find: (a) 95% CI for true proportion (b) 99% CI for true proportion
Problem 4
What sample size would reduce the margin of error in Problem 1 by half at 95% confidence?
Problem 5
Interpret this statement: “The 95% confidence interval for mean income is (42,000, 58,000).”
Summary
| Type | When to Use | Formula |
|---|---|---|
| z-interval (mean) | σ known | $\bar{x} \pm z^* \frac{\sigma}{\sqrt{n}}$ |
| t-interval (mean) | σ unknown | $\bar{x} \pm t^* \frac{s}{\sqrt{n}}$ |
| z-interval (prop.) | Proportions | $\hat{p} \pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ |
Next Topic
In the next chapter, we will study Determining Sample Size - how to calculate the required sample size to achieve a desired margin of error.
