How to Use This Page
1. Solve each exercise problem on your own first — attempt every step before looking here. 2. Compare your method with the step-by-step solution shown. 3. If your answer differs, rework from the step where you diverged. 4. Pay special attention to the Common Mistakes section — these are the errors that cost marks in exams. 5. Use the Exam Checklist before your test to ensure you have covered every topic.
Key Concepts (Quick Recap)
Sampling
- Population: The entire group under study
- Sample: A subset of the population selected for analysis
- Census: Study of every unit in the population
- Parameter: Numerical characteristic of a population (μ, σ², P)
- Statistic: Numerical characteristic computed from a sample (x̄, s², p̂)
Types of Sampling
- Simple Random Sampling: Every unit has equal probability of selection
- Stratified Sampling: Population divided into strata, then random samples from each stratum
- Systematic Sampling: Select every kth unit after a random start (k = N/n)
- Cluster Sampling: Population divided into clusters, entire clusters are randomly selected
Sampling Distribution and Standard Error
- Sampling Distribution: The probability distribution of a statistic computed from all possible samples of size n
- Standard Error (SE): Standard deviation of the sampling distribution
- SE of mean: SE(x̄) = σ/√n (known σ) or s/√n (estimated)
- SE of proportion: SE(p̂) = √(PQ/n) where Q = 1 − P
- SE of difference of means: SE(x̄₁ − x̄₂) = √(σ₁²/n₁ + σ₂²/n₂)
Central Limit Theorem
- For large n (n ≥ 30), the sampling distribution of x̄ approaches normal with mean μ and variance σ²/n
- This holds regardless of the population distribution shape
Estimation
- Point estimate: Single value used to estimate a parameter (e.g., x̄ estimates μ)
- Interval estimate (Confidence Interval): Range of values likely to contain the parameter
- Confidence Interval for mean (large sample): x̄ ± zₐ/₂ · σ/√n
- Confidence Interval for mean (small sample): x̄ ± tₐ/₂ · s/√n
- Confidence Interval for proportion: p̂ ± zₐ/₂ · √(p̂q̂/n)
Hypothesis Testing
- Null Hypothesis (H₀): The hypothesis to be tested (status quo)
- Alternative Hypothesis (H₁): What we conclude if H₀ is rejected
- Test statistic: z = (x̄ − μ₀)/(σ/√n) for large sample mean test
- Level of significance (α): Probability of Type I error
- Critical region: Values of test statistic that lead to rejection of H₀
- Type I error (α): Rejecting H₀ when true
- Type II error (β): Accepting H₀ when false
- Power: 1 − β (probability of correctly rejecting a false H₀)
Critical Values
- 95% confidence: z = 1.96
- 99% confidence: z = 2.58
- 90% confidence: z = 1.645
- 5% level (two-tailed): z = 1.96
- 1% level (two-tailed): z = 2.58
I. Choose the Best Answer (MCQs)
1. The standard error of the sample mean when σ is known is: (a) σ/n (b) σ²/n (c) σ/√n (d) s/√n
Answer(c) σ/√n Explanation: SE(x̄) = σ/√n when the population standard deviation σ is known. Option (d) uses s instead of σ, which applies when σ is unknown.
2. As sample size increases, the standard error: (a) Increases (b) Decreases (c) Remains same (d) Becomes zero
Answer(b) Decreases Explanation: SE = σ/√n. As n increases, √n increases, so SE decreases. This means larger samples give more precise estimates.
3. A Type I error occurs when we: (a) Accept H₀ when it is true (b) Reject H₀ when it is true (c) Accept H₀ when it is false (d) Reject H₀ when it is false
Answer(b) Reject H₀ when it is true Explanation: Type I error = rejecting a true null hypothesis. Its probability is denoted α, the level of significance.
4. The 95% confidence interval for the mean using a large sample is: (a) x̄ ± 1.645 · SE (b) x̄ ± 1.96 · SE (c) x̄ ± 2.58 · SE (d) x̄ ± 3.00 · SE
Answer(b) x̄ ± 1.96 · SE Explanation: For 95% confidence, z₀.₀₂₅ = 1.96. The CI is x̄ ± 1.96 · (σ/√n).
5. In stratified sampling, the population is divided into: (a) Clusters (b) Strata (c) Quotas (d) Systematic groups
Answer(b) Strata Explanation: In stratified sampling, the population is divided into homogeneous groups called strata, and then random samples are drawn from each stratum.
6. If a sample of size 64 is taken from a population with σ = 16, the standard error of the mean is: (a) 4 (b) 2 (c) 16 (d) 0.25
Answer(b) 2 Explanation: SE(x̄) = σ/√n = 16/√64 = 16/8 = 2.
7. The power of a test is: (a) α (b) β (c) 1 − β (d) 1 − α
Answer(c) 1 − β Explanation: Power = P(rejecting H₀ when H₁ is true) = 1 − β. Higher power means a better test.
8. For a small sample (n < 30) from a normal population, we use the: (a) z-test (b) t-test (c) chi-square test (d) F-test
Answer(b) t-test Explanation: When n < 30 and the population is normal with unknown σ, we use the t-test with n − 1 degrees of freedom.
II. Fill in the Blanks
1. The standard error of the sample proportion is √(PQ/n) where P is the population proportion and Q = 1 − P.
2. The Central Limit Theorem states that for large n, the sampling distribution of x̄ is approximately normal with mean μ and variance σ²/n.
3. If the level of significance is 5%, then the confidence level is 95%.
4. The probability of Type II error is denoted by β.
5. The critical value of z for a two-tailed test at 1% level of significance is 2.58.
6. In systematic sampling, if N = 1000 and n = 100, then the sampling interval k = 10.
7. The number of degrees of freedom for a t-test with sample size n is n − 1.
8. As the confidence level increases, the width of the confidence interval increases.
III. True or False
1. Statement: The sample mean x̄ is an unbiased estimator of the population mean μ.
True. E(x̄) = μ, so the sample mean is an unbiased estimator of the population mean.
2. Statement: Standard error increases as sample size increases.
False. Correction: Standard error = σ/√n decreases as sample size n increases. Larger samples give more precise estimates.
3. Statement: A census is always preferable to a sample survey.
False. Correction: A census may not be feasible due to cost, time, or destructiveness of testing. Sampling is often preferred for practical reasons, and with proper sampling techniques, results can be very accurate.
4. Statement: Decreasing the level of significance α will decrease Type I error but may increase Type II error.
True. There is a trade-off between α and β. Making it harder to reject H₀ (smaller α) reduces Type I error but increases Type II error.
5. Statement: In cluster sampling, all units in selected clusters are studied.
True. In cluster sampling, entire clusters are selected randomly, and all units within those clusters are included in the study.
6. Statement: The t-distribution approaches the standard normal distribution as degrees of freedom increase.
True. As n → ∞, the t-distribution with n − 1 degrees of freedom approaches the standard normal distribution N(0, 1).
IV. Match the Following
| Column A | Column B |
|---|
| 1. Simple random sampling | (a) Every kth unit selected |
| 2. Stratified sampling | (b) Equal probability for each unit |
| 3. Systematic sampling | (c) Divided into homogeneous groups |
| 4. Standard error of mean | (d) Rejecting true H₀ |
| 5. Type I error | (e) σ/√n |
Answer1 → (b): In simple random sampling, every unit has equal probability of selection. 2 → (c): In stratified sampling, the population is divided into homogeneous groups (strata). 3 → (a): In systematic sampling, every kth unit is selected after a random start. 4 → (e): SE of mean = σ/√n. 5 → (d): Type I error = rejecting H₀ when it is true.
V. Answer Briefly
1. Explain the differences between census and sampling.
Answer
| Feature | Census | Sampling | |---|---|---| | Coverage | Entire population | Subset of population | | Cost | Very high | Relatively low | | Time | Very long | Much shorter | | Accuracy | No sampling error | Has sampling error | | Practicality | Not feasible for large populations | Feasible for any population size | | Use | When population is small or 100% accuracy needed | When population is large or testing is destructive |
2. What are the characteristics of a good estimator?
Answer
A good estimator should be: 1. Unbiased: E(estimator) = parameter. Example: E(x̄) = μ 2. Consistent: As n → ∞, the estimator converges to the parameter 3. Efficient: Has the smallest variance among all unbiased estimators 4. Sufficient: Uses all relevant information in the sample
3. Define standard error. What are its uses?
Answer
Standard error is the standard deviation of the sampling distribution of a statistic. For the mean: SE(x̄) = σ/√n. Uses: 1. To construct confidence intervals for population parameters 2. To test hypotheses about population parameters 3. To determine the required sample size for a desired precision 4. To compare different estimators — smaller SE means more precise estimation
4. What is the Central Limit Theorem and why is it important?
Answer
The Central Limit Theorem states that for a population with mean μ and variance σ², the sampling distribution of the sample mean x̄ approaches a normal distribution with mean μ and variance σ²/n as the sample size n becomes large (n ≥ 30), regardless of the population distribution. Importance: 1. Allows us to use normal distribution theory for inference even when the population is not normal 2. Forms the basis for confidence intervals and hypothesis tests for means 3. Justifies the use of z-tests for large samples
5. Distinguish between Type I and Type II errors.
Answer
| | H₀ is True | H₀ is False | |---|---|---| | Reject H₀ | Type I error (α) | Correct decision (1 − β = Power) | | Do not reject H₀ | Correct decision (1 − α) | Type II error (β) | Type I error: Rejecting a true null hypothesis. Probability = α (level of significance). Type II error: Failing to reject a false null hypothesis. Probability = β. Both errors cannot be reduced simultaneously without increasing sample size.
6. What is a confidence interval? Interpret a 95% confidence interval.
Answer
A confidence interval is a range of values constructed from sample data that is likely to contain the population parameter with a specified confidence level. A 95% confidence interval means: if we construct 100 such intervals from 100 different samples, approximately 95 of them would contain the true population parameter. It does NOT mean there is a 95% probability that the parameter lies in this specific interval — the parameter is fixed, the interval varies.
VI. Answer in Detail
1. A random sample of 100 students has a mean height of 165 cm with a population standard deviation of 10 cm. Find: (i) standard error of mean, (ii) 95% confidence interval for the population mean, (iii) 99% confidence interval for the population mean.
Givenn = 100, x̄ = 165 cm, σ = 10 cm (i) Standard Error SE(x̄) = σ/√n = 10/√100 = 10/10 = 1 cm (ii) 95% Confidence Interval CI = x̄ ± z₀.₀₂₅ · SE = 165 ± 1.96 × 1 = 165 ± 1.96 = (163.04, 166.96) cm Interpretation: We are 95% confident that the true population mean height lies between 163.04 cm and 166.96 cm. (iii) 99% Confidence Interval CI = x̄ ± z₀.₀₀₅ · SE = 165 ± 2.58 × 1 = 165 ± 2.58 = (162.42, 167.58) cm Note: The 99% CI is wider than the 95% CI. Higher confidence requires a wider interval.
2. In a sample of 400 bulbs, 60 were found defective. Find: (i) sample proportion, (ii) standard error of proportion, (iii) 95% confidence interval for the population proportion.
Givenn = 400, number of defectives = 60 (i) Sample proportion p̂ = 60/400 = 0.15 q̂ = 1 − p̂ = 0.85 (ii) Standard Error of Proportion SE(p̂) = √(p̂q̂/n) = √(0.15 × 0.85/400) = √(0.1275/400) = √(0.000319) = 0.01786 (iii) 95% Confidence Interval for Population Proportion CI = p̂ ± 1.96 × SE(p̂) = 0.15 ± 1.96 × 0.01786 = 0.15 ± 0.035 = (0.115, 0.185) We are 95% confident that the true proportion of defective bulbs is between 11.5% and 18.5%.
3. A manufacturer claims that the average life of his bulbs is 2000 hours. A random sample of 100 bulbs has a mean life of 1980 hours with a standard deviation of 200 hours. Test the claim at 5% level of significance.
Answer
Step 1: State hypotheses H₀: μ = 2000 (claim is true) H₁: μ ≠ 2000 (claim is not true) Step 2: Choose the test Since n = 100 (large), use z-test. Two-tailed test at α = 0.05. Step 3: Compute test statistic z = (x̄ − μ₀)/(s/√n) = (1980 − 2000)/(200/√100) = −20/20 = −1.0 Step 4: Critical value For two-tailed test at 5% level: z₀.₀₂₅ = ±1.96 Step 5: Decision |z| = 1.0 < 1.96, so z falls in the acceptance region. We do not reject H₀. Step 6: Conclusion There is no significant evidence at 5% level to reject the manufacturer's claim. The average life of bulbs could be 2000 hours.
4. Two samples of students from two colleges gave the following results. Test whether the difference between the means is significant at 5% level.
Answer
| College A | College B |
|---|
| Mean | 75 | 72 |
| SD | 10 | 8 |
| Sample size | 50 | 60 |
Step 1: State hypotheses H₀: μ₁ = μ₂ (no difference between means) H₁: μ₁ ≠ μ₂ (there is a difference) Step 2: Compute SE of difference of means SE(x̄₁ − x̄₂) = √(s₁²/n₁ + s₂²/n₂) = √(100/50 + 64/60) = √(2 + 1.067) = √3.067 = 1.751 Step 3: Compute test statistic z = (x̄₁ − x̄₂)/SE = (75 − 72)/1.751 = 3/1.751 = 1.713 Step 4: Critical value For two-tailed test at 5% level: z₀.₀₂₅ = ±1.96 Step 5: Decision |z| = 1.713 < 1.96, so we do not reject H₀. Conclusion: The difference between the two sample means is not statistically significant at the 5% level.
5. A sample of 25 items has a mean of 46 and a standard deviation of 5. Test whether the population mean is 50 at 5% level of significance.
Answer
Step 1: State hypotheses H₀: μ = 50 H₁: μ ≠ 50 Step 2: Choose the test Since n = 25 (small sample < 30), use t-test with df = n − 1 = 24. Step 3: Compute test statistic t = (x̄ − μ₀)/(s/√n) = (46 − 50)/(5/√25) = −4/1 = −4.0 Step 4: Critical value For two-tailed test at 5% level with df = 24: t₀.₀₂₅,₂₄ = ±2.064 Step 5: Decision |t| = 4.0 > 2.064, so we reject H₀. Conclusion: There is significant evidence that the population mean differs from 50 at the 5% level.
6. What sample size is needed to estimate the population mean within ±2 units with 95% confidence, if the population standard deviation is 12?
Answer
Step 1: Identify the formula For 95% CI: n = (zₐ/₂ · σ/E)² where E = margin of error Step 2: Substitute values z₀.₀₂₅ = 1.96, σ = 12, E = 2 n = (1.96 × 12/2)² = (23.52/2)² = (11.76)² = 138.3 Step 3: Round up n = 139 (always round up to the next integer) We need a sample of at least 139 items to estimate the mean within ±2 units with 95% confidence.
Common Mistakes to Avoid
1. Confusing standard deviation with standard error — SD measures variability in the population, SE measures variability in the sampling distribution. SE = σ/√n is always smaller than σ for n > 1.
2. Using z-test for small samples — When n < 30 and σ is unknown, use the t-test with n − 1 degrees of freedom, not the z-test. The t-test accounts for the extra uncertainty from estimating σ.
3. Forgetting continuity correction — When approximating a discrete distribution (like binomial proportions) with a normal distribution for large samples, apply continuity correction where appropriate.
4. Incorrect hypotheses — H₀ always contains the equality (=, ≤, ≥). H₁ contains the alternative (> , < , ≠). The claim being tested can be either H₀ or H₁ depending on the question.
5. Not converting proportions correctly — When working with proportions, remember: p̂ = x/n (successes/sample size), q̂ = 1 − p̂, and SE = √(p̂q̂/n).
6. Misinterpreting confidence level — A 95% CI does NOT mean there is a 95% probability that the parameter lies in the interval. It means 95% of such intervals constructed from repeated sampling would contain the parameter.
Exam Checklist
- [ ] I can distinguish between population, sample, parameter, and statistic
- [ ] I can identify appropriate sampling methods (simple random, stratified, systematic, cluster)
- [ ] I can calculate standard error for mean and proportion
- [ ] I can construct confidence intervals for mean and proportion at 95% and 99% levels
- [ ] I can state hypotheses correctly (H₀ and H₁)
- [ ] I can perform z-test for large sample means and proportions
- [ ] I can perform t-test for small sample means
- [ ] I can determine sample size for a desired precision level
- [ ] I can distinguish between Type I and Type II errors
- [ ] I can apply the Central Limit Theorem to sampling distributions
