Unit 4: Solved Numerical Problems

This section contains 15+ fully solved problems covering estimation techniques, confidence intervals, and sample size determination.

Section A: Point Estimation and Sampling Distribution

Problem 1: Standard Error of Mean

Question: A population has standard deviation σ = 24. Calculate the standard error of the mean for samples of size: a) n = 16 b) n = 64 c) n = 144

Click to reveal solution **Formula:** $SE = \frac{\sigma}{\sqrt{n}}$ **Part (a):** n = 16 $$SE = \frac{24}{\sqrt{16}} = \frac{24}{4} = 6$$ **Part (b):** n = 64 $$SE = \frac{24}{\sqrt{64}} = \frac{24}{8} = 3$$ **Part (c):** n = 144 $$SE = \frac{24}{\sqrt{144}} = \frac{24}{12} = 2$$ **Answers:** a) **SE = 6**, b) **SE = 3**, c) **SE = 2** **Note:** As sample size increases, standard error decreases (by a factor of √n).

Problem 2: Standard Error of Proportion

Question: In a population, the proportion with a certain characteristic is p = 0.40. Calculate the standard error for samples of size n = 100.

Click to reveal solution **Formula:** $SE_p = \sqrt{\frac{p(1-p)}{n}}$ $$SE_p = \sqrt{\frac{0.40 \times 0.60}{100}} = \sqrt{\frac{0.24}{100}} = \sqrt{0.0024} = 0.049$$ **Answer:** **SE = 0.049 or 4.9%**

Problem 3: Sampling Distribution of Mean

Question: IQ scores have mean μ = 100 and SD σ = 15. For a sample of 36 students: a) What is the standard error? b) What is the probability that the sample mean exceeds 104? c) Between what values will 95% of sample means fall?

Click to reveal solution **Part (a):** Standard Error $$SE = \frac{15}{\sqrt{36}} = \frac{15}{6} = 2.5$$ **Part (b):** P(X̄ > 104) $$Z = \frac{104 - 100}{2.5} = \frac{4}{2.5} = 1.6$$ $$P(Z > 1.6) = 1 - 0.9452 = 0.0548$$ **Part (c):** 95% interval for sample means For 95%, Z = ±1.96 $$\bar{X} = \mu \pm 1.96 \times SE = 100 \pm 1.96 \times 2.5$$ $$= 100 \pm 4.9 = (95.1, 104.9)$$ **Answers:** a) **SE = 2.5**, b) **0.0548**, c) **(95.1, 104.9)**

Section B: Confidence Intervals for Mean

Problem 4: CI for Mean (σ Known, Large Sample)

Question: A random sample of 100 employees shows mean monthly salary of Rs. 45,000. The population standard deviation is known to be Rs. 8,000. Construct: a) 95% confidence interval b) 99% confidence interval

Click to reveal solution **Given:** n = 100, x̄ = 45,000, σ = 8,000 **Formula:** $\bar{x} \pm Z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$ **Part (a):** 95% CI (Z = 1.96) **Step 1:** Calculate standard error $$SE = \frac{8000}{\sqrt{100}} = \frac{8000}{10} = 800$$ **Step 2:** Calculate margin of error $$ME = 1.96 \times 800 = 1568$$ **Step 3:** Construct interval $$95\% \text{ CI} = 45000 \pm 1568 = (43432, 46568)$$ **Part (b):** 99% CI (Z = 2.576) $$ME = 2.576 \times 800 = 2060.8$$ $$99\% \text{ CI} = 45000 \pm 2060.8 = (42939.2, 47060.8)$$ **Answers:** a) **95% CI: (Rs. 43,432, Rs. 46,568)** b) **99% CI: (Rs. 42,939, Rs. 47,061)** **Interpretation:** We are 95% confident that the true mean salary lies between Rs. 43,432 and Rs. 46,568.

Problem 5: CI for Mean (σ Unknown, Large Sample)

Question: A sample of 64 transactions shows mean processing time of 15 minutes with sample standard deviation of 4 minutes. Construct a 95% confidence interval for the population mean.

Click to reveal solution **Given:** n = 64, x̄ = 15, s = 4 **Note:** Since n ≥ 30, we can use Z-distribution even with s instead of σ. **Step 1:** Calculate SE using sample SD $$SE = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{64}} = \frac{4}{8} = 0.5$$ **Step 2:** For 95% CI, Z = 1.96 $$ME = 1.96 \times 0.5 = 0.98$$ **Step 3:** Construct interval $$95\% \text{ CI} = 15 \pm 0.98 = (14.02, 15.98)$$ **Answer:** **95% CI: (14.02 minutes, 15.98 minutes)**

Problem 6: CI for Mean (Small Sample, t-Distribution)

Question: A random sample of 16 items has mean weight 52 kg and standard deviation 8 kg. Assuming normal distribution, construct a 95% confidence interval for the population mean.

Click to reveal solution **Given:** n = 16, x̄ = 52, s = 8, df = 15 **Step 1:** Find t-value For 95% CI with df = 15: t = 2.131 **Step 2:** Calculate SE $$SE = \frac{8}{\sqrt{16}} = 2$$ **Step 3:** Calculate margin of error $$ME = t \times SE = 2.131 \times 2 = 4.262$$ **Step 4:** Construct interval $$95\% \text{ CI} = 52 \pm 4.262 = (47.74, 56.26)$$ **Answer:** **95% CI: (47.74 kg, 56.26 kg)**

Problem 7: Comparing CI Widths

Question: A sample of size 25 has mean 80 and SD 10. Calculate 90%, 95%, and 99% confidence intervals and compare their widths.

Click to reveal solution **Given:** n = 25, x̄ = 80, s = 10, df = 24 $$SE = \frac{10}{\sqrt{25}} = 2$$ | Confidence Level | t-value | ME | CI | Width | | ---------------- | ------- | ---- | -------------- | ----- | | 90% | 1.711 | 3.42 | (76.58, 83.42) | 6.84 | | 95% | 2.064 | 4.13 | (75.87, 84.13) | 8.26 | | 99% | 2.797 | 5.59 | (74.41, 85.59) | 11.18 | **Conclusion:** As confidence level increases, the interval width increases (trade-off between confidence and precision).

Section C: Confidence Intervals for Proportion

Problem 8: CI for Population Proportion

Question: In a survey of 400 voters, 180 supported a particular candidate. Construct a 95% confidence interval for the true proportion of supporters.

Click to reveal solution **Step 1:** Calculate sample proportion $$\hat{p} = \frac{180}{400} = 0.45$$ **Step 2:** Calculate standard error $$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.45 \times 0.55}{400}}$$ $$= \sqrt{\frac{0.2475}{400}} = \sqrt{0.000619} = 0.0249$$ **Step 3:** For 95% CI, Z = 1.96 $$ME = 1.96 \times 0.0249 = 0.0488$$ **Step 4:** Construct interval $$95\% \text{ CI} = 0.45 \pm 0.0488 = (0.401, 0.499)$$ **Answer:** **95% CI: (40.1%, 49.9%)** or **(0.401, 0.499)** **Interpretation:** We are 95% confident that between 40.1% and 49.9% of all voters support this candidate.

Problem 9: CI for Proportion (Different Confidence Levels)

Question: In a quality check, 24 out of 200 items were found defective. Construct: a) 90% CI b) 95% CI c) 99% CI for the defect rate

Click to reveal solution **Step 1:** Calculate sample proportion $$\hat{p} = \frac{24}{200} = 0.12$$ **Step 2:** Calculate SE $$SE = \sqrt{\frac{0.12 \times 0.88}{200}} = \sqrt{0.000528} = 0.0230$$ **Part (a):** 90% CI (Z = 1.645) $$CI = 0.12 \pm 1.645 \times 0.0230 = 0.12 \pm 0.038$$ $$= (0.082, 0.158) \text{ or } (8.2\%, 15.8\%)$$ **Part (b):** 95% CI (Z = 1.96) $$CI = 0.12 \pm 1.96 \times 0.0230 = 0.12 \pm 0.045$$ $$= (0.075, 0.165) \text{ or } (7.5\%, 16.5\%)$$ **Part (c):** 99% CI (Z = 2.576) $$CI = 0.12 \pm 2.576 \times 0.0230 = 0.12 \pm 0.059$$ $$= (0.061, 0.179) \text{ or } (6.1\%, 17.9\%)$$ **Answers:** a) **(8.2%, 15.8%)**, b) **(7.5%, 16.5%)**, c) **(6.1%, 17.9%)**

Section D: Sample Size Determination

Problem 10: Sample Size for Mean Estimation

Question: A researcher wants to estimate the mean income with 95% confidence and margin of error of Rs. 2,000. If the population SD is estimated to be Rs. 15,000, what sample size is needed?

Click to reveal solution **Formula:** $n = \left(\frac{Z \times \sigma}{E}\right)^2$ **Given:** - E (margin of error) = 2,000 - σ = 15,000 - Z = 1.96 (for 95%) **Calculation:** $$n = \left(\frac{1.96 \times 15000}{2000}\right)^2 = \left(\frac{29400}{2000}\right)^2$$ $$= (14.7)^2 = 216.09$$ **Answer:** **n = 217** (always round up)

Problem 11: Sample Size for Proportion Estimation

Question: A researcher wants to estimate voter preference with 95% confidence and margin of error of 3%. a) If no prior estimate exists, what sample size is needed? b) If a pilot study showed p ≈ 0.40, what sample size is needed?

Click to reveal solution **Formula:** $n = \frac{Z^2 \times p(1-p)}{E^2}$ **Part (a):** No prior estimate (use p = 0.5 for maximum sample size) $$n = \frac{(1.96)^2 \times 0.5 \times 0.5}{(0.03)^2} = \frac{3.8416 \times 0.25}{0.0009}$$ $$= \frac{0.9604}{0.0009} = 1067.11$$ **Answer for (a):** **n = 1068** **Part (b):** With p = 0.40 $$n = \frac{(1.96)^2 \times 0.40 \times 0.60}{(0.03)^2} = \frac{3.8416 \times 0.24}{0.0009}$$ $$= \frac{0.922}{0.0009} = 1024.4$$ **Answer for (b):** **n = 1025**

Problem 12: Sample Size with Different Confidence Levels

Question: Compare sample sizes needed to estimate a mean with margin of error Rs. 500, σ = 3000, for: a) 90% confidence b) 95% confidence c) 99% confidence

Click to reveal solution | Confidence | Z-value | n = (Z × 3000/500)² | Required n | | ---------- | ------- | ------------------- | ---------- | | 90% | 1.645 | (9.87)² = 97.4 | **98** | | 95% | 1.96 | (11.76)² = 138.3 | **139** | | 99% | 2.576 | (15.46)² = 238.9 | **239** | **Conclusion:** Higher confidence requires larger sample sizes.

Problem 13: Sample Size for Desired Width

Question: A 95% CI for a mean has width Rs. 1,000. What sample size would reduce the width to Rs. 500?

Click to reveal solution **Key insight:** Width = 2 × ME, and ME ∝ 1/√n To halve the width, we need to quadruple the sample size. **If current n = n₁, new n₂ = 4n₁** **Alternative approach using formula:** Current: Width = 2 × Z × σ/√n₁ = 1000 New: Width = 2 × Z × σ/√n₂ = 500 $$\frac{\sqrt{n_2}}{\sqrt{n_1}} = \frac{1000}{500} = 2$$ $$\frac{n_2}{n_1} = 4$$ **Answer:** The new sample size must be **4 times** the original sample size.

Section E: Comprehensive Problems

Problem 14: Complete Estimation Problem

Question: A government agency wants to estimate average waiting time at service centers. A pilot study of 36 customers showed mean waiting time of 25 minutes with SD of 12 minutes.

a) Construct 95% CI for mean waiting time b) How many customers should be sampled to estimate mean within 2 minutes with 95% confidence? c) If 90% confidence is acceptable, how does the required sample size change?

Solution:

Part (a): 95% CI \(SE = \frac{12}{\sqrt{36}} = 2\) \(95\% \text{ CI} = 25 \pm 1.96 \times 2 = 25 \pm 3.92\) \(= (21.08, 28.92) \text{ minutes}\)

Part (b): Sample size for E = 2, 95% confidence \(n = \left(\frac{1.96 \times 12}{2}\right)^2 = (11.76)^2 = 138.3\) n = 139 customers

Part (c): Sample size for E = 2, 90% confidence \(n = \left(\frac{1.645 \times 12}{2}\right)^2 = (9.87)^2 = 97.4\) n = 98 customers

Savings: 139 - 98 = 41 fewer customers needed

Problem 15: Confidence Interval Interpretation

Question: A 95% CI for mean household income is (Rs. 35,000, Rs. 45,000). Which statements are correct?

a) 95% of households have income between Rs. 35,000 and Rs. 45,000 b) The probability that μ is in this interval is 0.95 c) If we repeat sampling many times, about 95% of such intervals will contain μ d) The sample mean is Rs. 40,000 e) The margin of error is Rs. 5,000

Solution:

a) INCORRECT - CI is about the population mean, not individual values

b) TECHNICALLY INCORRECT - μ is fixed; it either is or isn’t in the interval. The 95% refers to the method’s reliability.

c) CORRECT - This is the correct frequentist interpretation

d) CORRECT - Sample mean = (35,000 + 45,000)/2 = 40,000

e) CORRECT - ME = 45,000 - 40,000 = 5,000

Answer: Statements c, d, and e are correct.

Problem 16: Finite Population Correction

Question: A company has 500 employees. A random sample of 100 employees shows mean satisfaction score of 72 with SD 15. Construct a 95% CI for the population mean.

Click to reveal solution **Step 1:** Check if finite population correction (FPC) is needed $$\frac{n}{N} = \frac{100}{500} = 0.20 > 0.05$$ FPC is needed since we're sampling more than 5% of the population. **Step 2:** Calculate SE with FPC $$SE = \frac{s}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}}$$ $$= \frac{15}{\sqrt{100}} \times \sqrt{\frac{500-100}{499}}$$ $$= 1.5 \times \sqrt{\frac{400}{499}} = 1.5 \times 0.895 = 1.34$$ **Step 3:** Construct 95% CI $$CI = 72 \pm 1.96 \times 1.34 = 72 \pm 2.63$$ $$= (69.37, 74.63)$$ **Without FPC:** $$SE = 1.5, \quad CI = 72 \pm 2.94 = (69.06, 74.94)$$ **Answer:** With FPC: **(69.37, 74.63)**, Without FPC: (69.06, 74.94) The interval is narrower with FPC because we're sampling a significant portion of the population.

Practice Problems

  1. A sample of 49 items has mean 85 and SD 14. Construct 90% and 99% CIs for the population mean.

  2. In a survey of 500 people, 280 preferred option A. Construct a 95% CI for the true proportion.

  3. What sample size is needed to estimate a population mean within 5 units with 95% confidence if σ = 20?

  4. A researcher wants to estimate a proportion within 4% with 99% confidence. What sample size is needed?

  5. A sample of 25 measurements has mean 150 and SD 20. Construct a 95% CI using t-distribution.

Summary of Formulas

Parameter Confidence Interval Sample Size
Mean (σ known) $\bar{x} \pm Z \times \frac{\sigma}{\sqrt{n}}$ $n = \left(\frac{Z\sigma}{E}\right)^2$
Mean (σ unknown, large n) $\bar{x} \pm Z \times \frac{s}{\sqrt{n}}$ Use s for σ
Mean (σ unknown, small n) $\bar{x} \pm t \times \frac{s}{\sqrt{n}}$ df = n - 1
Proportion $\hat{p} \pm Z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ $n = \frac{Z^2 p(1-p)}{E^2}$

Critical Values Reference

Confidence Level Z-value t (df=10) t (df=20) t (df=30)
90% 1.645 1.812 1.725 1.697
95% 1.960 2.228 2.086 2.042
99% 2.576 3.169 2.845 2.750