Unit 3 Cheat Sheet: Sampling & Estimation
This cheat sheet provides a practical mapping framework to translate raw exam questions into correct parameters, select the appropriate formula, and compute exact confidence intervals.
1. Quick-Glance Decision Rules
Proportions (p p p ): Always use the Z Z Z -distribution (S E = p ^ q ^ n SE = \sqrt{\frac{\hat{p}\hat{q}}{n}} S E = n p ^ q ^ ).
Means (μ \mu μ ):
If n ≥ 30 n \ge 30 n ≥ 30 or σ \sigma σ is known ⟹ \implies ⟹ Z Z Z -distribution (S E = σ n SE = \frac{\sigma}{\sqrt{n}} S E = n σ ).
If n < 30 n < 30 n < 30 and σ \sigma σ is unknown ⟹ \implies ⟹ t t t -distribution with d f = n − 1 df = n - 1 df = n − 1 (S E = s n SE = \frac{s}{\sqrt{n}} S E = n s ).
Finite Population Correction (FPC): If total population N N N is given and n N > 0.05 \frac{n}{N} > 0.05 N n > 0.05 (5 % 5\% 5% ) ⟹ \implies ⟹ Multiply S E SE S E by N − n N − 1 \sqrt{\frac{N - n}{N - 1}} N − 1 N − n .
2. Problem-Clue Mapping Matrix
Use this table to map keywords in exam questions directly to mathematical symbols:
Question Keyword / Clue Mathematical Meaning Extracted Symbol Action Required "Sample of 36 employees..." Sample size n = 36 n = 36 n = 36 If n ≥ 30 n \ge 30 n ≥ 30 , use Z Z Z -distribution. "Sample of 15 students..." Small sample size n = 15 n = 15 n = 15 If σ \sigma σ unknown, use t t t -distribution (d f = 14 df = 14 df = 14 ). "Population standard deviation of 6..." Known population variance σ = 6 \sigma = 6 σ = 6 Use Z = x ˉ − μ σ / n Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} Z = σ / n x ˉ − μ . "Sample standard deviation of 10..." Sample standard deviation only s = 10 s = 10 s = 10 Use t = x ˉ − μ s / n t = \frac{\bar{x} - \mu}{s/\sqrt{n}} t = s / n x ˉ − μ if n < 30 n < 30 n < 30 . "180 reported satisfied out of 300..." Sample proportion successes x = 180 , n = 300 x = 180, \, n = 300 x = 180 , n = 300 Compute p ^ = 180 300 = 0.60 , q ^ = 0.40 \hat{p} = \frac{180}{300} = 0.60, \, \hat{q} = 0.40 p ^ = 300 180 = 0.60 , q ^ = 0.40 . "Assuming a total of 1000 customers..." Finite population size N = 1000 N = 1000 N = 1000 Check n N = 300 1000 = 0.30 > 0.05 ⟹ \frac{n}{N} = \frac{300}{1000} = 0.30 > 0.05 \implies N n = 1000 300 = 0.30 > 0.05 ⟹ Apply FPC. "At a 95% confidence level..." Confidence level (1 − α 1 - \alpha 1 − α ) 1 − α = 0.95 1 - \alpha = 0.95 1 − α = 0.95 Lookup Z α / 2 = 1.960 Z_{\alpha/2} = 1.960 Z α /2 = 1.960 or t 0.025 , d f t_{0.025, df} t 0.025 , df .
3. The "Golden" Critical Values Table
Memorize these two-tailed critical values for instant exam calculations:
Confidence Level (1 − α 1 - \alpha 1 − α ) Alpha (α \alpha α ) Tail Area (α / 2 \alpha/2 α /2 ) Normal Critical Value (Z crit Z_{\text{crit}} Z crit ) 90% 0.10 0.10 0.10 0.05 0.05 0.05 ± 1.645 \pm 1.645 ± 1.645 95% 0.05 0.05 0.05 0.025 0.025 0.025 ± 1.960 \pm 1.960 ± 1.960 99% 0.01 0.01 0.01 0.005 0.005 0.005 ± 2.576 \pm 2.576 ± 2.576
(Note: For t t t -values, always locate the column α / 2 \alpha/2 α /2 matching your row d f = n − 1 df = n - 1 df = n − 1 . Refer to the Statistical Tables Guide or view the Statistical Tables PDF .)
4. Worked Problem Archetypes
Every Unit 3 problem maps to one of the following three templates.
Archetype 1: Population Mean with Known σ \sigma σ or Large Sample (Z Z Z -Interval)
Exam Scenario: A random sample of 36 36 36 employees spends an average of 22 hours 22\text{ hours} 22 hours on training with a population standard deviation of 6 hours 6\text{ hours} 6 hours . Construct a 95 % 95\% 95% confidence interval for the true mean training time.
How to Map the Problem
Extract Parameters:
x ˉ = 22 hours \bar{x} = 22\text{ hours} x ˉ = 22 hours
σ = 6 hours \sigma = 6\text{ hours} σ = 6 hours (population standard deviation is known)
n = 36 n = 36 n = 36 (n ≥ 30 ⟹ n \ge 30 \implies n ≥ 30 ⟹ Normal Z Z Z -distribution applies)
95 % confidence ⟹ Z 0.025 = 1.960 95\%\text{ confidence} \implies Z_{0.025} = \mathbf{1.960} 95% confidence ⟹ Z 0.025 = 1.960
Calculate Standard Error (S E SE S E ):
S E = σ n = 6 36 = 6 6 = 1.0 hour SE = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{36}} = \frac{6}{6} = \mathbf{1.0\text{ hour}} S E = n σ = 36 6 = 6 6 = 1.0 hour
Calculate Margin of Error (M E ME M E ):
M E = Z α / 2 × S E = 1.960 × 1.0 = 1.960 hours ME = Z_{\alpha/2} \times SE = 1.960 \times 1.0 = \mathbf{1.960\text{ hours}} M E = Z α /2 × S E = 1.960 × 1.0 = 1.960 hours
Construct Confidence Interval:
CI = x ˉ ± M E = 22 ± 1.960 = [ 20.040 , 23.960 ] hours \text{CI} = \bar{x} \pm ME = 22 \pm 1.960 = \mathbf{[20.040, \, 23.960]\text{ hours}} CI = x ˉ ± M E = 22 ± 1.960 = [ 20.040 , 23.960 ] hours
Archetype 2: Population Mean with Small Sample & Unknown σ \sigma σ (t t t -Interval)
Exam Scenario: A researcher tests 15 15 15 students using a new study method. The sample mean test score is 78 78 78 with a sample standard deviation of 10 10 10 . Construct a 95 % 95\% 95% confidence interval for the true average score.
How to Map the Problem
Extract Parameters:
x ˉ = 78 \bar{x} = 78 x ˉ = 78
s = 10 s = 10 s = 10 (σ \sigma σ is unknown)
n = 15 n = 15 n = 15 (n < 30 ⟹ n < 30 \implies n < 30 ⟹ Small sample, requires t t t -distribution)
Degrees of Freedom: d f = n − 1 = 15 − 1 = 14 df = n - 1 = 15 - 1 = \mathbf{14} df = n − 1 = 15 − 1 = 14
Lookup t t t -Critical Value:
For 95 % 95\% 95% confidence (α / 2 = 0.025 \alpha/2 = 0.025 α /2 = 0.025 ) with d f = 14 df = 14 df = 14 :
t 0.025 , 14 = 2.145 t_{0.025, 14} = \mathbf{2.145} t 0.025 , 14 = 2.145
Calculate Standard Error (S E SE S E ):
S E = s n = 10 15 ≈ 10 3.873 ≈ 2.582 SE = \frac{s}{\sqrt{n}} = \frac{10}{\sqrt{15}} \approx \frac{10}{3.873} \approx \mathbf{2.582} S E = n s = 15 10 ≈ 3.873 10 ≈ 2.582
Calculate Margin of Error (M E ME M E ):
M E = t c × S E = 2.145 × 2.582 ≈ 5.538 ME = t_c \times SE = 2.145 \times 2.582 \approx \mathbf{5.538} M E = t c × S E = 2.145 × 2.582 ≈ 5.538
Construct Confidence Interval:
CI = 78 ± 5.538 = [ 72.46 , 83.54 ] \text{CI} = 78 \pm 5.538 = \mathbf{[72.46, \, 83.54]} CI = 78 ± 5.538 = [ 72.46 , 83.54 ]
Archetype 3: Proportion with Finite Population Correction (p p p + FPC)
Exam Scenario: Out of 1 , 000 1,000 1 , 000 registered retail customers, a random sample of 300 300 300 is surveyed. 180 180 180 report being satisfied. Construct a 95 % 95\% 95% confidence interval for the true proportion of satisfied customers.
How to Map the Problem
Extract Parameters:
N = 1000 N = 1000 N = 1000 (Total population)
n = 300 n = 300 n = 300 (Sample size)
x = 180 x = 180 x = 180 (Success count)
p ^ = 180 300 = 0.60 , q ^ = 1 − 0.60 = 0.40 \hat{p} = \frac{180}{300} = \mathbf{0.60}, \quad \hat{q} = 1 - 0.60 = \mathbf{0.40} p ^ = 300 180 = 0.60 , q ^ = 1 − 0.60 = 0.40
95 % confidence ⟹ Z 0.025 = 1.960 95\%\text{ confidence} \implies Z_{0.025} = \mathbf{1.960} 95% confidence ⟹ Z 0.025 = 1.960
Check the FPC Condition:
n N = 300 1000 = 0.30 ( 30 % ) > 0.05 ⟹ Must apply FPC! \frac{n}{N} = \frac{300}{1000} = 0.30\text{ (}30\%\text{)} > 0.05 \implies \text{\textbf{Must apply FPC!}} N n = 1000 300 = 0.30 ( 30% ) > 0.05 ⟹ Must apply FPC!
Calculate FPC Multiplier:
FPC = N − n N − 1 = 1000 − 300 1000 − 1 = 700 999 ≈ 0.837079 \text{FPC} = \sqrt{\frac{N - n}{N - 1}} = \sqrt{\frac{1000 - 300}{1000 - 1}} = \sqrt{\frac{700}{999}} \approx \mathbf{0.837079} FPC = N − 1 N − n = 1000 − 1 1000 − 300 = 999 700 ≈ 0.837079
Calculate Unadjusted and Adjusted Standard Error:
S E = p ^ q ^ n = 0.60 × 0.40 300 = 0.0008 ≈ 0.028284 SE = \sqrt{\frac{\hat{p}\hat{q}}{n}} = \sqrt{\frac{0.60 \times 0.40}{300}} = \sqrt{0.0008} \approx \mathbf{0.028284} S E = n p ^ q ^ = 300 0.60 × 0.40 = 0.0008 ≈ 0.028284
S E adj = S E × FPC = 0.028284 × 0.837079 ≈ 0.023676 SE_{\text{adj}} = SE \times \text{FPC} = 0.028284 \times 0.837079 \approx \mathbf{0.023676} S E adj = S E × FPC = 0.028284 × 0.837079 ≈ 0.023676
Calculate Margin of Error & Interval:
M E = 1.960 × 0.023676 ≈ 0.0464 ME = 1.960 \times 0.023676 \approx \mathbf{0.0464} M E = 1.960 × 0.023676 ≈ 0.0464
CI = 0.60 ± 0.0464 = [ 0.5536 , 0.6464 ] ( [ 55.36 % , 64.64 % ] ) \text{CI} = 0.60 \pm 0.0464 = \mathbf{[0.5536, \, 0.6464]}\quad (\mathbf{[55.36\%, \, 64.64\%]}) CI = 0.60 ± 0.0464 = [ 0.5536 , 0.6464 ] ( [ 55.36% , 64.64% ] )
5. Universal Execution Checklist for Exams
When solving any estimation problem, write down these 4 numbered steps on your exam script to ensure full method marks:
Step 1: Identify Parameters: List n , x ˉ (or p ^ ) , σ (or s ) , N (if given) n, \bar{x}\text{ (or } \hat{p}), \sigma\text{ (or } s), N\text{ (if given)} n , x ˉ (or p ^ ) , σ (or s ) , N (if given) .
Step 2: Justify Distribution Choice: State "Since n ≥ 30 n \ge 30 n ≥ 30 , we use the Z Z Z -distribution" OR "Since n < 30 n < 30 n < 30 and σ \sigma σ is unknown, we use the t t t -distribution with d f = n − 1 df = n - 1 df = n − 1 " .
Step 3: State & Compute S E SE S E and M E ME M E : Show the standard error formula and multiply by the critical value.
Step 4: State Final Interval with Units: Lower Limit = Estimate − M E \text{Lower Limit} = \text{Estimate} - ME Lower Limit = Estimate − M E , Upper Limit = Estimate + M E \text{Upper Limit} = \text{Estimate} + ME Upper Limit = Estimate + M E .
6. Exam Traps & Pro-Tips
Common Traps to Avoid
Standard Deviation vs. Standard Error: Never use σ \sigma σ or s s s directly as your margin of error. You must always divide by n \sqrt{n} n to obtain the Standard Error (S E SE S E ) before multiplying by the critical value.
Forgetting FPC: If a question explicitly specifies a total population size (e.g., "Assuming a total of 1000 customers..." ), it is an intentional signal to test whether you calculate and apply the FPC factor.
Hypothesis Testing via Confidence Interval (Duality): When an exam question asks whether a claimed value μ 0 \mu_0 μ 0 is supported by the sample data:
If μ 0 \mu_0 μ 0 falls inside the ( 1 − α ) (1 - \alpha) ( 1 − α ) interval ⟹ \implies ⟹ Fail to reject H 0 H_0 H 0 (Claim is plausible).
If μ 0 \mu_0 μ 0 falls outside the ( 1 − α ) (1 - \alpha) ( 1 − α ) interval ⟹ \implies ⟹ Reject H 0 H_0 H 0 (Claim is significantly different).