1. Problem 1: Calculate the sample size for a population of 500 using Krejcie & Morgan formula with confidence level 95% and margin of error 5%. The formula is: $$s=\frac{X^2NP(1-P)}{d^2(N-1)+X^2P(1-P)}$$ Where: $X^2=3.841$ (chi-square value for 95% confidence), $N=500$, $P=0.5$, $d=0.05$ 2. Calculate numerator: $$3.841 \times 500 \times 0.5 \times (1-0.5) = 3.841 \times 500 \times 0.25 = 480.125$$ 3. Calculate denominator: $$0.05^2 \times (500-1) + 3.841 \times 0.5 \times (1-0.5) = 0.0025 \times 499 + 3.841 \times 0.25 = 1.2475 + 0.96025 = 2.20775$$ 4. Calculate sample size $s$: $$s= \frac{480.125}{2.20775} \approx 217.48$$ Rounded to nearest whole number, sample size $s=217$. 5. Problem 2: Population $N=1200$ with same confidence and error level. Calculate numerator: $$3.841 \times 1200 \times 0.5 \times 0.5 = 3.841 \times 1200 \times 0.25 = 1152.3$$ Calculate denominator: $$0.0025 \times (1200-1) + 3.841 \times 0.25 = 0.0025 \times 1199 + 0.96025 = 2.9975 + 0.96025 = 3.95775$$ Sample size: $$s= \frac{1152.3}{3.95775} \approx 291.24$$ Rounded, $s=291$. 6. Problem 3: Population $N=300$, calculate sample sizes using Isaac & Michael formula for different error rates. Isaac & Michael formula: $$n=\frac{N}{1 + N e^2}$$ Using $e$ values: a. $e=0.01 (1\%)$: $$n=\frac{300}{1 + 300(0.01)^2} = \frac{300}{1+300(0.0001)}=\frac{300}{1+0.03}=\frac{300}{1.03}\approx 291.26$$ b. $e=0.05 (5\%)$: $$n=\frac{300}{1 + 300(0.05)^2} = \frac{300}{1 + 300(0.0025)}=\frac{300}{1 + 0.75}=\frac{300}{1.75} \approx 171.43$$ c. $e=0.10 (10\%)$: $$n=\frac{300}{1 + 300(0.1)^2} = \frac{300}{1 + 300(0.01)}=\frac{300}{1 + 3} = \frac{300}{4} = 75$$ Check these results against Isaac & Michael’s table for validation. 7. Problem 4: Population $N=850$, margin error $d=0.05$ for Isaac & Michael formula: Calculate sample size: $$n=\frac{850}{1 + 850 (0.05)^2} = \frac{850}{1 + 850 (0.0025)}=\frac{850}{1 + 2.125} = \frac{850}{3.125} \approx 272$$ 8. Problem 5: Population $N=1000$, error $e=0.10$, use Slovin’s formula: Slovin’s formula: $$n=\frac{N}{1 + N e^2}$$ Calculate: $$n=\frac{1000}{1+1000(0.10)^2} = \frac{1000}{1 + 1000(0.01)} = \frac{1000}{1+10} = \frac{1000}{11} \approx 90.91$$ Rounded result $n=91$.