1. **Problem 1: Variance of sample distributions** Given population $S = \{1, 2, 3, 4, 5\}$, population size $N=5$, sample size $n=2$. **Sampling with replacement:** - Each sample of size 2 is drawn with replacement. - Population mean $\mu = \frac{1+2+3+4+5}{5} = 3$. - Population variance $\sigma^2 = \frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5} = \frac{4+1+0+1+4}{5} = 2$. - Variance of sample mean with replacement: $Var(\bar{X}) = \frac{\sigma^2}{n} = \frac{2}{2} = 1$. **Sampling without replacement:** - Variance of sample mean without replacement is $Var(\bar{X}) = \frac{\sigma^2}{n} \times \frac{N-n}{N-1} = 1 \times \frac{5-2}{5-1} = 1 \times \frac{3}{4} = 0.75$. **Interpretation:** - Sampling without replacement reduces variance because samples are more representative. - Sampling with replacement has higher variance due to possible repeated elements. 2. **Problem 2a: Hypothesis testing for spot remover** - Null hypothesis $H_0: p=0.7$, alternative $H_a: p>0.7$. - Sample size $n=10$. - Reject $H_0$ if number of spots removed $\geq 6$. I. **Type I error probability ($\alpha$):** - Type I error is rejecting $H_0$ when $p=0.7$. - $\alpha = P(X \geq 6 | p=0.7) = 1 - P(X \leq 5)$ where $X \sim Binomial(10,0.7)$. - Calculate $P(X \leq 5) = \sum_{k=0}^5 \binom{10}{k} 0.7^k 0.3^{10-k}$. - Using binomial CDF, $P(X \leq 5) \approx 0.1503$. - So, $\alpha = 1 - 0.1503 = 0.8497$. II. **Power of the test when $p=0.5$:** - Power = $P(\text{reject } H_0 | p=0.5) = P(X \geq 6 | p=0.5)$. - Calculate $P(X \geq 6) = 1 - P(X \leq 5)$ for $X \sim Binomial(10,0.5)$. - $P(X \leq 5) \approx 0.6230$. - Power $= 1 - 0.6230 = 0.3770$. 3. **Problem 2b: Testing newborn babies weights** - Data: $4.0, 3.7, 2.5, 4.1, 6.7, 6.0, 2.8, 5.0, 2.35$. - Sample size $n=9$. **Step 1: Calculate sample mean $\bar{x}$** $$\bar{x} = \frac{4.0 + 3.7 + 2.5 + 4.1 + 6.7 + 6.0 + 2.8 + 5.0 + 2.35}{9} = \frac{37.15}{9} \approx 4.128$$ **Step 2: Calculate sample standard deviation $s$** $$s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$$ Calculate squared deviations: $(4.0-4.128)^2=0.0164$, $(3.7-4.128)^2=0.183$, $(2.5-4.128)^2=2.646$, $(4.1-4.128)^2=0.0008$, $(6.7-4.128)^2=6.635$, $(6.0-4.128)^2=3.504$, $(2.8-4.128)^2=1.768$, $(5.0-4.128)^2=0.761$, $(2.35-4.128)^2=3.168$. Sum = 18.682 $$s = \sqrt{\frac{18.682}{8}} = \sqrt{2.335} \approx 1.528$$ **Step 3: Hypothesis test for equal weights** - Null hypothesis: weights are the same (mean equals some value, e.g., population mean or test mean). - Without population mean given, test if mean differs significantly from a hypothesized mean (e.g., 4). - Test statistic: $$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$$ - For $\mu_0=4$, $$t = \frac{4.128 - 4}{1.528/\sqrt{9}} = \frac{0.128}{0.509} = 0.251$$ - Degrees of freedom $df=8$. - Critical t-values for two-tailed test at 5%: $\pm 2.306$, at 1%: $\pm 3.355$. - Since $|t|=0.251 < 2.306$, fail to reject null at 5% and 1% significance. **Step 4: Confidence intervals for mean** - 95% CI: $$\bar{x} \pm t_{0.025,8} \times \frac{s}{\sqrt{n}} = 4.128 \pm 2.306 \times 0.509 = (2.91, 5.35)$$ - 99% CI: $$4.128 \pm 3.355 \times 0.509 = (2.43, 5.83)$$ **Interpretation:** - The test shows no significant difference from hypothesized mean 4. - Confidence intervals provide range of plausible means at 95% and 99% levels.