1. **Problem statement:** Given sales data of stores in June 2022 with sales values and number of stores, assume sales follow an approximately normal distribution. We need to calculate sample mean, sample standard deviation, proportions, confidence intervals, hypothesis tests, and sample size. 2. **Data:** Sales values $x_i$: 11, 11.5, 12, 12.5, 13, 13.5 (in million VND) Number of stores $f_i$: 10, 15, 20, 30, 15, 10 Total stores $n = \sum f_i = 10+15+20+30+15+10 = 100$ 3. **Step 1: Calculate sample mean $\bar{x}$ and sample standard deviation $s$** - Sample mean formula: $$\bar{x} = \frac{\sum f_i x_i}{n}$$ - Calculate $\sum f_i x_i = 10\times11 + 15\times11.5 + 20\times12 + 30\times12.5 + 15\times13 + 10\times13.5 = 110 + 172.5 + 240 + 375 + 195 + 135 = 1227.5$$ - So, $$\bar{x} = \frac{1227.5}{100} = 12.275$$ million VND - Sample variance formula: $$s^2 = \frac{\sum f_i (x_i - \bar{x})^2}{n-1}$$ - Calculate each squared deviation: - $(11 - 12.275)^2 = 1.625625$ - $(11.5 - 12.275)^2 = 0.600625$ - $(12 - 12.275)^2 = 0.075625$ - $(12.5 - 12.275)^2 = 0.050625$ - $(13 - 12.275)^2 = 0.525625$ - $(13.5 - 12.275)^2 = 1.496625$ - Weighted sum: $$\sum f_i (x_i - \bar{x})^2 = 10\times1.625625 + 15\times0.600625 + 20\times0.075625 + 30\times0.050625 + 15\times0.525625 + 10\times1.496625 = 16.25625 + 9.009375 + 1.5125 + 1.51875 + 7.884375 + 14.96625 = 50.1475$$ - Sample variance: $$s^2 = \frac{50.1475}{99} \approx 0.50654$$ - Sample standard deviation: $$s = \sqrt{0.50654} \approx 0.7117$$ million VND 4. **Step 2: Calculate proportion of stores with sales above 12 million** - Stores with sales > 12 million: sales values 12.5, 13, 13.5 with frequencies 30, 15, 10 - Total stores above 12 million: $30 + 15 + 10 = 55$ - Proportion: $$\hat{p} = \frac{55}{100} = 0.55$$ 5. **Step 3: Find symmetric 95% confidence interval for proportion $p$** - Confidence level 95%, $z_{\alpha/2} = 1.96$ - Confidence interval formula: $$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ - Calculate margin of error: $$1.96 \times \sqrt{\frac{0.55 \times 0.45}{100}} = 1.96 \times 0.0497 = 0.0975$$ - Interval: $$0.55 \pm 0.0975 = (0.4525, 0.6475)$$ 6. **Step 4: Sample size needed for estimating proportion with accuracy 0.08 and 95% confidence** - Formula: $$n = \frac{z_{\alpha/2}^2 \hat{p}(1-\hat{p})}{E^2}$$ where $E=0.08$ - Calculate: $$n = \frac{1.96^2 \times 0.55 \times 0.45}{0.08^2} = \frac{3.8416 \times 0.2475}{0.0064} = \frac{0.9504}{0.0064} = 148.5$$ - Additional stores needed: $$148.5 - 100 = 48.5 \approx 49$$ stores 7. **Step 5: 95% confidence upper bound for mean sales** - Use $t$-distribution with $n-1=99$ degrees of freedom, $t_{0.05,99} \approx 1.66$ - Confidence interval for mean: $$\bar{x} \pm t_{\alpha} \frac{s}{\sqrt{n}}$$ - Calculate margin: $$1.66 \times \frac{0.7117}{10} = 1.66 \times 0.07117 = 0.1181$$ - Upper bound: $$12.275 + 0.1181 = 12.3931$$ million VND 8. **Step 6: Hypothesis test for mean sales > 12.13 million** - a) Hypotheses: - $H_0: \mu \leq 12.13$ - $H_1: \mu > 12.13$ - b) Test statistic: $$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{12.275 - 12.13}{0.7117/10} = \frac{0.145}{0.07117} = 2.037$$ - c) Critical value at 5% significance, one-tailed, $t_{0.05,99} = 1.66$ - Since $2.037 > 1.66$, reject $H_0$; accept the claim mean sales > 12.13 million 9. **Step 7: Hypothesis test for variance < 0.53** - a) Hypotheses: - $H_0: \sigma^2 \geq 0.53$ - $H_1: \sigma^2 < 0.53$ - b) Test statistic: $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{99 \times 0.50654}{0.53} = 94.56$$ - Critical value for left tail at 5% significance with 99 df: $$\chi^2_{0.05,99} = 73.361$$ - Rejection region: $$\chi^2 < 73.361$$ - Since $94.56 > 73.361$, do not reject $H_0$; no evidence variance is less than 0.53 **Final answers:** - Sample mean $\bar{x} = 12.275$ - Sample standard deviation $s = 0.7117$ - Proportion above 12 million $\hat{p} = 0.55$ - 95% CI for proportion: $(0.4525, 0.6475)$ - Additional stores needed: 49 - 95% upper bound mean sales: 12.3931 - Hypothesis test mean: reject $H_0$, accept mean > 12.13 - Hypothesis test variance: do not reject $H_0$