1. **Problem 1: Testing if proportions of shoppers preferring stores A, B, C, D, and E are the same** Given data: Total shoppers = 1100 Store counts: A=262, B=234, C=204, D=190, E=210 1. a. Null hypothesis $H_0$: The proportions of shoppers preferring each store are equal, i.e., $p_A = p_B = p_C = p_D = p_E = \frac{1}{5}$. b. Alternative hypothesis $H_a$: At least one store's proportion differs from the others. c. Significance level $\alpha = 0.05$. d. Degrees of freedom $df = k - 1 = 5 - 1 = 4$ where $k$ is the number of categories. e. Test statistic: Use the chi-square goodness-of-fit test. Expected frequency for each store $E = \frac{1100}{5} = 220$. Calculate $$\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(262-220)^2}{220} + \frac{(234-220)^2}{220} + \frac{(204-220)^2}{220} + \frac{(190-220)^2}{220} + \frac{(210-220)^2}{220}$$ Calculate each term: $$\frac{(42)^2}{220} = \frac{1764}{220} = 8.018$$ $$\frac{(14)^2}{220} = \frac{196}{220} = 0.891$$ $$\frac{(-16)^2}{220} = \frac{256}{220} = 1.164$$ $$\frac{(-30)^2}{220} = \frac{900}{220} = 4.091$$ $$\frac{(-10)^2}{220} = \frac{100}{220} = 0.455$$ Sum: $8.018 + 0.891 + 1.164 + 4.091 + 0.455 = 14.619$ f. Critical value from chi-square table at $df=4$ and $\alpha=0.05$ is approximately $9.488$. g. Decision: Since $14.619 > 9.488$, reject $H_0$. There is enough evidence to conclude the proportions are not all equal. 2. **Problem 2: Testing if a 6-sided die is unbiased based on 240 rolls** Given data: Total rolls = 240 Frequencies: 1=34, 2=44, 3=30, 4=46, 5=51, 6=35 1. a. Null hypothesis $H_0$: The die is unbiased, so each face has probability $\frac{1}{6}$. b. Alternative hypothesis $H_a$: The die is biased, i.e., probabilities are not all equal. c. Significance level $\alpha = 0.05$. d. Degrees of freedom $df = 6 - 1 = 5$. e. Test statistic: Expected frequency for each face $E = \frac{240}{6} = 40$. Calculate $$\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(34-40)^2}{40} + \frac{(44-40)^2}{40} + \frac{(30-40)^2}{40} + \frac{(46-40)^2}{40} + \frac{(51-40)^2}{40} + \frac{(35-40)^2}{40}$$ Calculate each term: $$\frac{(-6)^2}{40} = \frac{36}{40} = 0.9$$ $$\frac{(4)^2}{40} = \frac{16}{40} = 0.4$$ $$\frac{(-10)^2}{40} = \frac{100}{40} = 2.5$$ $$\frac{(6)^2}{40} = \frac{36}{40} = 0.9$$ $$\frac{(11)^2}{40} = \frac{121}{40} = 3.025$$ $$\frac{(-5)^2}{40} = \frac{25}{40} = 0.625$$ Sum: $0.9 + 0.4 + 2.5 + 0.9 + 3.025 + 0.625 = 8.35$ f. Critical value from chi-square table at $df=5$ and $\alpha=0.05$ is approximately $11.070$. g. Decision: Since $8.35 < 11.070$, fail to reject $H_0$. There is not enough evidence to conclude the die is biased. **Final answers:** Problem 1: Reject $H_0$, proportions differ. Problem 2: Fail to reject $H_0$, no evidence of bias.