1. **State the Problem:** Test if male athletes in Fiji are taller than non-athletes at 5% significance using P-value. 2. **Given Data:** Athletes: $n_1=20$, $\bar{x}_1=68.2$, $\sigma_1=2.5$ Non-athletes: $n_2=20$, $\bar{x}_2=67.5$, $\sigma_2=2.8$ 3. **Formulate Hypotheses:** $H_0: \mu_1 \leq \mu_2$, $H_a: \mu_1 > \mu_2$ 4. **Calculate Test Statistic:** $$z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} = \frac{68.2 - 67.5}{\sqrt{\frac{2.5^2}{20} + \frac{2.8^2}{20}}} = \frac{0.7}{\sqrt{0.3125 + 0.392}} = \frac{0.7}{\sqrt{0.7045}} = \frac{0.7}{0.8393} \approx 0.8337$$ 5. **Find P-value:** For $z=0.8337$, from standard normal tables, $p = 1 - 0.7977 = 0.2023$. 6. **Conclusion:** Since $p=0.2023 > 0.05$, we fail to reject $H_0$. There is insufficient evidence that athletes are taller than non-athletes. --- 1. **State the Problem:** Check difference in cholesterol levels between two age groups at $\alpha=0.01$ using traditional and confidence interval methods. 2. **Given Data:** Group 1: $n_1=30$, $\bar{x}_1=223$, $s_1=6.1$ Group 2: $n_2=25$, $\bar{x}_2=229$, $s_2=5.8$ 3. **Hypotheses:** $H_0: \mu_1 = \mu_2$, $H_a: \mu_1 \neq \mu_2$ 4. **Calculate Test Statistic:** Degrees of freedom using Welch's formula: $$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{\left(\frac{6.1^2}{30} + \frac{5.8^2}{25}\right)^2}{\frac{\left(\frac{37.21}{30}\right)^2}{29} + \frac{\left(\frac{33.64}{25}\right)^2}{24}}$$ Calculate numerator and denominator: $$\frac{s_1^2}{n_1} = \frac{37.21}{30} = 1.2403, \quad \frac{s_2^2}{n_2} = \frac{33.64}{25} = 1.3456$$ $$df = \frac{(1.2403 + 1.3456)^2}{\frac{(1.2403)^2}{29} + \frac{(1.3456)^2}{24}} = \frac{(2.5859)^2}{\frac{1.538}{29} + \frac{1.811}{24}} = \frac{6.688}{0.053 + 0.075} = \frac{6.688}{0.128} \approx 52.25$$ 5. **Test statistic:** $$t = \frac{223 - 229}{\sqrt{\frac{6.1^2}{30} + \frac{5.8^2}{25}}} = \frac{-6}{\sqrt{1.2403 + 1.3456}} = \frac{-6}{\sqrt{2.5859}} = \frac{-6}{1.608} \approx -3.73$$ 6. **Critical t-values for two-tailed $\alpha=0.01$ and $df=52$:** $t_{crit} = \pm 2.68$ 7. **Decision:** Since $|t|=3.73 > 2.68$, reject $H_0$. Significant difference exists. 8. **Confidence Interval (CI):** $$CI = (\bar{x}_1 - \bar{x}_2) \pm t_{crit} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = -6 \pm 2.68 \times 1.608 = -6 \pm 4.31$$ 9. **CI Interval:** $$(-10.31, -1.69)$$ Since CI does not contain 0, it supports rejecting $H_0$. --- 1. **State the Problem:** Test if there is a salary difference between chemists in Washington State and New Mexico at $\alpha=0.02$. 2. **Given Data:** Washington: $n_1=35$, $\bar{x}_1=39420$, $s_1=1659$ New Mexico: $n_2=40$, $\bar{x}_2=30215$, $s_2=4116$ 3. **Hypotheses:** $H_0: \mu_1 = \mu_2$, $H_a: \mu_1 \neq \mu_2$ 4. **Calculate degrees of freedom (Welch's):** $$df = \frac{\left(\frac{1659^2}{35} + \frac{4116^2}{40}\right)^2}{\frac{\left(\frac{1659^2}{35}\right)^2}{34} + \frac{\left(\frac{4116^2}{40}\right)^2}{39}}$$ Calculate terms: $$\frac{1659^2}{35} = \frac{2750881}{35} = 78653.74$$ $$\frac{4116^2}{40} = \frac{16941456}{40} = 423536.4$$ $$df = \frac{(78653.74 + 423536.4)^2}{\frac{(78653.74)^2}{34} + \frac{(423536.4)^2}{39}} = \frac{502190.1^2}{\frac{6.185 \times 10^9}{34} + \frac{1.7938 \times 10^{11}}{39}} = \frac{2.5219 \times 10^{11}}{1.819 \times 10^8 + 4.598 \times 10^9} = \frac{2.5219 \times 10^{11}}{4.7809 \times 10^9} \approx 52.73$$ 5. **Calculate test statistic:** $$t = \frac{39420 - 30215}{\sqrt{78653.74 + 423536.4}} = \frac{9205}{\sqrt{502190.1}} = \frac{9205}{708.64} \approx 12.99$$ 6. **Critical t-value for two-tailed $\alpha=0.02$ and $df=53$: ** $t_{crit} \approx \pm 2.40$ 7. **Decision:** Since $12.99 > 2.40$, reject $H_0$. Salaries differ significantly. 8. **Confidence Interval (98%):** $$CI = (39420 - 30215) \pm 2.40 \times 708.64 = 9205 \pm 1700.7$$ $$CI = (7504.3, 10905.7)$$ Since CI does not contain 0, it supports rejecting $H_0$. --- 1. **State the Problem:** Test if families in metropolitan East Coast city have higher income than rural Midwest families at $\alpha=0.05$. 2. **Given Data:** City: $n_1=15$, $\bar{x}_1=62456$, $s_1=9652$ Rural: $n_2=51$, $\bar{x}_2=60213$, $s_2=2009$ 3. **Hypotheses:** $H_0: \mu_1 \leq \mu_2$, $H_a: \mu_1 > \mu_2$ 4. **Calculate degrees of freedom (Welch's):** Calculate terms: $$\frac{9652^2}{15} = \frac{93166504}{15} = 6211100.3$$ $$\frac{2009^2}{51} = \frac{4036081}{51} = 79119.63$$ $$df = \frac{(6211100.3 + 79119.63)^2}{\frac{(6211100.3)^2}{14} + \frac{(79119.63)^2}{50}} = \frac{6280219.93^2}{2.757 \times 10^{12} + 1.251 \times 10^{10}} \approx \frac{3.9445 \times 10^{13}}{2.7695 \times 10^{12}} \approx 14.25$$ 5. **Calculate test statistic:** $$t = \frac{62456 - 60213}{\sqrt{6211100.3 + 79119.63}} = \frac{2243}{\sqrt{6280219.93}} = \frac{2243}{2506.03} \approx 0.895$$ 6. **Critical t-value (one-tailed) for $\alpha=0.05$ and $df=14$: ** $t_{crit} = 1.761$ 7. **Decision:** Since $t=0.895 < 1.761$, fail to reject $H_0$. No sufficient evidence city families have higher income. --- **Final answers:** 1. No evidence athletes are taller ($p=0.2023 > 0.05$). 2A. Significant cholesterol difference ($|t|=3.73 > 2.68$) at 1% level. 2B. 99% CI for difference: $(-10.31, -1.69)$ supports difference. 3A. Significant salary difference ($t=12.99 > 2.40$) at 2% level. 3B. 98% CI: $(7504.3, 10905.7)$ supports difference. 4. No sufficient evidence that city families earn more ($t=0.895 < 1.761$).