A random sample of 4100 person from a Large population 120 are female. Can you say pernale and fepnale are in ratio \( 5: 3 \) in the population. Using \( 1 \% \) leuel of significan

We want to test whether the ratio of males to females in the population is 5:3. In a 5:3 ratio, the proportion of females is   3/(5 + 3) = 3/8 = 0.375  (37.5%). The null hypothesis (H₀) is that the population proportion of females (p) is 0.375. The alternative hypothesis (H₁) is that it is not 0.375. Given:  • Sample size, n = 4100  • Number of females observed = 120  • Sample proportion, p̂ = 120/4100 ≈ 0.0293 (about 2.93%) Step 1. Compute the standard error (SE) under H₀:   SE = √[p₀(1 - p₀) / n] where p₀ = 0.375. Thus:   SE = √[0.375 × 0.625 / 4100]    = √[0.234375 / 4100]    ≈ √(0.0000572)    ≈ 0.007563 Step 2. Compute the z-test statistic:   z = (p̂ - p₀) / SE   = (0.0293 - 0.375) / 0.007563   = (-0.3457) / 0.007563   ≈ -45.74 Step 3. Determine the critical z-values for a 1% level of significance.  For a two-tailed test at the 1% level, the critical values are approximately ±2.5758.  Because our computed z-value of -45.74 is far below -2.5758, it falls in the rejection region. Step 4. Conclusion: Since |z| = 45.74 is extremely large compared to the critical value and the p-value is nearly 0, we reject the null hypothesis at the 1% significance level. This means the sample provides overwhelming evidence that the true proportion of females is not 37.5% as would be expected if the male and female counts were in the ratio 5:3. Final Answer: No, based on the sample data (with only 120 females out of 4100 persons), the evidence strongly rejects the claim that males and females are in a 5:3 ratio in the population.