Find a \( 95 \% \) confidence interval for the proportion of graduating students who took Chemistry if \( 35 \% \) of a simple random sample of 870 graduates took Chemistry. \( (0.02927,0.05119) \) \( (0.02717,0.05329) \) \( (0.31887,0.38228) \) \( (0.32397,0.37718) \)
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To find the \( 95\% \) confidence interval for the proportion of graduating students who took Chemistry, we can use the formula: \[ \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] where: - \(\hat{p} = 0.35\) (the sample proportion), - \(n = 870\) (the sample size), - \(z\) is the z-score corresponding to the desired confidence level. For \(95\%\), \(z \approx 1.96\). Calculating the standard error: \[ \text{Standard Error} = \sqrt{\frac{0.35 \times (1 - 0.35)}{870}} \approx \sqrt{\frac{0.35 \times 0.65}{870}} \approx \sqrt{0.000263} \approx 0.01624 \] Now calculate the margin of error: \[ \text{Margin of Error} = z \times \text{Standard Error} \approx 1.96 \times 0.01624 \approx 0.03181 \] Now, create the confidence interval: \[ \hat{p} \pm \text{Margin of Error} \Rightarrow 0.35 \pm 0.03181 \] Thus, the confidence interval is: \[ (0.35 - 0.03181, 0.35 + 0.03181) \Rightarrow (0.31819, 0.38181) \] Looking at the options you provided, the one that comes closest to this interval is: \( (0.31887, 0.38228) \). So the answer is: \( (0.31887, 0.38228) \).
