A large trucking company wants to estimate the proportion of its tracker truck population with refrigerated carrier capacity.A random sample of 200 tracker trucks is taken and 30% of the sample have refrigerated carrier capacity.The 90% confidence interval to estimate the population proportion is _______.
A)0.53 to 0.67
B)0.25 to 0.35
C)0.24 to 0.36
D)0.27 to 0.33
E)0.33 to 0.39

Question

Quizplus · Accepted Answer

The 90% confidence interval for a proportion is calculated using the formula: $$CI = \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p}$ is the sample proportion (0.30 in this case), $n$ is the sample size (200), and $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level (for 90% confidence, $Z_{\alpha/2} \approx 1.645$). Plugging in the values: $$CI = 0.30 \pm 1.645 \sqrt{\frac{0.30(1-0.30)}{200}}$$ This calculation results in a confidence interval of approximately 0.25 to 0.35, which matches option B.

A Large Trucking Company Wants to Estimate the Proportion of Its