Question

wi-fi access a survey of 53 students in grades 4 through 12 found 64% have classroom wi-fi access. find the 99% confidence interval of the population proportion. round all intermediate and final answers to at least three decimal places. \\(\square < p < \square\\)

Explanation:

Step1: Identify key values

Sample size $n=53$, sample proportion $\hat{p}=0.64$, confidence level 99%.

Step2: Find z-score

For 99% confidence, $z_{\alpha/2}=2.576$.

Step3: Calculate standard error

$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.64\times0.36}{53}}\approx\sqrt{0.00434}=0.0659$.

Step4: Compute margin of error

$ME=z_{\alpha/2}\times SE=2.576\times0.0659\approx0.170$.

Step5: Find confidence interval

Lower bound: $0.64 - 0.170=0.470$; Upper bound: $0.64 + 0.170=0.810$.

Answer:

0.470 < p < 0.810