99% interval for mean page load time

Question

A monitoring system has established that page load time has a known population standard deviation of $\sigma = 0.5$ seconds. A random sample of $n = 100$ page loads averages $\bar{x} = 3.2$ seconds.

Construct a 99% confidence interval for the true mean load time.

Accepted Answer

Setup. With \sigma known, use \bar{x} \pm z^{}\,\sigma/\sqrt{n}. For 99% confidence the critical value is z^{} = 2.576. Compute. The standard error is \frac{0.5}{\sqrt{100}} = \frac{0.5}{10} = 0.05. Margin of error = 2.576 	imes 0.05 = 0.1288, so 3.2 \pm 0.1288 = [3.071,\ 3.329]. Interpret. Raising confidence from 95% to 99% uses a larger critical value (2.576 instead of 1.96), which widens the interval. More confidence, less precision. Takeaway. The confidence level enters only through z^{}. Everything else being equal, a higher level means a wider interval.

99% interval for mean page load time

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