Confidence interval for a mean when sigma is known

Question

A bottling machine has a long-run fill variability that is known from years of records: the population standard deviation is $\sigma = 12$ grams. You pull a random sample of $n = 36$ bottles and find a sample mean of $\bar{x} = 505$ grams.

Construct a 95% confidence interval for the true mean fill weight.

Accepted Answer

Setup. Because the population standard deviation \sigma is known and we are estimating a mean, the interval uses a z critical value: \bar{x} \pm z^{}\,\frac{\sigma}{\sqrt{n}}. For 95% confidence, z^{} = 1.96. Compute. The standard error is \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2. The margin of error is 1.96 	imes 2 = 3.92. So the interval is 505 \pm 3.92 = [501.08,\ 508.92]. Interpret. The method that produced this interval captures the true mean fill weight in 95% of repeated samples. It is not a statement that any single bottle weighs in this range; it is about th…

Confidence interval for a mean when sigma is known

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