Question

Construct a​ 95% confidence interval for the population standard deviation sigma of a random sample of 15 crates which have a mean weight of 165.2 pounds and a standard deviation of 12.4 pounds. Round to the nearest thousandth. Assume the population is normally distributed.

Answer 1

Answer:

There is a 95% confidence that the sample has a mean between 158.92 pounds and 171.48 pounds

Step-by-step explanation:

Given that mean (μ) = 165.2 pounds, standard deviation (σ) = 12.4 pounds, sample size (n) = 15 crates. Confidence (C) = 95% = 0.95

α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

The z score of α/2 corresponds to the z score of 0.475 (0.5 - 0.025) which is 1.96. $z_{\frac{\alpha}{2} }=1.96$

The margin of error (E) is given by:

$E=z_{\frac{\alpha}{2} }\frac{\sigma}{\sqrt{n} } =1.96*\frac{12.4}{\sqrt{15} }= 6.28$

The confidence interval = μ ± E = 165.2 ± 6.28 = (158.92, 171.48)

The confidence interval is between 158.92 pounds and 171.48 pounds. There is a 95% confidence that the sample has a mean between 158.92 pounds and 171.48 pounds