A confidence interval for the population mean is x̄ ± z · σ/√n, with z = 1.96 for 95% and z = 2.575 for 99%. Its width is 2z · σ/√n, so it gets narrower when the sample is larger and wider when the confidence level is higher.
(x̄ − 1.96σ/√n, x̄ + 1.96σ/√n)(x̄ − 2.575σ/√n, x̄ + 2.575σ/√n)2z · σ/√nσ by s when n ≥ 30n ≥ 30), by the Central Limit Theorem.n ≥ 30), using the sample standard deviation s in place of σ.A 95% confidence interval does not mean the population mean has a 95% chance of lying in this particular interval. It means that if intervals were built from many samples in the same way, about 95% of them would contain the population mean.
A random sample of 64 items from a population with standard deviation 8 has mean 52. Find a 95% and a 99% confidence interval for the population mean.
σ/√n = 8/√64 = 152 ± 1.96 × 1, i.e. (50.04, 53.96)52 ± 2.575 × 1, i.e. (49.425, 54.575)Answer: 95%: (50.04, 53.96); 99%: (49.425, 54.575)
For the same population (σ = 8), find the least sample size so that the width of a 95% confidence interval is at most 2.
2 × 1.96 × 8/√n ≤ 2√n ≥ 15.68n ≥ 245.86Answer: The least sample size is 246.
n ≥ 245.86 means n = 246. Rounding down to 245 gives an interval slightly wider than allowed.x̄ ± z·σ/√n, so its full width is 2z·σ/√n. Questions about "width" need the factor 2.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.