Question

A bottling plant needs to know how to report its success in filling 2-liter (2 L) bottles. From previous work, they know that the population standard deviation is ±0.05 L.

a. For this part, assume an “infinite” population. They sample 35 bottles and obtain a sample mean of 1.99 L. For 95% confidence, what is the estimate of the range of the population mean?
b. For this part, assume an “infinite” population. If they desire an estimate of the interval for the population mean that is ±0.01 L, how many samples do they need to examine (with 95% confidence).

Answer 1

a)

sample mean, xbar = 1.99
sample standard deviation, σ = 0.05
sample size, n = 35

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 0.05/sqrt(35)
ME = 0.017

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (1.99 - 1.96 * 0.05/sqrt(35) , 1.99 + 1.96 * 0.05/sqrt(35))
CI = (1.9734 , 2.0066)

b)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 0.01, σ = 0.05

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 0.05/0.01)^2
n = 96.04

Therefore, the sample size needed to satisfy the condition n >= 96.04 and it must be an integer number, we conclude that the minimum required sample size is n = 97
Ans : Sample size, n = 97