Question

Let Y1​,Y2​,…,Yn​ denote a random sample of size n from a normal distribution with mean μ and variance σ2⋅[Yˉ−z2a​​n​s​,Yˉ+z2a​​n​s​] is the 100(1−α)% confidence interval for μ when σ is known σ is unknown and n is large σ is unknown and n is small σ is known and n is large

Answer 1

The
`100(1−α)%` confidence interval for μ when σ is unknown and n is large is
`[Yˉ−zα/2​(s/√n), Yˉ+ zα/2​(s/√n)].`

Step-by-step explanation:

When dealing with the confidence interval for the population mean μ, the formula varies based on whether the population standard deviation σ is known or unknown and if the sample size (n) is large or small. In this scenario where σ is unknown and n is large, we use the t-distribution due to the sample's size. The formula for the confidence interval in this case is:
`[Yˉ−tα/2​(s/√n), Yˉ+ tα/2​(s/√n)]`, which estimates μ.

However, as n becomes large, the t-distribution approaches the standard normal distribution. Therefore, in the limit of a large sample size, the t-distribution converges to the standard normal distribution, and the confidence interval formula becomes [
`Yˉ−zα/2​(s/√n), Yˉ+ zα/2​(s/√n)]`, where z represents the critical value for a standard normal distribution corresponding to the desired level of confidence.

For this particular scenario, where σ is unknown and n is large, the formula
`[Yˉ−zα/2​(s/√n), Yˉ+ zα/2​(s/√n)]`represents the
`100(1−α)%`confidence interval for estimating μ. Here, Yˉ denotes the sample mean, s represents the sample standard deviation, n is the sample size, and cis the critical value obtained from the standard normal distribution corresponding to the level of significance
`α/2` for a two-tailed test. This interval provides an estimation of the true population mean μ with a specified level of confidence based on the sample statistics.