Question

Suppose that each of two statisticians, A and B, independently takes a random sample of 20 observations from the normal distribution with unknown mean μ and known variance 4. Suppose also that statistician A finds the sample variance in his random sample to be 3.8, and statistician B finds the sample variance in her random sample to be 9.4. For which random sample is the sample mean likely to be closer to the unknown value of μ?

Short answer

Both are equally close to the unknown value of μ.

Step-by-step solution

Given information

Two statisticians, A and B, independently takes a random sample of 20 observations from the normal distribution with unknown mean μ and known variance 4 and statistician A finds the sample variance in his random sample to be 3.8, and statistician B finds the sample variance in her random sample to be 9.4. Need to check which random sample is the sample mean likely to be closer to the unknown value of μ.

Checking of which random sample is the sample mean likely to be closer to the unknown value of μ.

Theorem: Independence of Sample mean and sample variance:

Let \({X_1},...,{X_n}\) be a random sample from the normal distribution with mean μ and variance \({\sigma ^2}\). Then the sample mean \({\bar X_n} = \frac{1}{n}\sum\limits_{i = 1}^n {{X_i}} \) and sample variance \({\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - {{\bar X}_n}} \right)}^2}} \) are independent random variables.\(\bar X\) has the normal distribution with mean \(\mu \) and sample variance \(\frac{{{\sigma ^2}}}{n}\)

The theorem indicates the sample mean and sample variance are independent here .So, irrespective of the value of the sample variance the result is the same.