Question

At a large corporation, the distribution of years of employment for the employees has mean 20.6 years and standard deviation 5.3 years. A random sample of 100 employees was selected and surveyed about employee satisfaction. The sample of employees had a mean 20.3 years and standard deviation 6.1 years. Remy claims that the mean of the sampling distribution of the sample mean for samples of size 100 is 20.6 years. Is Remy’s claim correct?

Answer 1

Answer:

Yes

Explanation:

The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma}\\\\where\ \mu=mean.\sigma=standard\ deviation, x=raw\ score.\\\\For\ a \ sample\ size\ n:\\\\z=\frac{x-\mu}{\sigma/\sqrt{n} }$

For a normal distribution with a mean of μ and standard deviation σ, For any sample size (n) drawn from this population, the sample mean is normally distributed, with mean $\mu_x=\mu$ and standard deviation $\sigma_x=\frac{\sigma}{\sqrt{n } }$.

Hence given that a mean of μ = 20.6 years and standard deviation σ = 5.3 years, for a sample size of 100:

mean $\mu_x=\mu=20.6\ years$ and standard deviation $\sigma_x=\frac{\sigma}{\sqrt{n } } =\frac{5.3}{\sqrt{100} }=0.53\ years$.