Question

A sample ofn = 25 scores has a mean of M = 83 and a standard deviation ofs = 15.

Answer 1

Answer:

a) The Sample Standard deviation "measures the spread of a data distribution. It measures the typical distance between each data point and the mean"

b) $SE=\frac{15}{\sqrt{25}}=3$

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the scores. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =83,\sigma =15)$

We take a sample of n=25 nails.  

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Part a

The Sample Standard deviation "measures the spread of a data distribution. It measures the typical distance between each data point and the mean"

Part b

The standard error is given by:

$SE=\frac{\sigma}{\sqrt{n}}$

And replacing we got:

$SE=\frac{15}{\sqrt{25}}=3$