Question

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 33 minutes. From past experience, it is known that the population standard deviation equals 10 minutes.

Compute the standard error of the mean.

Answer 1

We select a sample size of n = 49 >30 large enough to use the central limit theorem.

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

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And on this case the standard error of the mean is given by:

`SE = (\sigma)/(√(n))= (10)/(√(49))= 1.429`

Explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

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".

Solution to the problem

Let X the random variable that represent the amount of time the customers stayed in the resturant of a population, and for this case we know the following info:

Where
`\bar X=33` and
`\sigma=10`

We select a sample size of n = 49 >30 large enough to use the central limit theorem.

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

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And on this case the standard error of the mean is given by:

`SE = (\sigma)/(√(n))= (10)/(√(49))= 1.429`

Answer 2

The standard error of the mean is 1.43.

Explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation(which is also called standard error of the mean)
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this proble, we have that:

`\sigma = 10, n = 49`

So

`s = (10)/(√(49)) = 1.43`

The standard error of the mean is 1.43.