Question

A research firm conducted a survey to determine the mean amount Americans spend on coffee during a week. They found the distribution of weekly spending followed the normal distribution with a population standard deviation of $5. A sample of 49 Americans revealed that x = $20.

What is the point estimate of the population mean?

Answer 1

By the Central Limit Theorem, the point estimate of the population mean is $20.

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
`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 problem, we have that:

The mean of the sample is $20.

So, by the Central Limit Theorem, the point estimate of the population mean is $20.

Answer 2

`X \sim N(\mu,5)`

Where
`\mu` and
`\sigma=5`

For this case we select a sample of n =49 observations and we got a sample mean os :

`\bar X= 20`

From the definition of mean we have that :

`\bar X = (\sum_(i=1)^n X_i)/(n)`

And if we find the expected value of this estimator we got this:

`E(\bar X) = E( (\sum_(i=1)^n X_i)/(n)) = (1)/(n) n \mu = \mu`

So then the best estimator unbiased for the population mean is the sample mean:

`\hat \mu = \bar X = 20`

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".

Solution to the problem

Let X the random variable that represent the weekly spending of a population, and for this case we know the distribution for X is given by:

`X \sim N(\mu,5)`

Where
`\mu` and
`\sigma=5`

For this case we select a sample of n =49 observations and we got a sample mean os :

`\bar X= 20`

From the definition of mean we have that :

`\bar X = (\sum_(i=1)^n X_i)/(n)`

And if we find the expected value of this estimator we got this:

`E(\bar X) = E( (\sum_(i=1)^n X_i)/(n)) = (1)/(n) n \mu = \mu`

So then the best estimator unbiased for the population mean is the sample mean:

`\hat \mu = \bar X = 20`