Question

How to solve the best point estimate for the population mean and calculate the margin of error? Wherein: A random sample of n= 75 observations from a quantitative population produced a mean of 29.7 and s= 3.286

Answer 1

The margin of error is `~~0.7560`.

Explanation:

The best point estimate for a population mean `mu` is the sample mean `barx`. In this case, we'd have the point estimate

`hatmu = barx=29.7`

The margin of error is a maximum value of how far `hatmu` will be from `mu,` based on a confidence level `alpha.` For example, if `alpha = 0.05,` then there is a 95% chance our `hatmu` will be within the margin of error of the actual population `mu.`

The formula for a margin of error (M.E.) for a sample mean is:

`"ME"=t_(alpha//2, n-1)xxs/sqrtn`

or, if `n` is large enough:

`"ME"=z_(alpha//2)xxs/sqrtn`

(This is because, as `nrarroo`, the `t` distribution with `n-1` degrees of freedom approaches the standard normal distribution `Z`.)

Using the first option with `alpha = 0.05`, we get:

`"ME"=t_(0.05//2,"  " 75-1)xx3.286/sqrt75`

`color(white)"ME"~~t_(0.025,74)xx3.286/(8.6603)`

`color(white)"ME"~~1.9925xx0.3794`

`color(white)"ME"~~0.7560`

Using the second option (again, with `alpha=0.05`), we get:

`"ME"=z_(0.05//2)xx3.286/sqrt75`

`color(white)"ME"~~z_(0.025)xx3.286/(8.6603)`

`color(white)"ME"~~1.9600xx0.3794`

`color(white)"ME"~~0.7437`

As you can see, both methods give almost the same value (0.7560 and 0.7437 are about 0.013 apart). This is why we often just use the second formula, since it's easier to find values for `z_(alpha//2).` However, the first option is more accurate, since the distribution of `barX` is closer to `t` than to `Z,` and it will always give a wider margin of error, and thus it's a bit safer.