Question

Suppose the germination periods, in days, for grass seed are normally distributed. If the population standard deviation is 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer.

Answer 1

The minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean is 25.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.9)/(2) = 0.05`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.05 = 0.95`, so
`z = 1.645`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

If the population standard deviation is 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean?

This is n when
`\sigma = 3, M = 1`. So

`M = z*(\sigma)/(√(n))`

`1 = 1.645*(3)/(√(n))`

`√(n) = 3*1.645`

`(√(n))^(2) = (3*1.645)^(2)`

`n = 24.3`

Rouding up to the nearest integer, 25.

The minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean is 25.