Anonymous

suppose a sample of 75 1-square-meter plots, randomly chosen in north americas boreal forest, produced a mean biomass of 4.2 kilograms per square meter (kg/m2), with a standard deviation of 1.5 kg/m2.

estimate the average biomas for the boreal forest of north america and find the margine of eror for the estimation.

thank you!

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• Merlyn
Lv 7

This is a simple example of propagation of error.

Let X1, X2, ... Xn, be a simple random sample of size n from a population with mean μ and variance σ².

Let Xbar be the mean of the n samples. the mean, or expectation of Xbar is:

E(Xbar)

= E( ∑X / n)

= 1/n * ∑ E(X)

= 1/n * ∑ μ

= 1/n * nμ

= μ

Variance(Xbar)

= Var(∑X / n)

= 1/n^2 Var(∑X)

= 1/n^2 * ∑ Var(X) ... this is only valid because the Xi's are independent

= 1/n^2 * ∑ σ²

= 1/n^2 * nσ²

= σ²/n

so you have:

the expectation of the average of the 75 samples is 4.2 and the variance is (1.5^2) / 75 = 0.03. The standard deviation is the square root of the variance = sqrt(0.03) = 0.1732051

This is known as propagation of error.

margin of error=(test statistic)x(standard error)

standard error=standard deviation/sqrt(sample size)

........................=1.5/sqrt(75)

........................=0.17

In order to choose a test statistic, we need to know the sample distribution. Since the sample size is large (>=30), we can apply the central limit theorem and assume a normal distribution. Now, that narrows things down to either a Z- or t-statistic. Our next question is, whether or not we know the population standard deviation. Since in this case, the standard deviation was calculated from a sample, we can say the population standard deviation is unknown. Thus, we would use the t-statistic with n-1=75-1=74 degrees of freedom.

Take alpha=0.05 (most common):

1-alpha/2=1-0.05/2=0.975

So, t(0.975;d.f.=74)=~1.99

Thus, margin of error=1.99x0.17=0.34, or 0.3.