please help me on this statistics?
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.
- MerlynLv 71 decade agoFavorite Answer
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( ∑X / n)
= 1/n * ∑ E(X)
= 1/n * ∑ μ
= 1/n * nμ
= 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
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.
- 1 decade ago
margin of error=(test statistic)x(standard error)
standard error=standard deviation/sqrt(sample size)
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):
Thus, margin of error=1.99x0.17=0.34, or 0.3.