Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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.

thank you!

2 Answers

  • Merlyn
    Lv 7
    1 decade ago
    Favorite 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

    = μ


    = 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.

  • 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):


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

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

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