Probability and Statistics?

Question

Studies indicate that the biomass for tropical woodlands, thought to be about 35kg per kg/m^2 may in fact be too high and that tropical biomass values vary regionally from about 5-55kg/m^2. Suppose you measure the tropical biomass in 400 randomly selected square meter plots.

A. Approximate the standard deviation of the biomass measurements.

B. What is the probablility that your sample average is within 2 units of the true average tropical biomass?

C. If your sample average is equal to 31.75, what would you conclude about the overestimation that concerns the scientists?

The answers are:

A. Approx. 12.5
B. .9986
C. They are probably correct.

How does one obtain those answers? Can someone explain it to me please?

Thank you,

Sue

Answer 1

A. Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55. So we can approximate this standard deviation by saying

(55-5)/4 = 12.5

B. Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625. Since the sample size is large, then the sample mean is approximately normal. To be within 2 units, then you would have to be within mu-2 and mu+2. So we can standardize.

z = (mu-2-mu)/0.625 = -3.2
z = mu+2-mu)/0.625 = 3.2

The area between these two values would then be approximately 0.9986.

C. In this case, our hypothesis test would look like this.

H0: mu = 35
HA: mu < 35

Our test statistic would be (31.75-35)/0.625 = -5.2

We would then reject the null hypothesis. So we would conclude that the scientists are correct.

Answer 2

A. You can't really answer this. Apparently you are supposed to assume that the 'about 5 - 55 kg/m^2' means that the distance from the midrange (30) to 55 is twice the standard deviation. This is not a valid assumption without knowing something about the shape of the distribution.

B. If the standard deviation is 12.5 as they assert in part a, then the standard error of the sample mean is 12.5 / √400, which is about 0.625. Two units from the mean is a z-score of +/- (2 / 0.625) or +/- 3.2. From the z-table or your calculator this corresponds to and area of 0.9986. (From a table you would have to subtract the p-value for -3.2 from the value for 3.2. Thi is 0.999313 - 0.0006872 = 0.9986.

C. If a random sample of 400 plots gives a mean measurement of 31.75, the plausible values for the population mean are within two units of this. 35 is outside this range.

I hope this helps.