Question

1. Studies indicate that the earth's vegetative mass, or biomass for tropical woodlands, thought to be about 35 kilograms per square meter (kg/m2), may in fact be too high and that tropical biomass values vary regionally—from about 5 to 55 kg/m2. Suppose you measure the tropical biomass in 400 randomly selected square-meter plots.
(a) Approximate σ, the standard deviation of the biomass measurements (in kg/m2).
σ = kg/m2

(b) What is the probability that your sample average is within two units of the true average tropical biomass? (Use your answer from part (a) in your calculations. Round your answer to four decimal places.)

(c) If your sample average is x = 30.75, what would you conclude about the overestimation that concerns the scientists?

a. Because z is less than 3 standard deviations below the mean, it is not likely that the mean is an overestimate of the mean biomass for tropical woodlands.

b. Because z is more than 3 standard deviations below the mean, it is more likely that the scientists are correct in assuming that the mean is an overestimate of the mean biomass for tropical woodlands.

c. Because z is negative, it is not likely that the mean is an overestimate of the mean biomass for tropical woodlands.

d. Because z is less than 3 standard deviations below the mean, it is more likely that the scientists are correct in assuming that the mean is an overestimate of the mean biomass for tropical woodlands.

e. Because z is more than 3 standard deviations below the mean, it is not likely that the mean is an overestimate of the mean biomass for tropical woodlands.

Answer 1

a.)

Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 Standard Deviations of the mean, then that would mean that there are a total of 4 Standard Deviationsbetween 5 and 55.

So we can approximate this standard deviation by saying,

SD = (55-5) / 4 = 12.5

b.)

Sample size, n = 400

So, standard error of mean, SE = SD / SQRT(n) = 12.5/20 = 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 Range Mean - 2 and Mean + 2

So we can standardize.

z = (Mean - x) / SE = (Mean - (Mean - 2)) / 0.625 = 3.2

z = (Mean - x) / SE = (Mean - (Mean + 2)) / 0.625 = -3.2

So, -3.2 < z < 3.2

From normal distrobution graph and z distribution calculator Area between these two values is 0.9986

So, Probability = 0.9986

c.)

Given sample average, x = 30.75

The Hypotheses becomes,

Null, Ho: Mean = 35

Alternate, Ha: Mean < 35

Test statistic, z = (30.75 - 35) / 0.625 = -4.25 / 0.625 = -6.8

Since Z is more than 3 standard deviations below the mean, we would reject the NULL HYPOTHESES.

So, Option B is correct