Ecologists wanted to estimate the mean biomass (amount of vegetation) of a certain forested region. The ecologists divided the region into plots measuring 1 square meter each, and they selected a random sample of 9 plots. The mean biomass of the 9 plots was 4.3 kilograms per square meter (kg/m2) and the standard deviation was 1.5 kg/m2 . Assuming all conditions for inference are met, which of the following is a 95 percent confidence interval for the population mean biomass, in kg/m2?
A) 4.3±1.96 (underroot 1.5/3).
B) 4.3±1.96 (1.5/3).
C) 4.3 ± 2.306 (underroot 1.5/9).
D) 4.3±2.3064 (1.5/9).
E) 4.3+2.30/15 (1.5/3).

Answer:

[tex]4.3 \pm 2.306\frac{1.5}{\sqrt{9}}[/tex], option c

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 9 - 1 = 8

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.306

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 2.306\frac{1.5}{\sqrt{9}}[/tex]

In which s is the standard deviation of the sample and n is the size of the sample.

Confidence interval:

The confidence interval is the sample mean plus/minus the margin of error. So

[tex]4.3 \pm 2.306\frac{1.5}{\sqrt{9}}[/tex]

The correct answer is given by option c.

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