Question

A forestry researcher wants to estimate the average height of trees in a forest near Atlanta, Georgia. She takes a random sample of 18 trees from this forest. The researcher found that the average height was 4.8 meters with a standard deviation of 0.55 meters. Assume that the distribution of the heights of these trees is normal. For this sample what is the margin of error for her 99% confidence interval

Answer 1

Answer:

The margin of error for her confdence interval is of 0.3757.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used 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 = 18 - 1 = 17

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.8982

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

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

Standard deviation of 0.55 meters.

This means that $s = 0.55$

What is the margin of error for her 99% confidence interval?

$M = T\frac{s}{\sqrt{n}}$

$M = 2.8982\frac{0.55}{\sqrt{18}}$

$M = 0.3757$

The margin of error for her confdence interval is of 0.3757.

Answer 2

Margin of error is the distance between the mean and the limit of confidence intervals. The margin of error for the given condition is 3.28 approximately.

What is the margin of error for small samples?

Suppose that we have:

Sample size n  < 30

• Sample standard deviation = s
• Population standard deviation = $\sigma$
• Level of significance = $\alpha$
• Degree of freedom = n-1

Then the margin of error(MOE) is obtained as

• Case 1: Population standard deviation is known

Margin of Error = $MOE = T_{c}\dfrac{\sigma}{\sqrt{n}}$

• Case 2: Population standard deviation is unknown

$MOE = T_{c}\dfrac{s}{\sqrt{n}}$

where $T_{c}$ is critical value of the test statistic at level of significance

For the given case, taking the random variable X to be tracking the height of trees in the sample taken of trees from the considered forest.

Then, by the given data, we get:

$\overline{x} = 4.8$, $s = 4.8$, n = 18

The degree of freedom is n-1 = 17

Level of significance = 100% - 99% = 1% = 0.01

The critical value of t at level of significance 0.01 with degree of freedom 17 is obtained as T = 2.90 (from the t critical values table)

Thus, margin of error for 99% confidence interval for considered case is:

$MOE = T_{c}\dfrac{s}{\sqrt{n}}\\\\MOE = 2.9 \times \dfrac{4.8}{\sqrt{18}} \approx 3.28$

Thus, the margin of error for the given condition is 3.28 approximately.

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