Question

1messageJosh wants to estimate the mean age in a population of trees. He'll sample n trees and build a 95% confidence interval for the mean age. He doesn't want the margin of error to exceed 3 years. Preliminary data suggests that the standard deviation for the ages of trees in this population is 16 years.Which of these is the smallest approximate sample size required to obtain the desired margin of error?11 trees110 trees144 trees43 trees87 trees

Answer 1

The formula for the margin of error is

`\begin{gathered} \text{MOE}=SE* critical\text{ value} \\ SE\Rightarrow\text{standard error} \end{gathered}`

The critical value for 95% confidence interval is 1.96

The standard error is

`\begin{gathered} SE=\frac{\sigma}{\sqrt[]{n}} \\ \sigma\Rightarrow standard\text{ deviation} \\ n\Rightarrow\text{sample size} \end{gathered}`

To know which of the list of options will yield approximately the desired margin of error, we will have to calculate the margin of error with the following values

For 110 trees, n= 110, standard dev. =16

`\begin{gathered} SE=\frac{16}{\sqrt[]{110}}=1.53 \\ \text{MOE}=1.53*1.96=2.9988 \end{gathered}`

For 144 trees, n=144, standard dev=16

`\begin{gathered} SE=\frac{16}{\sqrt[]{144}}=(16)/(12)=1.33 \\ \text{MOE}=1.33*1.96=2.61 \end{gathered}`

For 43 trees, n=43 , standard dev.=16

`\begin{gathered} SE=\frac{16}{\sqrt[]{43}}=2.44 \\ \text{MOE}=2.44*1.96=4.782 \end{gathered}`

For 87 trees, n= 87, standard dev=16

`\begin{gathered} SE=\frac{16}{\sqrt[]{87}}=1.72 \\ \text{MOE}=1.72*1.96=3.3712 \end{gathered}`

Hence, from the above calculations, the most prefered is the sample size of 110 trees (because the margin of error of 2.9988 is approximately 3 )

Answer is 110 trees