Question

Forest rangers wanted to better understand the rate of growth for younger trees in the park. They took measurements of a random sample of 50 young trees in 2009 and again measured those same trees in 2019. The data below summarize their measurements, where the heights are in feet:

Answer 1

Answer:

$12.5 - 2.68 \frac{7.2}{\sqrt{50}}=9.771$

$12.5 +2.68 \frac{7.2}{\sqrt{50}}=15.229$

The 99% confidence interval for the difference is given by (9.771;15.229)

And we can conclude that at 99% of confidence the average growth is between 9.771 and 15.229 feet.

Step-by-step explanation:

We assume the following data:

$\bar X_{2009} =12.0 , \bar X_{2019}=24.5$

$s_{2009}= 3.5 , s_{2019}= 9.5$

$n_{2009}= n_{2019}=50$

We also have the information about the differences defined as $d = \bar X_{2019} -\bar X_{2009}$

$\bar d = 12.5$

$s_d = 7.2$

$n= 50$

Notation and definitions

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$. The degrees of freedom are given by:  

$df=n-1=50-1=49$  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.005,49)" for $t_{\alpha/2}=-2.68$  

"=T.INV(1-0.005,49)" for $t_{1-\alpha/2}=2.68$  

The critical value $tc=\pm 2.68$

Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s_d}{\sqrt{n}}$

$m=2.68 \frac{7.2}{\sqrt{50}}=2.729$

Calculate the confidence interval  

The interval for the mean is given by this formula:

$\bar d \pm t_{c} \frac{s_d}{\sqrt{n}}$

And calculating the limits we got:

$12.5 - 2.68 \frac{7.2}{\sqrt{50}}=9.771$

$12.5 +2.68 \frac{7.2}{\sqrt{50}}=15.229$

The 99% confidence interval for the difference is given by (9.771;15.229)

And we can conclude that at 99% of confidence the average growth is between 9.771 and 15.229 feet.