Question

The data represent a random sample of the number of home fire started by candles for the past several years (data are from the national fire protection association.) Find 99% confidence interval for the mean number of home fires started by candles each year. interpret the result!5460 5900 6090 6310 7160 8440 9930

Answer 1

Answer:

$7041.43-3.71\frac{1610.27}{\sqrt{7}}=4783.43$  

$7041.43+3.71\frac{1610.27}{\sqrt{7}}=9299.43$  

So on this case the 99% confidence interval would be given by (4783.43;9299.43)  

Step-by-step explanation:

Previous concepts  

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".  

Solution to the problem

For this case we have the following data:

5460 5900 6090 6310 7160 8440 9930

And for this case we can calculate the mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$\bar X=7041.43$ represent the sample mean  

$\mu$ population mean (variable of interest)  

s=1610.27 represent the sample standard deviation  

n=7 represent the sample size  

Calculate the confidence interval

The confidence interval for the mean is given by the following formula:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=7-1=6$  

Since the confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,6)".And we see that $t_{\alpha/2}=3.71$

Now we have everything in order to replace into formula (1):  

$7041.43-3.71\frac{1610.27}{\sqrt{7}}=4783.43$  

$7041.43+3.71\frac{1610.27}{\sqrt{7}}=9299.43$  

So on this case the 99% confidence interval would be given by (4783.43;9299.43)