Question

A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of 0.81 second and a sample standard deviation of 0.34 second. (a) Calculate a 99% confidence interval for the true average echo duration. (b) This sample data is used as a pilot study, and now the investigator would like to design a new study to construct a 99% confidence interval with width 0.1. What is the necessary sample size?

Answer 1

Answer:

a) $0.81-2.62\frac{0.34}{\sqrt{110}}=0.725$    

$0.81+2.62\frac{0.34}{\sqrt{110}}=0.895$    

So on this case the 99% confidence interval would be given by (0.725;0.895)    

b) $n=(\frac{2.58(0.34)}{0.1})^2 =76.95 \approx 77$

So the answer for this case would be n=77 rounded up to the nearest integer

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".

$\bar X=0.81$ represent the sample mean

$\mu$ population mean (variable of interest)

s=0.34 represent the sample standard deviation

n=110 represent the sample size  

Part a

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:

$df=n-1=110-1=109$

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(0.005,119)".And we see that $t_{\alpha/2}=2.62$

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

$0.81-2.62\frac{0.34}{\sqrt{110}}=0.725$    

$0.81+2.62\frac{0.34}{\sqrt{110}}=0.895$    

So on this case the 99% confidence interval would be given by (0.725;0.895)    

Part b

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigmas}{\sqrt{n}}$    (a)

And on this case we have that ME =0.1 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

We can assume the following estimator for the population deviation $\hat \sigma =s =0.34$

The critical value for 99% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.005;0;1)", and we got $z_{\alpha/2}=2.58/tex], replacing into formula (b) we got: [tex]n=(\frac{2.58(0.34)}{0.1})^2 =76.95 \approx 77$

So the answer for this case would be n=77 rounded up to the nearest integer