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?

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".

[tex]\bar X=0.81[/tex] represent the sample mean

[tex]\mu[/tex] 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:

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

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

[tex]df=n-1=110-1=109[/tex]

Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], 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 [tex]t_{\alpha/2}=2.62[/tex]

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

[tex]0.81-2.62\frac{0.34}{\sqrt{110}}=0.725[/tex]

[tex]0.81+2.62\frac{0.34}{\sqrt{110}}=0.895[/tex]

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:

[tex] ME=z_{\alpha/2}\frac{\sigmas}{\sqrt{n}}[/tex] (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:

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

We can assume the following estimator for the population deviation [tex]\hat \sigma =s =0.34[/tex]

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 [tex]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[/tex]

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