Question

Confidence Interval for Lightning Deaths Use the sample values given in Cumulative Review Exercise 1 to construct a 99% confidence interval estimate of the population mean. Assume that the population has a normal distribution. Write a brief statement that interprets the confidence interval.

Short answer

The 99% confidence interval is equal to (29.2 deaths, 45.0 deaths).

There is 99% confidence that the population’s mean number of deaths due to lightning strikes will fall in the interval (29.2 deaths, 45.0 deaths).

Step-by-step solution

Given information

Data are given on the number of deaths that have occurred due to lightning strikes for a series of 14 consecutive years.

Confidence interval

The following formula is used to compute the confidence interval estimate of the population mean:

\(CI = \left( {\bar x - E,\bar x + E} \right)\)where

\(\bar x\)is the sample mean number of deaths

E is the margin of error

The formula to compute the value of E is shown below:

\(E = {t_{\frac{\alpha }{2}}}\frac{s}{{\sqrt n }}\)

Here, s is the sample standard deviation, and n is the sample size.

\({t_{\frac{\alpha }{2}}}\) is the value of the t-distribution at \(\frac{\alpha }{2}\) level of significance with \(\left( {n - 1} \right)\) degrees of freedom.

Sample size, sample mean, and sample standard deviation

The sample size (n) is equal to.

The sample mean is computed below:

\(\begin{array}{c}\bar x = \frac{{51 + 44 + ........ + 23}}{{14}}\\ = 37.1\end{array}\)

The sample standard deviation is computed below:

\(\begin{array}{c}s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {51 - 37.1} \right)}^2} + {{\left( {44 - 37.1} \right)}^2} + ....... + {{\left( {23 - 37.1} \right)}^2}}}{{14 - 1}}} \\ = 9.8\end{array}\)

Value of \({t_{\frac{\alpha }{2}}}\)

The confidence level is given as equal to 99%. Thus, the corresponding value of the level of significance \(\left( \alpha \right)\) is equal to 0.01.

The value of the degrees of freedom is computed below:

\(\begin{array}{c}df = n - 1\\ = 14 - 1\\ = 13\end{array}\)

The value of \({t_{\frac{\alpha }{2}}}\) with 13 degrees of freedom is equal to 3.0123.

Calculation of confidence interval

The margin of error is equal to:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2}}}\frac{s}{{\sqrt n }}\\ = 3.0123\frac{{9.8}}{{\sqrt {14} }}\\ = 7.8897\end{array}\)

The 99% confidence interval estimate of the population mean is equal to

\(\begin{array}{c}CI = \left( {\bar x - E,\bar x + E} \right)\\ = \left( {37.1 - 7.8897,37.1 + 7.8897} \right)\\ = \left( {29.2,45.0} \right)\end{array}\)

Thus, the 99% confidence interval estimate of the population mean is equal to (29.2 deaths, 45.0 deaths).

There is 99% confidence that the population’s mean number of deaths due to lightning strikes will lie between the values 29.2 and 45.0.