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:

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

$\overline{x}$ 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:

$$\begin{gathered} \overline{x}=\frac{51+44+........+23}{14} \\ =37.1 \end{gathered}$$

The sample standard deviation is computed below:

$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(51-37.1\right)}^2+{\left(44-37.1\right)}^2+.......+{\left(23-37.1\right)}^2}{14-1}} \\ =9.8 \end{gathered}$$

Value of   tα2

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:

$$\begin{gathered} df=n-1 \\ =14-1 \\ =13 \end{gathered}$$

The value of with 13 degrees of freedom is equal to 3.0123.

Calculation of confidence interval

The margin of error is equal to:

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\frac{s}{\sqrt{n}} \\ =3.0123\frac{9.8}{\sqrt{14}} \\ =7.8897 \end{gathered}$$

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

$$\begin{gathered} CI=\left(\overline{x}-E,\overline{x}+E\right) \\ =\left(37.1-7.8897,37.1+7.8897\right) \\ =\left(29.2,45.0\right) \end{gathered}$$

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.