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Chapter 8: Q3CRE (page 356)

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

Expert verified

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

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

Step by step solution

01

Given information

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

02

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.

03

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:

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

04

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.

05

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鈥檚 mean number of deaths due to lightning strikes will lie between the values 29.2 and 45.0.

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