Question

Earthquake Magnitudes Listed below are Richter scale magnitudes of randomly selected earthquakes.

a. Identify the best point estimate of the population mean $\mu$.

b. Construct a 95% confidence interval estimate of the mean magnitude of the population of earthquakes.

c. Write a statement that interprets the confidence interval.

3.09 2.76 2.65 3.44 3.01 2.94 3.45 2.72 2.69 2.89 2.71 2.76

Short answer

a. The sample mean $\left(\overline{x}\right)$equal to 2.926 is the best point estimate of the population mean.

b.The 95% confidence interval estimate of the population mean magnitude of earthquakes is equal to (2.749, 3.102).

c.The 95% confidence interval estimate explains that there is 95% confidence that the true population mean magnitude of earthquakes will lie between 2.749 and 3.102.

Step-by-step solution

Given information

A sample of magnitudes of earthquakes measured on the Richter scale is given.

Point estimate of population mean

a.

It is known that the best point estimate of the population mean is the value of the sample mean.

Here, the sample mean magnitude of the earthquakes is computed below:

$$\begin{gathered} \overline{x}=\frac{\sum x}{n} \\ =\frac{3.09+2.76+.......+2.76}{12} \\ =2.9258 \\ \approx 2.926 \end{gathered}$$

Therefore, the sample mean $\left(\overline{x}\right)$ is approximately equal to 2.926 is the best point estimate of the population mean.

Formula of confidence interval

b.

The formula for computing the confidence interval estimate of the population mean magnitude of earthquakes is written below:

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

Where

$E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$

Sample standard deviation

The sample standard deviation of the magnitude of earthquakes is computed below:

$$\begin{gathered} s=\sqrt{\frac{{\sum }_{i=1}^n{(x_i-\overline{x})}^2}{n-1}} \\ =\sqrt{\frac{{\left(3.09-2.926\right)}^2+{\left(2.76-2.926\right)}^2+.......+{\left(2.76-2.926\right)}^2}{12-1}} \\ =0.278 \end{gathered}$$

Thus, s=0.278

Sample size and t-value

The sample size is given to be equal to n = 12.

The confidence level is given to be equal to 95%. Thus, the level of significance is equal to 0.05.

The degrees of freedom is equal to:

$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$

The value of $t_{\frac{\alpha }{2}}$ with $\alpha =0.05$ and degrees of freedom equal to 11 is equal to 2.201.

Calculation

The margin of error becomes:

$$\begin{gathered} E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}} \\ =2.201\times \frac{0.278}{\sqrt{12}} \\ =0.1766 \end{gathered}$$

The 95% confidence interval becomes as follows:

$$\begin{gathered} CI=\left(\overline{x}-E,\overline{x}+E\right) \\ =\left(2.9258-0.1766,2.9258+0.1766\right) \\ =\left(2.749,3.102\right) \end{gathered}$$

Therefore, the 95% confidence interval estimate of the population mean magnitude of earthquakes is equal to (2.749,3.102).

Interpretation

c.

The 95% confidence interval estimate explains that there is 95% confidence that the true population mean magnitude of earthquakes will lie between 2.749 and 3.102.