Question

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Earthquake Depths Data Set 21 “Earthquakes” in Appendix B lists earthquake depths, and the summary statistics are n = 600, x = 5.82 km, s = 4.93 km. Use a 0.01 significance level to test the claim of a seismologist that these earthquakes are from a population with a mean equal to 5.00 km.

Short answer

The hypotheses are:

\(\begin{array}{l}{H_0}:\mu = 5\\{H_1}:\mu \ne 5\end{array}\)

The test statistic is 4.074.

The critical values are -2.584 and 2.584.

The null hypothesis is rejected. There is enough evidence to conclude that the population mean depth in the earthquake is equal to 5.00 km.

Step-by-step solution

Given information

A sample is taken from the depth of earthquakes with a sample size of 600 with the claim that the mean depth of the earthquake is equal to 5.00km.

The significance level is 0.01.

Hypothesis criteria 

The null hypothesis,\({H_0}\)represents the population mean depth is equal to 5. Also, the alternate hypothesis,\({H_1}\)represents the population mean depth is not equal to 5.

Let\(\mu \)be the population mean depth of the earthquake.

State the null and alternate hypotheses.

\(\begin{array}{l}{H_0}:\mu = 5\\{H_1}:\mu \ne 5\end{array}\)

State the critical value

The degrees of freedom are obtained by using the formula\(df = n - 1\)where\(n = 600\).

\(\begin{array}{c}df = 600 - 1\\ = 599\end{array}\)

The critical value can be obtained using the t-distribution table with\({\bf{df = 599}}\)and the significance level\({\bf{\alpha = 0}}{\bf{.01}}\), for two tailed tests.

\({{\bf{t}}_{{\bf{0}}{\bf{.005,599}}}}{\bf{ = 2}}{\bf{.584}}\)

Thus, the critical values are -2.584 and 2.584.

Compute the observed test statistic

Apply the t-test to compute the test statistic using the formula,\(t = \frac{{x - \mu }}{{\frac{s}{{\sqrt n }}}}\).

Substitute the respective values in the above formula and simplify the equation as follows:

\(\begin{array}{c}t = \frac{{5.82 - 5}}{{\frac{{4.93}}{{\sqrt {600} }}}}\\ = 4.074\end{array}\)

Thus, the test statistic is 4.074.

state the decision

Reject null hypothesis when the absolute value of observed test statistics is greater than the critical value. Otherwise fail to reject the null hypothesis.

\(\begin{array}{c}t = \left| {4.074} \right|\\ = 4.074\\ > 2.584\\ > {t_{0.005,599}}\end{array}\)

The absolute value of the observed test statistic is greater than the critical value. This implies that the null hypothesis is rejected.

There is no sufficient evidence to conclude that the population mean depth of earthquake is equal to 5.