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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.

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.

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

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.

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.

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.