Answer:
Null hypothesis ![H_0:\mu=5.00km](https://tex.z-dn.net/?f=H_0%3A%5Cmu%3D5.00km)
Alternative hypothesis ![H_1:\mu\neq 5.00km](https://tex.z-dn.net/?f=H_1%3A%5Cmu%5Cneq%205.00km)
The p value is 0.000517, which is less than the significance level 0.01, therefore we reject the null hypothesis and conclude that population mean is not equal to 5.00.
Step-by-step explanation:
It is given that a data set lists earthquake depths. The summary statistics are
![n=300](https://tex.z-dn.net/?f=n%3D300)
![\overline{x}=5.89km](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D5.89km)
![s=4.44km](https://tex.z-dn.net/?f=s%3D4.44km)
Level of significance = 0.01
We need to test the claim of a seismologist that these earthquakes are from a population with a mean equal to 5.00.
Null hypothesis ![H_0:\mu=5.00km](https://tex.z-dn.net/?f=H_0%3A%5Cmu%3D5.00km)
Alternative hypothesis ![H_1:\mu\neq 5.00km](https://tex.z-dn.net/?f=H_1%3A%5Cmu%5Cneq%205.00km)
The formula for z-value is
![z=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z=\frac{5.89-5.00}{\frac{4.44}{\sqrt{300}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B5.89-5.00%7D%7B%5Cfrac%7B4.44%7D%7B%5Csqrt%7B300%7D%7D%7D)
![z=\frac{0.89}{0.25634351952}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B0.89%7D%7B0.25634351952%7D)
![z=3.4719](https://tex.z-dn.net/?f=z%3D3.4719)
The p-value for z=3.4719 is 0.000517.
Since the p value is 0.000517, which is less than the significance level 0.01, therefore we reject the null hypothesis and conclude that population mean is not equal to 5.00.