High School

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------------------------------------------------ A data set lists earthquake depths. The summary statistics are [tex]n = 500, \bar{x} = 6.75 \text{ km}, s = 4.56 \text{ km}[/tex]. 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 6.00 km. Assume that a simple random sample has been selected.

Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim.

What are the null and alternative hypotheses?

A. [tex]H_0: \mu = 6.00 \text{ km}[/tex]
[tex]H_1: \mu \neq 6.00 \text{ km}[/tex]

B. [tex]H_0: \mu = 6.00 \text{ km}[/tex]
[tex]H_1: \mu \textless 6.00 \text{ km}[/tex]

C. [tex]H_0: \mu \neq 6.00 \text{ km}[/tex]
[tex]H_1: \mu = 6.00 \text{ km}[/tex]

D. [tex]H_0: \mu = 6.00 \text{ km}[/tex]
[tex]H_1: \mu \textgreater 6.00 \text{ km}[/tex]

Determine the test statistic.
[tex]\square[/tex] (Round to two decimal places as needed.)

Answer :

We begin by testing the claim that the population mean earthquake depth is [tex]$6.00$[/tex] km against the alternative that it is not [tex]$6.00$[/tex] km. Since the claim is questioned (i.e., the seismologist is suggesting that the mean might differ), we use a two-tailed test with the following hypotheses:

[tex]$$
H_0: \mu = 6.00 \text{ km} \quad \text{(null hypothesis)}
$$[/tex]

[tex]$$
H_1: \mu \neq 6.00 \text{ km} \quad \text{(alternative hypothesis)}
$$[/tex]

Given the sample statistics:
- Sample size: [tex]$n = 500$[/tex]
- Sample mean: [tex]$\bar{x} = 6.75$[/tex] km
- Sample standard deviation: [tex]$s = 4.56$[/tex] km

and the hypothesized mean [tex]$\mu_0 = 6.00$[/tex] km, we first compute the standard error of the mean:

[tex]$$
\text{Standard Error} = \frac{s}{\sqrt{n}} = \frac{4.56}{\sqrt{500}} \approx 0.20393 \text{ km}
$$[/tex]

Next, the test statistic is calculated using the formula for a [tex]$z$[/tex]-test when the sample size is large:

[tex]$$
z = \frac{\bar{x} - \mu_0}{\text{Standard Error}} = \frac{6.75 - 6.00}{0.20393} \approx 3.68
$$[/tex]

After calculating the [tex]$z$[/tex]-statistic, we find the [tex]$P$[/tex]-value for a two-tailed test. The [tex]$P$[/tex]-value is given by:

[tex]$$
P\text{-value} = 2 \times \left(1 - \Phi(3.68)\right)
$$[/tex]

where [tex]$\Phi(z)$[/tex] is the cumulative distribution function of the standard normal distribution. The computed [tex]$P$[/tex]-value is approximately:

[tex]$$
P\text{-value} \approx 0.000235
$$[/tex]

Since the significance level is [tex]$\alpha = 0.01$[/tex], we compare the [tex]$P$[/tex]-value to [tex]$\alpha$[/tex]:

[tex]$$
0.000235 < 0.01
$$[/tex]

Because the [tex]$P$[/tex]-value is less than [tex]$\alpha$[/tex], we reject the null hypothesis.

In summary:

1. The hypotheses are:
- Null Hypothesis: [tex]$\boxed{H_0: \mu=6.00 \text{ km}}$[/tex]
- Alternative Hypothesis: [tex]$\boxed{H_1: \mu \neq 6.00 \text{ km}}$[/tex]

2. The test statistic is:
[tex]$$
z \approx 3.68
$$[/tex]

3. The P-value is:
[tex]$$
\text{P-value} \approx 0.000235
$$[/tex]

4. Final conclusion: At the [tex]$0.01$[/tex] significance level, there is sufficient evidence to reject [tex]$H_0$[/tex]. We conclude that the true mean earthquake depth is statistically different from [tex]$6.00$[/tex] km.