A data set lists earthquake depths. The summary statistics are n= 500, x = 6.41 km, s = 4.62 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 6.00. 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? O B. Ho: µ= 6.00 km H1: µ> 6.00 km O A. Ho: µ=6.00 km H1: µ<6.00 km O D. Ho: H#6.00 km H: µ= 6.00 km OC. Ho: u=6.00 km H: µ#6.00 km Determine the test statistic. 1.98 (Round to two decimal places as needed.) Determine the P-value. (Round to three decimal places as needed.)

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## Hypotheses Testing for Earthquake Depths

A data set lists earthquake depths. The summary statistics are:
- Sample size (n) = 500
- Sample mean (x̄) = 6.41 km
- Sample standard deviation (s) = 4.62 km

A seismologist claims that these earthquakes are from a population with a mean equal to 6.00 km. We will use a significance level (α) of 0.01 to test this claim. 

### Step 1: Identify the Null and Alternative Hypotheses
The null and alternative hypotheses represent statements about the population mean (μ):

**Option A:** 
- Null Hypothesis (H₀): μ = 6.00 km 
- Alternative Hypothesis (H₁): μ < 6.00 km 

**Option B:** 
- Null Hypothesis (H₀): μ = 6.00 km 
- Alternative Hypothesis (H₁): μ > 6.00 km

**Option C:** 
- Null Hypothesis (H₀): μ = 6.00 km 
- Alternative Hypothesis (H₁): μ ≠ 6.00 km

**Option D:** 
- Null Hypothesis (H₀): μ ≠ 6.00 km 
- Alternative Hypothesis (H₁): μ = 6.00 km

Given the claim that the mean is equal to 6.00 km, we test it against the possibility that it differs. Thus, the correct hypotheses are:
**Option C:** 
- Null Hypothesis (H₀): μ = 6.00 km 
- Alternative Hypothesis (H₁): μ ≠ 6.00 km

### Step 2: Determine the Test Statistic
We will use a t-test to determine the test statistic (t). Given the values:
\[ n = 500 \]
\[ \bar{x} = 6.41 \]
\[ s = 4.62 \]
\[ \mu = 6.00 \]

The test statistic is calculated using the formula:
\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

Plugging in the values:
\[ t = \frac{6.41 - 6.00}{4.62 / \sqrt{500}} \]
\[ t = \frac{0.
Transcribed Image Text:## Hypotheses Testing for Earthquake Depths A data set lists earthquake depths. The summary statistics are: - Sample size (n) = 500 - Sample mean (x̄) = 6.41 km - Sample standard deviation (s) = 4.62 km A seismologist claims that these earthquakes are from a population with a mean equal to 6.00 km. We will use a significance level (α) of 0.01 to test this claim. ### Step 1: Identify the Null and Alternative Hypotheses The null and alternative hypotheses represent statements about the population mean (μ): **Option A:** - Null Hypothesis (H₀): μ = 6.00 km - Alternative Hypothesis (H₁): μ < 6.00 km **Option B:** - Null Hypothesis (H₀): μ = 6.00 km - Alternative Hypothesis (H₁): μ > 6.00 km **Option C:** - Null Hypothesis (H₀): μ = 6.00 km - Alternative Hypothesis (H₁): μ ≠ 6.00 km **Option D:** - Null Hypothesis (H₀): μ ≠ 6.00 km - Alternative Hypothesis (H₁): μ = 6.00 km Given the claim that the mean is equal to 6.00 km, we test it against the possibility that it differs. Thus, the correct hypotheses are: **Option C:** - Null Hypothesis (H₀): μ = 6.00 km - Alternative Hypothesis (H₁): μ ≠ 6.00 km ### Step 2: Determine the Test Statistic We will use a t-test to determine the test statistic (t). Given the values: \[ n = 500 \] \[ \bar{x} = 6.41 \] \[ s = 4.62 \] \[ \mu = 6.00 \] The test statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] Plugging in the values: \[ t = \frac{6.41 - 6.00}{4.62 / \sqrt{500}} \] \[ t = \frac{0.
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