35 per year. What is the z-value rounded to the nearest hunc evidence to reject the claim? e is -0.54. There is not enough evidence to reject the claim. e is 0.54. There is enough evidence to reject the claim. e is -0.54. There is enough evidence to reject the claim. e is 0.54. There is not enough evidence to reject the claim.

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### Understanding Z-Value and Hypothesis Testing: An Example

**Context:**
An automobile insurance company asserts that its rates for teenage drivers are, on average, $470 less per year compared to another company's rates. To assess this claim, consider the following information:

- **Claimed Average Savings:** $470 per year
- **Sample Size (n):** 30 customers
- **Sample Average Savings:** $435 per year
- **Standard Deviation (σ):** $65 per year

**Question:**
What is the z-value rounded to the nearest hundredth? Is there sufficient evidence to reject the insurance company's claim?

**Options:**
1. ○ The z-value is -0.54. There is not enough evidence to reject the claim.
2. ○ The z-value is 0.54. There is enough evidence to reject the claim.
3. ○ The z-value is -0.54. There is enough evidence to reject the claim.
4. ○ The z-value is 0.54. There is not enough evidence to reject the claim.

**Explanation:**

The z-value is a measure that describes how many standard deviations an element is from the mean. It is calculated using the formula:

\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]

Where:
- \(\bar{x}\) is the sample mean
- \(\mu\) is the population mean
- \(\sigma\) is the standard deviation
- \(n\) is the sample size

**Calculation:**

Given:
- \(\bar{x} = 435\)
- \(\mu = 470\)
- \(\sigma = 65\)
- \(n = 30\)

\[ z = \frac{435 - 470}{\frac{65}{\sqrt{30}}} \]
\[ z = \frac{-35}{\frac{65}{\sqrt{30}}} \]
\[ z = \frac{-35}{11.87} \]
\[ z \approx -2.95 \]

**Interpreting the Z-Value:**
In hypothesis testing, comparing the z-value to critical values from the z-table helps determine the statistical significance. A z-value of -2.95 is quite large in magnitude, indicating that the sample mean is significantly different from the claimed mean at most common significance levels (e.g.,
Transcribed Image Text:### Understanding Z-Value and Hypothesis Testing: An Example **Context:** An automobile insurance company asserts that its rates for teenage drivers are, on average, $470 less per year compared to another company's rates. To assess this claim, consider the following information: - **Claimed Average Savings:** $470 per year - **Sample Size (n):** 30 customers - **Sample Average Savings:** $435 per year - **Standard Deviation (σ):** $65 per year **Question:** What is the z-value rounded to the nearest hundredth? Is there sufficient evidence to reject the insurance company's claim? **Options:** 1. ○ The z-value is -0.54. There is not enough evidence to reject the claim. 2. ○ The z-value is 0.54. There is enough evidence to reject the claim. 3. ○ The z-value is -0.54. There is enough evidence to reject the claim. 4. ○ The z-value is 0.54. There is not enough evidence to reject the claim. **Explanation:** The z-value is a measure that describes how many standard deviations an element is from the mean. It is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Where: - \(\bar{x}\) is the sample mean - \(\mu\) is the population mean - \(\sigma\) is the standard deviation - \(n\) is the sample size **Calculation:** Given: - \(\bar{x} = 435\) - \(\mu = 470\) - \(\sigma = 65\) - \(n = 30\) \[ z = \frac{435 - 470}{\frac{65}{\sqrt{30}}} \] \[ z = \frac{-35}{\frac{65}{\sqrt{30}}} \] \[ z = \frac{-35}{11.87} \] \[ z \approx -2.95 \] **Interpreting the Z-Value:** In hypothesis testing, comparing the z-value to critical values from the z-table helps determine the statistical significance. A z-value of -2.95 is quite large in magnitude, indicating that the sample mean is significantly different from the claimed mean at most common significance levels (e.g.,
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