**1.     The sample mean life of ten (10) car batteries is 101.7 months, with a standard deviation of 8.54 months. What are the 80% confidence limits for the mean and standard deviation of a pdf that represents these batteries?**

To calculate the 80% confidence limits for the mean and standard deviation of the car battery lifespan, we can use the following procedures.

### 1. **Confidence Interval for the Mean**:

For the sample mean, we use the formula for the confidence interval (CI) based on the t-distribution because the sample size is small (\(n = 10\)).

\[
CI = \bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}
\]

Where:
- \(\bar{x} = 101.7\) (sample mean),
- \(s = 8.54\) (sample standard deviation),
- \(n = 10\) (sample size),
- \(t_{\alpha/2, n-1}\) is the t-distribution critical value at 80% confidence level and \(n-1 = 9\) degrees of freedom.

For 80% confidence and 9 degrees of freedom, \(t_{\alpha/2, 9}\) can be found using statistical tables or a calculator, which gives approximately \(t_{0.10, 9} \approx 1.383\).

Now, let's calculate the margin of error (MoE) and the confidence interval for the mean:

\[
MoE = 1.383 \times \frac{8.54}{\sqrt{10}}
\]

\[
CI = 101.7 \pm MoE
\]

### 2. **Confidence Interval for the Standard Deviation**:

For the confidence interval of the standard deviation, we use the chi-square distribution. The confidence limits for the standard deviation are calculated using the following formula:

\[
\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right)
\]

Where:
- \(n = 10\),
- \(s = 8.54\),
- \(\chi^2_{\alpha/2}\) and \(\chi^2_{1-\alpha/2}\) are the chi-square critical values for 80% confidence and \(n-1 = 9\) degrees of freedom.

For 80% confidence, we split the tails: \(\alpha/2 = 0.10\) and \(1-\alpha/2 = 0.90\). From chi-square tables:
- \(\chi^2_{0.10, 9} \approx 3.325\),
- \(\chi^2_{0.90, 9} \approx 14.684\).

Now, we calculate the confidence interval for the variance and then take the square root to find the confidence interval for the standard deviation.

---

I'll calculate both intervals now.### 1. **Confidence Interval for the Mean:**
The 80% confidence interval for the mean lifespan of the car batteries is:
\[
(97.97 \text{ months}, 105.43 \text{ months})
\]

### 2. **Confidence Interval for the Standard Deviation:**
The 80% confidence interval for the standard deviation is:
\[
(6.69 \text{ months}, 14.05 \text{ months})
\]

This means we are 80% confident that the true mean of the battery life lies between 97.97 and 105.43 months, and the standard deviation lies between 6.69 and 14.05 months.

Do you have any questions or would you like more details on any part of this solution?

---

Here are 5 related questions:
1. How does increasing the sample size affect the confidence interval?
2. What if the confidence level was changed to 90% instead of 80%?
3. How can the t-distribution be approximated by the normal distribution for large samples?
4. How do you interpret confidence intervals in hypothesis testing?
5. How does the standard deviation impact the width of the confidence interval?

**Tip:** A larger sample size reduces the width of the confidence interval, leading to a more precise estimate.

The sample mean life of ten (10) car batteries is 101.7 months, with a standard deviation of 8.54 months. What are the 80% confidence limits for the mean and standard deviation of a pdf that represents these batteries?

This detailed solution calculates the 80% confidence limits for the mean and standard deviation of car battery life based on a sample mean of 101.7 months and a standard deviation of 8.54 months. Using the t-distribution and chi-square distribution, find the confidence intervals with clear steps.

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