Annual Precipitation The average annual precipitations (in inches) of a random sample of 30 years in San Francisco, California have a sample standard deviation of 8.18 inches. Use a $$ 98 \% $$ level of confidence. (Source: Golden Gate Weather Services)

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To construct a confidence interval for the average annual precipitation based on the given data, we will follow these steps:

1. **Identify the known values:**
   - Sample size ($$ n $$): 30 years
   - Sample standard deviation ($$ s $$): 8.18 inches
   - Confidence level: 98%

2. **Determine the critical value:**
   Since the sample size is less than 30, we will use the t-distribution. We need to find the critical t-value for a 98% confidence level with $$ n - 1 = 30 - 1 = 29 $$ degrees of freedom.

3. **Calculate the standard error (SE):**
   The standard error is calculated using the formula:
   $$
   SE = \frac{s}{\sqrt{n}}
   $$

4. **Calculate the margin of error (ME):**
   The margin of error is calculated using the formula:
   $$
   ME = t^* \times SE
   $$
   where $$ t^* $$ is the critical t-value.

5. **Construct the confidence interval:**
   The confidence interval is given by:
   $$
   \text{CI} = \bar{x} \pm ME
   $$
   where $$ \bar{x} $$ is the sample mean. Since the sample mean is not provided, we will express the confidence interval in terms of $$ \bar{x} $$.

Now, let's calculate the standard error and find the critical t-value for 29 degrees of freedom.

First, we will calculate the standard error.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{8.18}{\sqrt{30}}$$
- step1: Divide the terms:
  $$\frac{409}{50\sqrt{30}}$$
- step2: Multiply by the Conjugate:
  $$\frac{409\sqrt{30}}{50\sqrt{30}\times \sqrt{30}}$$
- step3: Multiply the numbers:
  $$\frac{409\sqrt{30}}{1500}$$
The standard error (SE) is approximately $$ 1.49 $$ inches.

Next, we need to find the critical t-value for a 98% confidence level with 29 degrees of freedom. Let's calculate that.
Evaluate by following steps:
- step0: Solution:
  $$t$$
It seems there was an issue retrieving the critical t-value directly. However, for a 98% confidence level and 29 degrees of freedom, the critical t-value is approximately $$ 2.462 $$.

Now, we can calculate the margin of error (ME) using the standard error and the critical t-value:

$$
ME = t^* \times SE
$$

Substituting the values we have:

$$
ME = 2.462 \times 1.49
$$

Let's calculate the margin of error.
Calculate the value by following steps:
- step0: Calculate:
  $$2.462\times 1.49$$
- step1: Multiply the numbers:
  $$3.66838$$
The margin of error (ME) is approximately $$ 3.67 $$ inches.

Now, we can construct the confidence interval. The confidence interval is given by:

$$
\text{CI} = \bar{x} \pm ME
$$

Since we do not have the sample mean ($$ \bar{x} $$), we can express the confidence interval as:

$$
\text{CI} = \bar{x} \pm 3.67
$$

Thus, the confidence interval for the average annual precipitation in San Francisco, California, at a 98% confidence level is:

$$
(\bar{x} - 3.67, \bar{x} + 3.67)
$$

If you have the sample mean, you can substitute it into this expression to get the specific confidence interval.
The 98% confidence interval for the average annual precipitation in San Francisco is $$ \bar{x} \pm 3.67 $$ inches.

Annual Precipitation The average annual precipitations (in inches) of a random sample of 30 years in San Francisco, California have a sample standard deviation of 8.18 inches. Use a \( 98 \% \) level of confidence. (Source: Golden Gate Weather Services)

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