The total of individual weights of garbage discarded by 15 households in one week is normally distributed with a mean of 32.1 lbs with a sample standard deviation of 14.3 lbs. Find the 99% confidence interval of the mean. Select an answer v < Do not round in between steps. Round answers to at least 4 decimal places.

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**Problem Statement:**

The total of individual weights of garbage discarded by 15 households in one week is normally distributed with a mean of 32.1 lbs and a sample standard deviation of 14.3 lbs. Find the 99% confidence interval of the mean.

**Solution:**

1. **Calculate the standard error (SE)**
   
   \[
   SE = \frac{\text{sample standard deviation}}{\sqrt{n}}
   \]
   
   Where:
   
   - Sample standard deviation (s) = 14.3 lbs
   - n (sample size) = 15

   \[
   SE = \frac{14.3}{\sqrt{15}}
   \]

2. **Find the critical value for the 99% confidence level**
   
   Since the sample size is small (< 30), we use the t-distribution. Using a t-table or calculator:

   Degrees of freedom (df) = n - 1 = 15 - 1 = 14

   For a 99% confidence level and 14 degrees of freedom, the critical value (t*) is approximately:

   \[
   t^* \approx 2.977
   \]

3. **Calculate the margin of error (ME)**

   \[
   ME = t^* \times SE
   \]

4. **Determine the confidence interval**

   The confidence interval is given by:

   \[
   \text{mean} \pm ME
   \]

**Note:**

- Do not round in between steps.
- Round final answers to at least 4 decimal places.

**Input Fields:**

\[ \left\langle \text{Lower Bound} \ \boxed{\text{Select an answer}} \ \text{Upper Bound} \right\rangle \]

**Rounding Instruction:**

Do not round in between steps. Round answers to at least 4 decimal places.
Transcribed Image Text:**Problem Statement:** The total of individual weights of garbage discarded by 15 households in one week is normally distributed with a mean of 32.1 lbs and a sample standard deviation of 14.3 lbs. Find the 99% confidence interval of the mean. **Solution:** 1. **Calculate the standard error (SE)** \[ SE = \frac{\text{sample standard deviation}}{\sqrt{n}} \] Where: - Sample standard deviation (s) = 14.3 lbs - n (sample size) = 15 \[ SE = \frac{14.3}{\sqrt{15}} \] 2. **Find the critical value for the 99% confidence level** Since the sample size is small (< 30), we use the t-distribution. Using a t-table or calculator: Degrees of freedom (df) = n - 1 = 15 - 1 = 14 For a 99% confidence level and 14 degrees of freedom, the critical value (t*) is approximately: \[ t^* \approx 2.977 \] 3. **Calculate the margin of error (ME)** \[ ME = t^* \times SE \] 4. **Determine the confidence interval** The confidence interval is given by: \[ \text{mean} \pm ME \] **Note:** - Do not round in between steps. - Round final answers to at least 4 decimal places. **Input Fields:** \[ \left\langle \text{Lower Bound} \ \boxed{\text{Select an answer}} \ \text{Upper Bound} \right\rangle \] **Rounding Instruction:** Do not round in between steps. Round answers to at least 4 decimal places.
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