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Let's go through each part of this problem step-by-step.

### Given Information:
- Sample size (\( n \)) = 16
- Population standard deviation (\( \sigma \)) = 7.23
- Data values (temperatures in Celsius) are provided.

1. **Calculate the Sample Mean (\( \bar{x} \))**:
   The sample mean is the average of the provided data points:
   \[
   \bar{x} = \frac{\sum x}{n}
   \]

   Data points:
   \[
   67, 69.1, 68.4, 69.3, 75.2, 85.8, 76.8, 80.8, 82.8, 71.3, 89.1, 77.2, 85.8, 78.5, 74.7, 86.9
   \]

2. **Standard Error of the Mean (SEM)**:
   The SEM is calculated as:
   \[
   \text{SEM} = \frac{\sigma}{\sqrt{n}}
   \]

3. **Solution Steps for Each Part**:

   **(a) 99% Confidence Interval for the Mean Temperature**:
   - For a 99% confidence interval, use the \( z \)-score corresponding to 99% confidence, which is approximately \( z = 2.576 \).
   - The confidence interval formula is:
     \[
     \bar{x} \pm z \cdot \text{SEM}
     \]

   **(b) 95% Confidence Single-Sided Lower-Bound on the Mean**:
   - For a one-sided 95% confidence interval, use the \( z \)-score corresponding to 95% confidence, which is \( z = 1.645 \).
   - The lower bound is:
     \[
     \bar{x} - z \cdot \text{SEM}
     \]

   **(c) Sample Size for Mean Error \( E < 1 \) with 95% Confidence**:
   - Given \( E = 1 \), we solve for \( n \) using:
     \[
     E = z \cdot \frac{\sigma}{\sqrt{n}}
     \]
     Rearranging to solve for \( n \):
     \[
     n = \left( \frac{z \cdot \sigma}{E} \right)^2
     \]
     Use \( z = 1.96 \) for a 95% confidence level.

I'll calculate each part and provide the answers next.Here are the solutions to each part:

1. **(a) 99% Confidence Interval for the Mean Temperature**:
   \[
   72.76 \leq \mu \leq 82.07
   \]

2. **(b) 95% Confidence Single-Sided Lower-Bound on the Mean**:
   \[
   \mu \geq 74.45
   \]

3. **(c) Sample Size for Mean Error \( E < 1 \) with 95% Confidence**:
   \[
   n = 201
   \]

Let me know if you need further details or have any questions. Here are some related questions you might find useful:

1. What would the confidence intervals be if the standard deviation were unknown?
2. How does changing the sample size affect the confidence interval width?
3. Why do we use the \( z \)-score in this problem instead of the \( t \)-score?
4. What does a one-sided confidence interval imply in practical terms?
5. How would the calculations change if we wanted a 90% confidence level?

**Tip:** Increasing the sample size decreases the standard error, which narrows confidence intervals and provides more precise estimates of the population mean.

An experiment yielded n = 16 temperature measurements with a known standard deviation of σ = 7.23. Calculate the following:
(a) The 99% two-sided confidence interval for the mean temperature.
(b) The 95% confidence single-sided lower-bound on the mean.
(c) The sample size needed for a 95% confidence level to ensure the estimated mean error E = |x̄ − μ| < 1.

Solve an undergraduate-level statistics problem involving confidence intervals for a set of temperature measurements. Given a sample size of 16, standard deviation of 7.23, and specific confidence levels, this problem demonstrates how to calculate a two-sided 99% confidence interval, a one-sided 95% lower bound, and the required sample size for a specified margin of error.

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