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To solve the problem, let us calculate the **margin of error** (ME) and the confidence interval (CI) for the population mean using the given data and standard normal distribution.

### Data provided:
- Sample size ($n$) = 13
- Sample mean ($\bar{x}$) = 70.00
- Standard deviation ($\sigma$) = 15.50
- Confidence level = 95%

### Step 1: Find the critical z-value for a 95% confidence level
For a 95% confidence level, the critical z-value ($z^*$) for a two-tailed test is approximately **1.96**.

### Step 2: Calculate the standard error (SE)
The formula for standard error when the population standard deviation is known is:
\[
SE = \frac{\sigma}{\sqrt{n}}
\]
Substitute the values:
\[
SE = \frac{15.50}{\sqrt{13}} \approx \frac{15.50}{3.606} \approx 4.30
\]

### Step 3: Calculate the margin of error (ME)
The formula for the margin of error is:
\[
ME = z^* \cdot SE
\]
Substitute the values:
\[
ME = 1.96 \cdot 4.30 \approx 8.43
\]

### Step 4: Construct the confidence interval (CI)
The confidence interval is calculated as:
\[
CI = \bar{x} \pm ME
\]
Substitute the values:
\[
CI = 70.00 \pm 8.43
\]
This gives:
\[
CI = (70.00 - 8.43, 70.00 + 8.43) = (61.57, 78.43)
\]

### Final Results:
- **Margin of Error**: $ \mathbf{8.43} $
- **Confidence Interval using z-distribution**: $ \mathbf{(61.57, 78.43)} $

### Comparison to t-distribution Confidence Interval
The t-distribution confidence interval provided in the problem is $(60.6, 79.4)$. The difference arises because the t-distribution accounts for the smaller sample size, providing slightly wider intervals than the z-distribution.

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Would you like further details or assistance with any related concepts?

Here are some related questions to deepen understanding:
1. What is the difference between z-distribution and t-distribution, and when is each used?
2. How do you interpret confidence intervals in the context of population means?
3. How does sample size affect the margin of error and confidence interval width?
4. Why does the margin of error increase with a higher confidence level?
5. Can the margin of error ever be zero? Under what conditions?

**Tip:** Always check whether the population standard deviation or sample standard deviation is used in calculations, as it determines whether z or t-distribution should be applied.

In a random sample of 13 microwave ovens, the mean repair cost was $70.00 and the standard deviation was $15.50. Using the standard normal distribution, find the margin of error and construct a 95% confidence interval for the population mean μ. A 95% confidence interval using the t-distribution was (60.6, 79.4). Compare the results.

Learn how to calculate the margin of error and 95% confidence interval for a population mean using the z-distribution. This problem involves a sample of 13 microwave ovens, a mean repair cost of $70.00, and a standard deviation of $15.50. Compare the results to the t-distribution confidence interval (60.6, 79.4).

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