To find the margin of error (E) and construct the 99% confidence interval for the population mean $\mu$, we'll use the formula for the margin of error in a t-distribution:

$E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$

Where:

• $t_{\alpha/2}$ is the t-value for the 99% confidence level (based on degrees of freedom $df = n - 1$),
• $s$ is the sample standard deviation,
• $n$ is the sample size.

Given dataTo find the margin of error (E) and construct the 99% confidence interval for the population mean $\mu$, we'll use the formula for the margin of error in a t-distribution:

$E = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$

Where:

• $t_{\alpha/2}$ is the t-value for the 99% confidence level (based on degrees of freedom $df = n - 1$),
• $s$ is the sample standard deviation,
• $n$ is the sample size.

Given

• Sample mean $\bar{x} = 402.50$ dollars,
• Sample standard deviation $s = 181.00$ dollars,
• Sample size $n = 6$,
• Confidence level = 99%, so $\alpha = 0.01$ and $\alpha/2 = 0.005$.

Step 1: Find the t-value

Since $n = 6$, degrees of freedom $df = 6 - 1 = 5$. From the t-distribution table, for $df = 5$ and a 99% confidence level (i.e., 0.005 in each tail), the critical value $t_{\alpha/2} \approx 4.032$.

Step 2: Calculate the margin of error

Using the formula:

$E = 4.032 \cdot \frac{181.00}{\sqrt{6}}$

First, calculate the denominator:

$\frac{181.00}{\sqrt{6}} = \frac{181.00}{2.4495} \approx 73.88$

Now calculate the margin of error:

$E = 4.032 \cdot 73.88 \approx 297.7$

Thus, the margin of error is approximately $297.7.

Part 2: Construct the confidence interval

The confidence interval is given by:

$\text{Confidence Interval} = \bar{x} \pm E$

$\text{Confidence Interval} = 402.50 \pm 297.7$

So, the confidence interval is:

$(402.50 - 297.7, 402.50 + 297.7) = (104.8, 700.2)$

Interpretation:

We are 99% confident that the true mean retail price of all cell phones is between $104.8 and $700.2.

Would you like more details or have any questions?

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Here are 5 questions that extend the topic:

1. How would the confidence interval change if the sample size increased?
2. What happens to the margin of error if we use a 95% confidence level instead?
3. How does the t-distribution differ from the normal distribution when constructing confidence intervals?
4. What assumptions must be met to use the t-distribution for this problem?
5. How does the standard deviation of the sample affect the width of the confidence interval?

Tip: A larger sample size reduces the margin of error, making your confidence interval more precise.