In a random sample of six cell​ phones, the mean full retail price was ​$402.50 and the standard deviation was ​$181.00. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 99​% confidence interval for the population mean mu. Interpret the results.
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Part 1
Identify the margin of error.

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▼
dollars
cell phones
square dollars
​(Round to one decimal place as​ needed.)

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.

In a random sample of six cell​ phones, the mean full retail price was ​$402.50 and the standard deviation was ​$181.00. Assume the population is normally distributed and use the​ t-distribution to find the margin of error and construct a 99​% confidence interval for the population mean mu. Interpret the results.

Learn how to calculate the margin of error and construct a 99% confidence interval for the mean retail price of cell phones. This problem uses t-distribution for a sample of 6 phones, with a sample mean of $402.50 and a standard deviation of $181.00. Ideal for college-level statistics students.

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