in a random sample of six cell phones, the mean full retail price was $579.00 and the standard deviation was $210.00. 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 μ. interpret the results. identify the margin of error. (round to one decimal place as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Determine the degrees of freedom
The degrees of freedom $df=n - 1$, where $n = 6$. So $df=6-1 = 5$.
## Step2: Find the $t$-value
For a $99\%$ confidence interval and $df = 5$, using the $t$-distribution table or calculator, $t_{\alpha/2}=4.032$.
## Step3: Calculate the margin of error
The formula for the margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$. Given $s = 210$, $n = 6$, and $t_{\alpha/2}=4.032$.
$$
\begin{align*}
E&=4.032\times\frac{210}{\sqrt{6}}\
&=4.032\times\frac{210}{2.449}\
&=4.032\times85.75\
&\approx346.8
\end{align*}
$$

# Answer:
$346.8$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Determine the degrees of freedom

Step2: Find the \(t\)-value

Step3: Calculate the margin of error

Answer: