Question

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.
dollars
(rou to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom and the t - value

The degrees of freedom \(df=n - 1\), where \(n = 6\), so \(df=6-1 = 5\).
For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
Using the t - distribution table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.005,5}=4.032\).

Step2: Calculate the margin of error formula

The formula for the margin of error \(E\) when using the t - distribution is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 210\) (sample standard deviation) and \(n = 6\) (sample size).
Substitute the values: \(E=4.032\times\frac{210}{\sqrt{6}}\).
First, calculate \(\sqrt{6}\approx2.449\).
Then \(\frac{210}{2.449}\approx85.75\).
Finally, \(E = 4.032\times85.75\approx345.8\).

Answer:

\(345.8\)