Question

in a random sample of seven cell phones, the mean full retail price was $463.00 and the standard deviation was $222.00. assume the population is normally distributed and use the t - distribution to find the margin of error and construct a 90% confidence interval for the population mean μ. interpret the results. identify the margin of error. (round to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom and t - value

The degrees of freedom \(df=n - 1\), where \(n = 7\), so \(df=7-1 = 6\).
For a \(90\%\) confidence interval, the significance level \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\).
Looking up the t - value in the t - distribution table with \(df = 6\) and \(\frac{\alpha}{2}=0.05\), we get \(t_{\frac{\alpha}{2}}=1.943\).

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 \(s = 222\) (sample standard deviation) and \(n = 7\) (sample size).
Substitute the values: \(E=1.943\times\frac{222}{\sqrt{7}}\).
First, calculate \(\frac{222}{\sqrt{7}}\approx\frac{222}{2.6458}\approx83.9\).
Then \(E = 1.943\times83.9\approx163.0\).

Answer:

\(163.0\)