Question

in a random sample of 8 cell phones, the mean full retail price was $535.10 and the standard deviation was $191.00. further research suggests that the population mean is $428.06. does the t - value for the original sample fall between - t_{0.99} and t_{0.99}? assume that the population of full retail prices for cell phones is normally distributed.
the t - value of t = fall between - t_{0.99} and t_{0.99} because t_{0.99} =.
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the t - value

The formula for the t - value is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Given \(\bar{x} = 535.10\), \(\mu=428.06\), \(s = 191.00\), \(n = 8\).
First, calculate \(s/\sqrt{n}=\frac{191}{\sqrt{8}}\approx\frac{191}{2.828}\approx67.53\)
Then \(t=\frac{535.10 - 428.06}{67.53}=\frac{107.04}{67.53}\approx1.59\)

Step2: Find the value of \(t_{0.99}\)

The degrees of freedom \(df=n - 1=8 - 1 = 7\).
Looking up the t - distribution table (or using a calculator with t - distribution functions), for a one - tailed probability of \(0.99\) and \(df = 7\), \(t_{0.99}\approx2.998\approx3.00\)

Answer:

The t - value of \(t = 1.59\) fall between \(-t_{0.99}\) and \(t_{0.99}\) because \(t_{0.99}=3.00\)