Question

in a random sample of 8 cell phones, the mean full retail price was $616.70 and the standard deviation was $189.00. further research suggests that the population mean is $431.80. 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 = 1.27 does fall between - t_{0.99} and t_{0.99} because t_{0.99} = 3.00 (round to two decimal places as needed.)

Explanation:

Step1: Calculate the t - value

The formula for the t - value in a one - sample t - test 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 \(\bar{x} = 616.70\), \(\mu=431.80\), \(s = 189.00\), and \(n = 8\).
First, calculate \(s/\sqrt{n}=\frac{189}{\sqrt{8}}\approx\frac{189}{2.828}\approx66.82\).
Then \(t=\frac{616.70 - 431.80}{66.82}=\frac{184.9}{66.82}\approx2.77\).

Step2: Determine the critical values

For a two - tailed test with \(\alpha=0.01\) and degrees of freedom \(df=n - 1=8 - 1 = 7\).
Looking up the t - distribution table, \(t_{0.005,7}=3.499\) (for a two - tailed test with \(\alpha = 0.01\)), but if we assume a one - tailed interpretation (since the problem is not clear on tails, but if we consider the range \(-t_{0.005}\) and \(t_{0.005}\)).
However, if we consider the problem may have some calculation errors in the initial thought (maybe a wrong formula application in the original problem's partial answer).
If we re - check with the formula \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), \(\bar{x} = 616.70\), \(\mu = 431.80\), \(s=189\), \(n = 8\)
\(t=\frac{616.70-431.80}{189/\sqrt{8}}=\frac{184.9}{189/2.828}=\frac{184.9\times2.828}{189}\approx\frac{523.9}{189}\approx2.77\)

Answer:

The t - value \(t\approx2.77\). Since \(t_{0.005,7} = 3.499\) (two - tailed, \(\alpha=0.01\)), \(|t|=2.77<3.499\), so the t - value does fall between \(-t_{0.005}\) and \(t_{0.005}\)