Question

in a random sample of 7 cell phones, the mean full retail price was $525.70 and the standard deviation was $185.00. further research suggests that the population mean is $431.69. does the t - value for the original sample fall between - t_{0.95} and t_{0.95}? assume that the population of full retail prices for cell phones is normally distributed.
the t - value of t = fall between - t_{0.95} and t_{0.95} because t_{0.95} =.
(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 \(\bar{x} = 525.70\), \(\mu=431.69\), \(s = 185\), \(n = 7\).

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Step2: Find \(t_{0.95}\)

The degrees of freedom \(df=n - 1=7-1 = 6\).
Looking up the t - distribution table for \(\alpha=1 - 0.95=0.05\) (two - tailed) and \(df = 6\), \(t_{0.95}=1.94\)

Answer:

The t - value of \(t = 1.34\) fall between \(-t_{0.95}\) and \(t_{0.95}\) because \(t_{0.95}=1.94\)