Answer: [tex]p-value \approx 0.01[/tex]
Step-by-step explanation:
Hypothesis testing
[tex]\\\left\{\begin{matrix}H_0 : \mu_0 =0\\ H_1: \mu_0 \neq 0\end{matrix}\right.\\[/tex]
For this problem, we need to use the t-student distribution to make inference about the data. We calculate the t-statistics as below:
[tex]\bar{X} = 20\:\mu m\\S.E. = 60\:\mu m\\n=82\\t_{stat} = \frac{\bar{X}-\mu_0}{S.E/\sqrt{n}}=\frac{20-0}{60/\sqrt{82}}=3.0185 [/tex]
Using a t-statistics table, or using the function TDIST in Microsoft Excel with [tex]t_{stat} = 3.0185,\: df (degree\: of\: freedom) = 14 - 1 = 13[/tex] with two-tailed distribution, we obtain [tex]p-value \approx 0.01[/tex].
