What is the p-value?

The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is $3.4$minutes. A random sample of $30$patients in the doctor’s office has a standard deviation of waiting times of $4.1$minutes.

We discovered that the average waiting time in a doctor's office changes, based on the data. In a doctor's office, the standard deviation of waiting periods is $3.4$minutes. The standard deviation of waiting times in a random sample of 30 patients in a doctor's office is$4.1$minutes. Because this is a right-tailed test, the single population variance chi-square test is

${\chi }^2=\frac{(n-1)s^2}{{\sigma }^2}$

Where

$n=30$is sample size

The sample standard deviation is$s=4.1$

The population standard deviation is $\sigma =3.4$

$df=30-1$degrees of freedom

localid="1650534389154" $$\begin{gathered} {\chi }^2=\frac{(30-1)(4.1{)}^2}{(3.4{)}^2} \\ \approx 42.1704 \end{gathered}$$

So, the p-value for the right-tailed test using the p-value table will be

localid="1650534395773" $$\begin{gathered} df=30-1 \\ =29 \end{gathered}$$

degrees of freedom