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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

${\mathrm{χ}}^2=\frac{(n−1)s^2}{{\mathrm{σ}}^2}$

Where

$n=30$is sample size

The sample standard deviation is$s=4.1$

The population standard deviation is $\mathrm{σ}=3.4$

$df=30-1$degrees of freedom

localid="1650534389154" $$\begin{gathered} {\mathrm{χ}}^2=\frac{(30−1)(4.1{)}^2}{(3.4{)}^2} \\ ≈42.1704 \end{gathered}$$

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