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Expert-verified Found in: Page 658 ### Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208 # What is the p-value?

The $p-$value is $p=0.054209$.

See the step by step solution

## Step 1: Given Information

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.

## Step 2: Explanation

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

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

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" ${\chi }^{2}=\frac{\left(30-1\right)\left(4.1{\right)}^{2}}{\left(3.4{\right)}^{2}}\phantom{\rule{0ex}{0ex}}\approx 42.1704$

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

localid="1650534395773" $df=30-1\phantom{\rule{0ex}{0ex}}=29$

degrees of freedom ### Want to see more solutions like these? 