The tutor shows a hypothesis testing example with a p-value.

In yesterday’s post I brought up p-values. Today I’ll give an example:

A population is believed to have mean μ₀=42.5; a sample of 30 is taken with mean x=41.6, standard devation s=28.14. Find the p-value that the population mean indeed is 42.5.

Solution:

With 30 or more in the sample, we can use the z table, even though we’re using the sample standard deviation:

z=(41.6-42.5)/(28.14/30^0.5)=-0.175

z=-0.175 leads to a table value of 0.43. However, a p-value measures the likelihood that z could be as far from 0 next time. When testing μ≠μ₀, it could show up on the other side, with equal probability, since the normal distribution is symmetric:

In this case, the p-value is 2(0.43) = 0.86. The TI-83 Plus agrees. As a p-value, 0.86 is very high: we must continue to believe that the population mean is indeed 42.5.

I’ll be talking more about statistics in future posts:)

Source:

Harnett, Donald L. and James L. Murphy. Statistical Analysis for Business and Economics, first Canadian edition. Don Mills: Addison-Wesley, 1993.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.