Find the P-values:

Question

Emissions of sulfur dioxide by industry set off chemical changes in the atmosphere that result in "acid rain".

The acidity of liquids is measured by pH on a scale of 0 to 14.

Distilled water has pH of 7.0, and lower pH values indicate acidity.

Normal rain is somewhat acidic, so acid rain is sometimes defined as rainfall with a pH below 5.0. Suppose that pH measurements of rainfall on different days in a Canadian forest follow a Normal distribution with standard deviation σ = 0.5.

A sample of n days finds that the mean pH is x = 4.8 (x-bar)

What are the P- Values for tests of sample sizes n = 5, n = 15 and n = 40?

n = 5 _____

n = 15 _____

n = 40 _____

Answer 1

Our null hypothesis is that the mean is 5, and the alternative hypothesis is that the mean is LOWER. And we know that this distribution is normal with that sigma= .5 . The standard error of the sample mean is sigma/sqrt(n) . So we need to find the Z score of 4.8 when the mean is 5 and the standard error is .5/sqrt(n) . So in each case we calculate:

(4.8-5)/(.5/sqrt(n)) . This is the Z-score, i.e. the number of standard deviations from the mean. The next thing we need to do is find the p-value, i.e. "what is the probability that we would get a result as extreme or more extreme than this value?" In this case "more extreme" means "lower" (see the alternative hypothesis). So when we get a Z score of -1.8 for exampe, we look that up on the Z-table and see .0359 for example. That would be that probability, which is the p-value for -1.8 .