Question

Listed below are amounts of strontium-90 (in millibecquerels, or mBq, per gram of calcium) in a simple random sample of baby teeth obtained from Pennsylvania residents and New York residents born after 1979 (based on data from “An Unexpected Rise in Strontium-90 in U.S. Deciduous Teeth in the 1990s,”) by Mangano, et al., Science of the Total Environment, Vol. 317). Use a 0.05 significance level to test the claim that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean amount from New York residents.

Pennsylvania:	155	142	149	130	151	163	151	142	156	133	138	161
New York:	133	140	142	131	134	129	128	140	140	140	137	143

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviation are equal. Test the given claim using the P-value method and critical value method.

Answer 1

ANS.

Using MINITAB , we get:

Two-Sample T-Test and CI: Pennsylvania, New York

Two-sample T for Pennsylvania vs New York

N Mean StDev SE Mean
Pennsylvania 12 147.6 10.6 3.1
New York 12 136.42 5.21 1.5

Difference = mu (Pennsylvania) - mu (New York)
Estimate for difference: 11.1667
95% lower bound for difference: 5.1714
T-Test of difference = 0 (vs >): T-Value = 3.27 P-Value = 0.003 DF = 15

Critical t-value:

Decision using P-value method:

Decision using Critical Value Method:

​​​​​​​