Question

Radiation in Baby Teeth 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). Use a 0.05 significance level to test the claim that the median amount of strontium-90 from Pennsylvania residents is the same as the median 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

Short answer

There isenough evidence to conclude that there is a difference in the median values of the amount of strontium in baby teeth between the cities of Pennsylvania and New York.

Step-by-step solution

Given information

Two samples show the amount of strontium (in millibecquerels) of baby teeth of two cities—Pennsylvania and New York.

Frame the hypotheses

The Wilcoxon rank-sum test is performed to test the difference in the medians of the given samples.

The null hypothesis is as follows:

There is no difference in the median values of the amount of strontium in baby teeth in the two cities.

The alternative hypothesis is as follows:

There is a difference in the median values of the amount of strontium in baby teeth in the two cities.

It is a two-tailed test.

Assigning ranks to the observations

The ranks are computed by combining the two samples and tagging each observation with the sample name/number it comes from. Here, the city name/ sample name is tagged with ranks. Let “P” indicate the residents ofPennsylvania and “N”indicatethe residents of New York.

The lowest observation is assigned rank1; the second-lowest observation is assigned rank 2, and so on.

If at least two observations have the same value, the mean of the ranks is assigned to them.

The following table shows the ranks:

Amount of Strontium	Sample Type	Ranks
155	P	21
142	P	15
149	P	18
130	P	3
151	P	19.5
163	P	24
151	P	19.5
142	P	15
156	P	22
133	P	5.5
138	P	9
161	P	23
133	N	5.5
140	N	11.5
142	N	15
131	N	4
134	N	7
129	N	2
128	N	1
140	N	11.5
140	N	11.5
140	N	11.5
137	N	8
143	N	17

Compute the sum of the ranks

Compute the sum of ranks for Pennsylvania.

\(21 + 15 + 18 + .... + 17 = 194.5\)

Thus, the sum of the ranks for Pennsylvania is 194.5 and is denoted by R.

Compute the mean and the standard deviation

Let \({n_1}\) be the sample size for Pennsylvania.

Let \({n_2}\) be the sample size for New York.

Here,

\(\begin{array}{l}{n_1} = 12\\{n_2} = 12\end{array}\)

Compute the mean value \(\left( {{\mu _R}} \right)\).

\(\begin{array}{c}{\mu _R} = \frac{{{n_1}\left( {{n_1} + {n_2} + 1} \right)}}{2}\\ = \frac{{12\left( {12 + 12 + 1} \right)}}{2}\\ = 150\end{array}\)

Compute the standard deviation \(\left( {{\sigma _R}} \right)\).

\(\begin{array}{c}{\sigma _R} = \sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} \\ = \sqrt {\frac{{12 \times 12\left( {12 + 12 + 1} \right)}}{{12}}} \\ = 17.32\end{array}\)

Calculate the test statistic

The test statistic is calculated below:

\(\begin{array}{c}z = \frac{{R - {\mu _R}}}{{{\sigma _R}}}\\ = \frac{{194.5 - 150}}{{17.32}}\\ = 2.57\end{array}\)

The absolute value of the z-score is 2.57.

Determine the critical value and the conclusion of the test

The critical value of z from the standard normal table for a two-tailed test with \(\alpha = 0.05\) is equal to 1.96.

As the test statistic value is greater than the critical value, the null hypothesis is rejected.

There isenough evidence to conclude that there is a difference in the median values of the amount of strontium in baby teeth between the cities of Pennsylvania and New York.