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SSPS 15 Please show your work Problem-1 In a study of factors thought to be resp

ID: 2959974 • Letter: S

Question

SSPS 15

Please show your work


Problem-1
In a study of factors thought to be responsible for the adverse effects of smoking on human reproduction, cadmium level determinations (in nanograms per gram) were made on placenta tissue of a sample of 14 mothers who were smokers and an independent random sample of 18 nonsmoking mothers. The results were as follows:
Nonsmokers: 10.0, 8.4, 12.8, 25.0, 11.8, 9.8, 12.5, 15.4, 23.5,
9.4, 25.1, 19.5, 25.5, 9.8, 7.5, 11.8, 12.2, 15.0
Smokers: 30.0, 30.1, 15.0, 24.1, 30.5, 17.8, 16.8, 14.8,
13.4, 28.5, 17.5, 14.4, 12.5, 20.4
Source: Adapted from Daniel, Wayne D. (2005). Biostatistics: A Foundation for Analysis in the Health Sciences.
NY: Wiley. Exercise 6.4.10: page 181
Instructions:
Is there significant evidence at the = 0.10 level that the placenta tissue of mothers who smoke contain, on average, a higher level of cadmium (in nanograms per gram) than the placenta tissue of mothers who do not smoke? Be sure to include your hypothesis and describe the appropriate analysis for this problem.
Specify/discuss the assumption(s) that needs to be met so that your inference is valid. In other words, can you trust the results of your inference? Check the assumption(s) with the appropriate SPSS output and discuss/interpret the results in your write-up.

Explanation / Answer

Nonsmokers: n1=18, xbar1=14.72, s1= 6.2
Smokers:n2=14, xbar2=20.41, s2=6.81

The test hypothesis is
Ho:1=2
Ha:1<2

The test statistic is

t=(xbar1-xbar2)/[s1^2/n1 + s2^2/n2]

=(14.72-20.41)/sqrt(6.2^2/18 + 6.81^2/14)

= -2.44

Given a=0.1, the criticial value is t(0.1, df=n1+n2-2=18+14-2 =30) = -1.31 (check student t table)

Since t=-2.44 < -1.31, we reject Ho.

SO we can conclude that the placenta tissue of mothers who smoke contain, on average, a higher level of cadmium (in nanograms per gram) than the placenta tissue of mothers who do not smoke.