Question

In a genetic inheritance study discussed by Margolin, blood was collected from samples of individuals from several ethnic groups, and mean sister chromatid exchange (MSCE) was measured for each individual. We wish to see if there's a difference in average MSCE for the groups labeled "Native American" and "Caucasian." Here's the data:

Native American: 8.50, 9.48, 8.65, 8.16, 8.83, 7.76, 8.63

Caucasian: 8.27, 8.20, 8.25, 8.14, 9.00, 8.10, 7.20, 8.32, 7.70

a. What's the experimental unit? What measurements are taken on the experimental units? Is this a problem with one or two independents samples?

b. Carefully define your parameter of interest and give null and alternative hypotheses for an appropriate t-test.

c. Perform an appropriate t-test, and give your t-statistic, degrees of freedom, and P-value. State any assumptions you make.

d. Give a 95% t-confidence interval for your parameter of interest.

e. The P-value you should get isn't that small. Should we then conclude that there's no difference in average MSCE between Native Americans and Caucasians? Explain why or why not.

Answer 1

(a) here experimental unit is native american and Caucasian.

Mean sister chromatid exchange (MSCE) is measured on experimental unit.

This is a problem of two independents samples.

(b) difference in average MSCE is the parameter of interest.

here we use t-test with

null hypothesis H0:mean1=mean2 and

alternate hypothesis H1:mean1≠mean2

(c)

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2)=0.4417/0.2562=1.7244

with df is n=n1+n2-2=7+9-2=14 and s_p²=((n1-1)s1²+(n2-1)s2²)/n

t-test sample mean s s2 n (n-1)s2 Native Americal(x1) 8.5729 0.5377 0.2891 7 1.7347 Caucasian(x2) 8.1311 0.4851 0.2353 9 1.8827 difference= 0.4417 0.5245 16 3.6174 sp2= 0.2584 sp= 0.5083 SE= 0.2562 t= 1.7244 two tailed p-value= 0.1066 critical t(0.05, 11) 2.1448

assumption: sample should come from normally distributed population

(d)

(1-alpha)*100% confidence interval for population mean difference=

=sample mean difference±t(alpha/2,n)*SE(difference)=

95% confidence interval =0.4417±2.1448*0.2562=0.4417±0.5495=(-0.1078,0.9912)

(e) we should conclude that there's no difference in average MSCE between Native Americans and Caucasians.

since p-value is greater than the alpha=level of significance=0.05, so we fail to reject ( or accept) H0.