ach me = 0.22 pear to differ with respect to aver- showed of 2. At say th nitrog

Question

ach me = 0.22 pear to differ with respect to aver- showed of 2. At say th nitroge and M details e shear strength, at the 1% ag o Significance level? study was conducted by the Florida Game d Fish Commission to assess the amounts 10.40 an of chemical residues found in the brain tissue 10.43 In the brown pelicans. For DDT, random sample of = 10 juveniles and n2 = 13 nestlings gave the following results (measurements in parts as no troubl ples the M corre and of A per million): Juveniles nl = 10 y = 0.041 SI = 0.017 Nestlings n2 13 ½ = 0.026 Nativ = 0.016 Non lest the hypothesis that there is no difference nd in iuve-

Explanation / Answer

Q.10

H0: 1 = 2

Ha : 1  2

Variances are about same and equal to each other.

pooled standard deviation sp = [{(n1-1)s12+ (n2-1)s22} / (n1+ n2-2)]

sp = [ (9 * 0.0172 + 12 * 0.0162) /21] = 0.01644

standard error of the difference se0= sp * sqrt[1/n1+ 1/n2] = 0.01644 * sqrt (1/10 + 1/13) = 0.00697

Test statistic

Z =(x1-x2)/  se0 = (0.041 - 0.026)/ 0.0069 = 2.152

Here for alpha level = 0.01

Z = 2.575

so here Z <  Zcriticalso we shall not reject the null hypothesis and there is no significant difference between DDT level for JUveniles and Nestlings.

Juveniles (1) Nestlings(2) n 10 13 y 0.041 0.026 s 0.017 0.016

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