Almost twothirds of iron in the body is found in hemoglobin

Almost two-thirds of iron in the body is found in hemoglobin, the protein in the red blood cells that carries oxygen to the tissues. Iron deficiency could lead to anemia, a condition characterized by less than the normal quantity of hemoglobin in the blood. The following data gives the hemoglobin values for 7 female patients at risk for anemia, before and after they followed a 3-month dietary iron intake program:

a) Was the program efficient in increasing the hemogloblin level? Justify your answer using a test of hypothesis for H₀: mu₁= mu₂ against H₁: mu₁< mu₂ , where mu₁ and mu₂ are the average hemoglobin levels before and after the program, respectively. Use the level alpha=0.01

b) What is the conclusion of a test of level alpha=0.01, if we assume (incorreclty) that the hemoglobin level after the program is independent of the hemoglobin level before the program? Assume that the two populations are normally distributed with equal variances.

Solution

a)

Set Up Hypothesis
Null, Ho: u1 > u2
Alternate, - H1: u1 < u2
Test Statistic
X(Mean)=11.6
Standard Deviation(s.d1)=1.1343 ; Number(n1)=7
Y(Mean)=12.9857
Standard Deviation(s.d2)=1.3643; Number(n2)=7
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =11.6-12.9857/Sqrt((1.28664/7)+(1.86131/7))
to =-2.07
| to | =2.07
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 6 d.f is 3.143
We got |to| = 2.06635 & | t | = 3.143
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Left Tail - Ha : ( P < -2.0664 ) = 0.04216
Hence Value of P0.01 < 0.04216,Here We Do not Reject Ho

b) WHEN POPULATION WITH EQUAL VARIANCE

X (Mean)=11.6; Standard Deviation (s.d1)=1.1343
Number(n1)=7
Y(Mean)=12.9857; Standard Deviation(s.d2)=1.3643
Number(n2)=7
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
S^2 = (6*1.2866 + 6*1.8613) / (14- 2 )
S^2 = 1.574
we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=11.6-12.9857/Sqrt((1.574( 1 /7+ 1/7 ))
to=-1.3857/0.6706
to=-2.0663
| to | =2.0663
Critical Value
The Value of |t | with (n1+n2-2) i.e 12 d.f is 1.782
We got |to| = 2.0663 & | t | = 1.782
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value: Left Tail - Ha : ( P < -2.0663 ) = 0.03054
Hence Value of P0.05 > 0.03054,Here we Reject Ho