Negative Confidence Interval

eamanfor

New Member
3. In two wards for elderly women in a nursing home the following levels of hemoglobin were found:

Ward A: 12.2, 11.1, 14.0, 11.3, 10.8, 12.5, 12.2, 11.9, 13.6, 12.7, 13.4, 13.7 grams/dl

Ward B: 11.9, 10.7, 12.3, 13.9, 11.1, 11.2, 13.3, 11.4, 12.0, 11.1 g/dl

Ward A Ward B
n = 12 n = 10
mean = 12.45 mean = 11.89
SD = 1.07 SD=1.03

(a) What is the difference between the mean hemoglobin level in the two wards, and what is the significance, if any, of the difference? State which test you used to determine the answer to this question, along with your hypotheses, and then input the data into SPSS to conduct the statistical test. You can create one variable for “hemoglobin” and one for “ward” to do this in SPSS (Hint: you will have to define the ‘grouping variable’ for the analysis so be sure to give your “ward” variable a numeric value; you cannot use “A” and “B”).

The test used is a two sided t-test.
Step 1.
Ward A Ward B
N1 = 12 N2 = 10
X1 = 12.45 X2 = 11.89
SD 1= 1.07 SD 2=1.03

[(12-1)(12.45)2 +(10-1)(11.89)2]/(12+10-2)
11*155=1705;9*141.37=1272.34;1705+1272.34=2977.34
12+10=22-2=20;2977.34/20=148.87

Step 2.
Sqrt(148.87/12 +148.87/10)=sqrt(12.40+14.887)=5.224

Step 3.
(11.89-12.45)/5.22=-0.107
Shows that at 20 degrees of freedom (that is (12-1)+(10-1)), t=0.107 does not lie between 2.086 and 2.528. Consequently, this degree probability is different from the conventional level. The null hypothesis that there is no difference between the means is therefore somewhat likely.

(b) What is the 95% confidence interval for the difference in treatments? Does this lead you to the same conclusion as your answer in part a?
12.45-11.89 2.08*5.224
0.56-10.87 to 0.56+10.87= -10.31,11.43
I included the formulas but they are not showing up. The negative answers are very confusing especially if the X1 and X2=mean gives you a positive answer.

hlsmith

Less is more. Stay pure. Stay poor.
I did not go through and check your work, but big picture, a confidence interval around a difference that has both a negative and positive value, just means it included 0. If your alternative hypotheseis was a difference (greater or less than 0, which "0" would be no difference), then you fail to reject the null hypothesis of no effect, since the confidence interval included "0".