Statistics- confidence interval

Question

n a randomized comparative experiment on the effect of calcium in the diet on blood pressure, researchers divided 54 healthy white males at random into two groups. One group received calcium; the other, a placebo. At the beginning of the study, the researchers measured many variables on the subjects. The paper reporting the study gives x = 114.9 and s = 9.3 for the seated systolic blood pressure of the 27 members of the placebo group.


Give a 95% confidence interval for the mean blood pressure in the population from which the subjects were recruited.
What conditions for the population and the study design are required by the procedure you used in (a)? Which of these conditions are important for the validity of the procedure in this case?

Answer 1

Hello,

 

Any confidence interval can be found using the following equation.

 

X ± t*SE

Where X = the mean, t equals the t critical value for your confidence level and SE is the standard error of the mean.

 

To find t, we need to consult the normal distribution table (your teacher should have giveny ou one or there should be one in your textbook). We need to find the value critical value that corresponds with an alpha level of 5% (or 0.05) and 26 degrees of freedom (27 - 1). This value is 2.056.

 

Next we need the SE. The SE = s/sqrt(n). In your case, it is 9.3/sqrt(27) = 1.789. Now we plug everything in.

 

114.9 + 2.056*1.789 = 118.58

114.9 - 2.056*1.789 = 111.22

 

So the confidence interval is 111.22 to 118.58.

 

We need to make a few assumptions in order for this to be valid. Some of the most important are 1) these 27 have to be a random sample from the population and 2) the variable must be normally distributed