Question

(23.01) In a randomized comparative experiment on the effect of dietary calcium on blood pressure, researchers divided 58 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⎯⎯⎯=113.7 and s = 9.1 for the seated systolic blood pressure of the 29 members of the placebo group. Give a 96% confidence interval for the mean blood pressure in the population from which the subjects were recruited is.

Answer 1

Answer:

$113.7-2.154\frac{9.1}{\sqrt{29}}=110.06$    

$113.7+2.154\frac{9.1}{\sqrt{29}}=117.34$    

And the 96% confidence is given by (110.06; 117.34)

Step-by-step explanation:

Information given

$\bar X=113.7$ represent the sample mean

$\mu$ population mean (variable of interest)

s=9.1 represent the sample standard deviation

n=29 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are given by:

$df=n-1=29-1=28$

The Confidence is 0.96 or 96%, the significance is $\alpha=0.04$ and $\alpha/2 =0.02$, and the critical value would be $t_{\alpha/2}=2.154$

Replacing the info we got:

$113.7-2.154\frac{9.1}{\sqrt{29}}=110.06$    

$113.7+2.154\frac{9.1}{\sqrt{29}}=117.34$    

And the 96% confidence is given by (110.06; 117.34)