Question

Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the differences between right and left arm blood pressure measurements.
Right Arm 102 101 94 79 79
Left Arm 175 169 182 146 144
a. Find the values of d and sd (you may use a calculator).
b. Construct a 90% confidence interval for the mean difference between all right and left arm blood pressure measurements.

Answer 1

Answer:

a) $\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2$

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311$

b) $63.331 < \mu_{left arm}-\mu_{right arm} <81.069$  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution

Let put some notation  

x=value for right arm , y = value for left arm

x: 102, 101,94,79,79

y: 175,169,182,146,144

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: 73, 68, 88, 67, 65

Part a

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311$

Part b

The next step is calculate the degrees of freedom given by:

$df=n-1=5-1=4$

Now we need to calculate the critical value on the t distribution with 4 degrees of freedom. The value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$, so we need a quantile that accumulates on each tail of the t distribution 0.05 of the area.

We can use the following excel code to find it:"=T.INV(0.05;4)" or "=T.INV(1-0.05;4)". And we got $t_{\alpha/2}=\pm 2.13$

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

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

Now we have everything in order to replace into formula (1):  

$72.2-2.13\frac{9.311}{\sqrt{5}}=63.331$  

$72.2+2.13\frac{9.311}{\sqrt{5}}=81.069$  

So on this case the 90% confidence interval would be given by (63.331;81.069).

$63.331 < \mu_{left arm}-\mu_{right arm} <81.069$