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

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

b) The 90% confidence interval would be given by (63.330;81.070)

$63.330 < \mu_{left}- \mu_{right}$  

Step-by-step explanation:

1) Previous concepts  and notation

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".

Let put some notation  

x=left arm , y = right arm  

x: 175 169 182 146 144  

y: 102 101 94 79 79

The first step is define the difference $d_i=x_i-y_i$, that is given so we have:

d:  73, 68, 88, 67, 65

The second step is calculate the mean difference  

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

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

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

2) Confidence interval

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=5-1=4$  

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,4)".And we see that $t_{\alpha/2}=2.13$  

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

$72.2-2.13\frac{9.311}{\sqrt{5}}=63.330$  

$72.2+2.13\frac{9.311}{\sqrt{5}}=81.070$  

So on this case the 90% confidence interval would be given by (63.330;81.070)

$63.330 < \mu_{left}- \mu_{right}$