*Please show work 13.4 – Blood Pressure Measurements. Listed below are systolic

Question

*Please show work

13.4 – Blood Pressure Measurements. Listed below are systolic blood pressure measurements (mmHg) obtained from the same woman (based on data from “Consistency of Blood Pressure Differences Between the Left and Right Arms,” by Eguchi, et al., Archives of Internal Medicine, Vol. 167). Is there sufficient evidence to that there is a linear correlation between right and left arm systolic blood pressure measurements?

a) Construct a scatterplot. We can use the TI-83/84 Plus to construct a scatter plot:

1. Press STAT & choose 1: Edit

2. Enter the data from right arm into L1 (press ENTER between each value)

3. Enter the data from left arm into L2 (press ENTER between each value)

4. Press 2ND and then Y= (STAT PLOT) and highlight 1: Plot 1…Off

5. Press ENTER to turn ON Plot1. Make sure that Xlist is L1 and Ylist is L2.

6. Select the first type of plot and make sure all other graphs are turned off.

7. To view the data points in the best window possible, press ZOOM & then highlight 9: ZoomStat & press ENTER

8. Draw or screenshot the resulting scatter plot window.

b) Find the value of the linear correlation coefficient r (using TI-83/84 Plus).

1. Press STAT & choose TESTS & select LinRegTTest & Press ENTER

2. Make sure that Xlist is L1.

3. Make sure that Ylist is L2

4. Make sure Freq is 1.

5. For the ? & ? highlight the ? 0 choice

6. Select “Calculate” and press ENTER.

7. Record r

c) Find the P-value or the critical values of r from Table A-6 using ? = 0.05.

d) Determine whether there is sufficient evidence to support a claim of a linear correlation between the two variables.

Explanation / Answer

If r is the observed correlation coefficient in a sample of n pairs of observations from a bivariate normal population, then under the null hypothesis, H0 : r = 0, i.e., population correlation coefficient is zero, the statistic

t = r*V(n-2)/(1-r2) follows Student's t-distribution with (n-2) df.

Where coefficient of correlation r = Cov (X, Y) / [var(X).Var(Y)]

= (NXY - XY) / [{NX2 - (X)2}*{NY2 - (Y)2}]

After calculating all the values of summations etc., we get

r = 0.867

It means X and Y are highly correlated.

Now t = r*V(n-2)/(1-r2) = 0.867 *V(5-2) /V(1- (0.867)2 )= 3.014

(c) & (d) At 0.05 level the value of t = 3.18 for 3 df

AS the calculated value of t is less than the tabulated value of t, there is sufficient evidence to support a claim of a linear correlation between the two variables.

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