Question

An article describes a study comparing two methods of measuring mean arterial blood pressure. The auscultatory method is based on listening to sounds in a stethoscope, while the oscillatory method is based on oscillations in blood flow. Following are measurements on six subjects in mmHg, consistent with means and standard deviations presented in the article. Auscultatory Oscillatory Difference 92.9 86.3 6.6 101.5 97.3 4.2 74.3 79.8 −5.5 95.0 98.1 −3.1 91.4 82.1 9.3 80.6 84.5 −3.9 Can you conclude that the mean reading is greater for the auscultatory method?

Answer 1

Answer:

$p_v =P(t_{(5)}>0.499) =0.319$

The p value is higher than any significance level given for example $\alpha=0.05,0.1$, so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations we can use it.  

Let put some notation  

x=Auscultatory method , y = Oscillatory method

x: 74 86 84 79 70 78 79 70

y: 70 85 90 110 71 80 69 74

The system of hypothesis for this case are:

Null hypothesis: $\mu_x- \mu_y \leq 0$

Alternative hypothesis: $\mu_x -\mu_y >0$

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

d: 6.6, 4.2, -5.5, -3.1, 9.3, -3.9

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=1.267$

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

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =6.215$

The fourth step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.267 -0}{\frac{6.215}{\sqrt{6}}}=0.499$

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

$df=n-1=6-1=5$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(5)}>0.499) =0.319$

The p value is higher than any significance level given for example $\alpha=0.05,0.1$, so then we can conclude that we FAIL to reject the null hypothesis. So we don't have enough evidence to conclude that the mean reading is greater for the auscultatory method at 5% or 10% of significance.