Two Dependent Samples (Matched Pairs) question

Question

 

Please help me with this question and help explain what the answer is and how you got it step-by-step. Thanks!!

 

Question/Problem:

 

"Before/After Treatment Results... Captopril is a drug designed to lower systolic blood pressure. When subjects were treated with this drug, their systolic blood pressure readings (in mm Hg) were measured before and after the drug was taken. Results are given in the accompanying table. Using a 0.01 significance level, is there sufficient evidence to support the claim that Captopril is effective in lowering systolic blood pressure?"

 

Here is the table: 

 

Subjects:

A --> Before: 200, After: 191

B --> Before: 174, After: 170

C --> Before: 198, After: 177

D --> Before: 170, After: 167

E --> Before: 179, After: 159

F --> Before: 182, After: 151

G --> Before: 193, After: 176

H --> Before: 209, After: 183

I --> Before: 185, After: 159

J --> Before: 155, After: 145

K --> Before: 169, After: 146 

L --> Before: 210, After: 177

Answer 1

I think the key difference in this problem is that the data are paired. In other words, a measurement is taken on each person twice: before and after treatment. Therefore, it isn't a two-sample t-test of independent samples. The samples are dependent. 

 

The t-test calculations are different for paired data. Here, we are looking at the difference values before and after:

 

A --> 200 - 191 = 9
B --> 174 - 170 = 4
C --> 198 - 177 = 21
D --> 170 - 167 = 3
E --> 179 - 159 = 20
F --> 182 - 151 = 31
G --> 193 - 176 = 17
H --> 209 - 183 = 26
I --> 185 - 159 = 26
J --> 155 - 145 = 10
K --> 169 - 146 = 23
L --> 210 - 177 = 33

 

The mean of the differences is 18.58. The SD = 10.1. The sample size is still n= 12. So, the t-statistic is:

 

18.58 / (10.1/sqrt(12) ) = 6.37

 

An alternative hypothesis could be Mean(before) > Mean(after).

 

Looking at the t-distribution for 6.37 and degrees of freedom = 11 (n - 1 => 12 - 1), the p-value = .0000246 or, more appropriately reported, p < .0001. Still significant, but a different t-test since the data are paired.

 

 

Answer 2

First, we look at the control group, and calculate the average. We get an average (which we will call A) of 185.3. The standard deviation was not given.

 

Then, we calculate the average of the sample group (which we will call a), which is 166.7 and its standard deviation (which we will call S) which is 14.7.

 

Then, we can use the t test, the formula for which is

t = (√n • |a - A| ) / S    (remember that the vertical lines | mean the absolute value which is always a positive number)

In this case, t = (√12 • |166.7 - 185.3|) / 14.7, which becomes

t = (3.46 • 18.6) / 14.7

t = 4.377

 

We then look for the t value in a table of t values to find its significance.

 

For a sample size of 12, the t value is greater than the value given at P = 0.01.

We conclude that, based on this sample, the results are not due to chance.