Note: Final answers are highlighted in colour.
In this question we have two samples in which observations in one sample(Systolic blood pressure Before) can be paired with observations in the other sample(Systolic blood pressure after). These samples are called dependent samples, and we can perform a paired t-test.
Subject | Before | After | Difference di(Before-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 |
Mean Difference() | AVERAGE(D2:D12)= 17.27272727 | ||
Standard Deviation(Sd) | STDEV.S(D2:D12)= 9.466688007 |
Hypothesis:
H0: (Where )
H1:
is the mean Systolic blood pressre before captopril drug, is the mean Systolic blood pressre after captopril drug.
A) As we are testing the alternate hypothesis of , this is going to be a Right tailed test.
B) Test statistic
t stat= =
= 6.05
C) Critical t - score at 0.01 significance level and degrees of freedom of 10(n-1) is 2.76(From t distribution table)
D) P value for t=6.05, degrees of freedom=10 is 0.000062
P - value > 0.00001 (Option v)
E) As p value(0.000062) < 0.01, we can reject the null hypothesis and conclude that Systolic blood pressure has reduced due to captopril drug.
Based on the statistical testing, Yes we conclude that the captopril drug lowers the blood pressure.
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