3) Here we have:
Sample mean of difference(x̅d )=
10.1
Sample standard deviation of difference(sd
)= 11.2
n = 13
Now
Hypothesis:
Ho : µd = 0
H1 : µd > 0
Test statistic:
T test
t = (x̅d)/(sd/√n) =10.1/
11.2/√13=3.25
At n=13 df = 13-1=12
P-value = TDIST( 3.25, 12, 1 ) = 0.003478 0.0035
Hence the p value is 0.0035 (Ans)
p-value < α, Reject the null hypothesis.
There is sufficient evidence to conclude that the hypertension medicine lowered blood pressure.
4) The statement is TRUE
A random sample of size 13 is selected from men with hypertension. For each person, systolic blood pressure was measured right before and one hour after taking the medicine. The mean reduction o...
A random sample of size 13 is selected from men with hypertension. For each person, systolic blood pressure was measured right before and one hour after taking the medicine. The mean reduction of the blood pressures was 10.1 and the standard deviation of the difference was 11.2. Is there sufficient evidence to conclude that the hypertension medicine lowered blood pressure? Test using Assume normal population.
Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.01 significance level to test for a difference between the measurements from the two arms. What can be concluded? Right arm 137 145 165 131 171 136 174 139 144 Left arm 137 In this example....
Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.10 significance level to test for a difference between the measurements from the two arms. What can be concluded? Right arm Left arm 148 183 142 179 126 187 136 145 130 143 In this example,...