Question

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 149 131 130 129 137 Left arm 174 161 173 137 139 Identify the test statistic. (round to two decimals) what is the P value? (round to three decimals) what is the conclusion based on the hypothesis test?

Answer 1

Solution:

State the hypotheses.

Let μ_d be the population mean difference between the measurements from the two arms

H0: μ_d = 0      
Ha: μ_d ≠ 0

From the paired data, the mean and standard deviation for the paired differences (d = Right arm – Left arm) are:
Mean, d-bar = -21.6
Standard deviation s_d = 16.6523
Sample size, n = 5

Compute the test statistic
t_STAT = (d-bar - μ_d)/(s_d/√n)
= (-21.6 - 0)/(16.6523/√5) = -2.90

Number of pairs, n = 5
Level of significance α=0.10
df = 5-1 = 4
P-value = 0.0441
Conclusion: P-value is less than α value, so fail to reject the null hypothesis.