Listed below are systolic blood pressure measurements (mm Hg) taken from the rig
ID: 3254228 • Letter: L
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
0.01
0.01 significance level to test for a difference between the measurements from the two arms. What can be concluded?
Right arm
152
152
135
135
140
140
133
133
129
129
Left arm
176
176
173
173
172
172
141
141
149
149
In this example, d is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis test?
A.H0: d0
H1:d=0
B. H0: d0
H1:d>0
C. H0: d=0
H1: d<0
D. H0: d=0
H1: d0
Identify the test statistic.
t=_____ (Round to two decimal places as needed.)
Identify the P-value.
P-value=____ (Round to three decimal places as needed.)
What is the conclusion based on the hypothesis test?
Since the P-value is
greater
less
than the significance level,
reject
fail to reject
the null hypothesis. There
is
is not
sufficient evidence to support the claim of a difference in measurements between the two arms.
Right arm
152
152
135
135
140
140
133
133
129
129
Left arm
176
176
173
173
172
172
141
141
149
149
Explanation / Answer
Solution:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = d = 0
Alternative hypothesis: 1 - 2 d 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(8.872/5) + (16.022/5)] = 8.19
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (8.872/5 + 16.022/5)2 / { [ (8.872 / 5)2 / (4) ] + [ (16.022 / 5)2 / (4) ] }
DF = 4497.51 / (61.90 + 658.64) = 6.24
t = [ (x1 - x2) - d ] / SE = [ (137.8 - 162.2) - 0 ] / 8.19 = -2.98
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 6.24 degrees of freedom is more extreme than -2.98; that is, less than -2.98 or greater than 2.98.
We use the t Distribution Calculator to find P(t < -2.98)
The P-Value is 0.023502.
The result is not significant at p < 0.01.
Interpret results. Since the P-value (0.024) is more than the significance level (0.01), we can accept the null hypothesis.
Conclusion. Fail to reject the null hypothesis. There is no difference in measurements between the two arms.
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