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 0.05 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

Right_arm(mm_Hg) Left_arm(mm_Hg)
149 166
136 179
129 190
137 148
139 138

Data was entered in SPSS using the paired t-test approach!!


a. In this​ example, μd 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?

b.) Identify the test statistic.

c.) Identify the​ P-value.

d.) What is the conclusion based on the hypothesis​ test?

Respuesta :

Answer:

There is a significant difference in the systolic blood pressure measurements between the two arms.

Step-by-step explanation:

The dependent t-test (also known as the paired t-test or paired samples t-test) compares the two means associated groups to conclude if there is a statistically significant difference amid these two means.

In this case a paired t-test is used to determine whether there is a difference in the systolic blood pressure measurements between the two arms.

The SPSS output is attached below.

(a)

The hypothesis for the test can be defined as follows:

H₀: There is no difference in the systolic blood pressure measurements between the two arms, i.e. d = 0.

Hₐ: There is a significant difference in the systolic blood pressure measurements between the two arms, i.e. d ≠ 0.

(b)

Consider the SPSS output.

The test statistic value is t = 0.871.

(c)

Consider the SPSS output.

The p-value of the test is:

p-value = 0.433.

(d)

The significance level of the test is, α = 0.05.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

p-value = 0.433 > α = 0.05

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

Thus, it can be concluded that there is a significant difference in the systolic blood pressure measurements between the two arms.

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