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?

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.

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation