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 significance level to test for a difference between the measurements from the two arms. What can be​ concluded? right arm: 148, 139, 121, 129, 129 left arm: 182, 173, 189, 150, 147 In this​ example, 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? Identify the test statistic. Identify the​ P-value. What is the conclusion based on the hypothesis​ test?

Answer 1

Sample 1 Sample 2 Difference = Sample 1 - Sample 2 148 182 -34 139 173 -34 121 189 -68 129 150 -21 129 147 -18 Average 133.2 168.2 -35 St. Dev. 10.45 18.887 19.849 n 5 5 5 From the sample data, it is found that the corresponding sample means are: X1 = 133.2 X2 = 168.2 Also, the provided sample standard deviations are: $1 = 10.45 S2 = 18.887 and the sample size is n = 5. For the score differences we have D = -35 SD = 19.849 (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: up = 0 Ha: pp 70 This corresponds to a two-tailed test, for which a t-test for two paired samples be used. (2) Rejection Region 0.05, and the Based on the information provided, the significance level is a = degrees of freedom are df = 4. Hence, it is found that the critical value for this two-tailed test is tc = 2.776, for a = 0.05 and df = 4. The rejection region for this two-tailed test is R = {t: \t| > 2.776}. (3) Test Statistics The t-statistic is computed as shown in the following formula: t D SD/m -35 19.849/5 -3.943 (4) Decision about the null hypothesis = 3.943 >te = 2.776, it is then concluded that the Since it is observed that t null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0169, and since p = 0.0169 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean ui is different than um, at the 0.05 significance level.

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