Q:

# Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What do you conclude? Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal.Right arm102101947979Left arm175169182146144

Accepted Solution

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Answer:The p value is lower than the significance level given $$\alpha=0.05$$, so then we can conclude that we can reject the null hypothesis that no difference between the two measures. So there is a significant difference between the measurements from the two arms.Step-by-step explanation:We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.  A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.   Let put some notation  x=test value with right arm , y = test value with left arm x: 102, 101, 94, 79, 79  y: 175, 169, 182, 146, 144  The system of hypothesis for this case are: Null hypothesis: $$\mu_y- \mu_x = 0$$ Alternative hypothesis: $$\mu_y -\mu_x \neq 0$$ The first step is calculate the difference $$d_i=y_i-x_i$$ and we obtain this: d: 73, 68, 88, 67, 65 The second step is calculate the mean difference  $$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2$$ The third step would be calculate the standard deviation for the differences, and we got: $$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311$$ The 4 step is calculate the statistic given by : $$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{72.2 -0}{\frac{9.311}{\sqrt{5}}}=17.339$$ The next step is calculate the degrees of freedom given by: $$df=n-1=5-1=4$$ Now we can calculate the p value, since we have a two tailed test the p value is given by: $$p_v =2*P(t_{(4)}>17.339) =6.49x10^{-5}$$ The p value is lower than the significance level given $$\alpha=0.05$$, so then we can conclude that we can reject the null hypothesis that no difference between the two measures. So there is a significant difference between the measurements from the two arms.