Question

1. listed below are systolic blood pressure measurements (mmhg) taken from the right and left arms of the same woman. use a 0.01 significance level to test for a difference between the measurements from the two arms. what do you conclude? right arm 105 101 124 98 119 left arm 102 110 112 106 124

Explanation:

Step1: Calculate the differences

Let \(d=\text{Right arm}-\text{Left arm}\).
For the given data:
\(d_1 = 105 - 102=3\), \(d_2=101 - 110=-9\), \(d_3=124 - 112 = 12\), \(d_4=98 - 106=-8\), \(d_5=119 - 124=-5\)

Step2: Calculate the mean of differences \(\bar{d}\)

\(\bar{d}=\frac{\sum_{i = 1}^{n}d_i}{n}\)
\(\sum_{i=1}^{5}d_i=3-9 + 12-8-5=-7\)
\(n = 5\)
\(\bar{d}=\frac{-7}{5}=-1.4\)

Step3: Calculate the standard deviation of differences \(s_d\)

$$s_d=\sqrt{\frac{\sum_{i = 1}^{n}(d_i-\bar{d})^2}{n - 1}}$$

\((d_1-\bar{d})=(3-(-1.4)) = 4.4\), \((d_2-\bar{d})=(-9-(-1.4))=-7.6\), \((d_3-\bar{d})=(12-(-1.4)) = 13.4\), \((d_4-\bar{d})=(-8-(-1.4))=-6.6\), \((d_5-\bar{d})=(-5-(-1.4))=-3.6\)
\(\sum_{i = 1}^{n}(d_i-\bar{d})^2=(4.4)^2+(-7.6)^2+(13.4)^2+(-6.6)^2+(-3.6)^2\)
\(=19.36 + 57.76+179.56+43.56 + 12.96=313.2\)
\(s_d=\sqrt{\frac{313.2}{4}}\approx\sqrt{78.3}\approx8.85\)

Step4: Calculate the test statistic \(t\)

The test statistic for a paired - t test is \(t=\frac{\bar{d}-\mu_d}{s_d/\sqrt{n}}\), where \(\mu_d = 0\) (null hypothesis \(H_0:\mu_d=0\), alternative hypothesis \(H_1:\mu_d
eq0\))
\(t=\frac{-1.4-0}{8.85/\sqrt{5}}\approx\frac{-1.4}{3.96}\approx - 0.35\)

Step5: Determine the critical value

The degrees of freedom \(df=n - 1=5-1 = 4\). For a two - tailed test with \(\alpha = 0.01\), the critical values are \(t_{\alpha/2,df}=t_{0.005,4}\)
From the t - distribution table, \(t_{0.005,4}=\pm5.598\)

Step6: Make a decision

Since \(\vert t\vert=\vert- 0.35\vert=0.35

Answer:

We fail to reject the null hypothesis \(H_0:\mu_d = 0\). There is not sufficient evidence at the \(0.01\) significance level to conclude that there is a difference between the systolic blood pressure measurements from the two arms.