The differences di=Xi−Yi d_i = X_i - Y_i di=Xi−Yi for each observation are calculated as follows:
d=[−4.3,−1.1,−1.3,−5.4,−1.4,−2.4,−2.2,−0.6] d = [-4.3, -1.1, -1.3, -5.4, -1.4, -2.4, -2.2, -0.6] d=[−4.3,−1.1,−1.3,−5.4,−1.4,−2.4,−2.2,−0.6]
The mean of the differences d‾ \overline{d} d is calculated using the formula:
d‾=∑i=1NdiN=−18.78=−2.337 \overline{d} = \frac{\sum_{i=1}^N d_i}{N} = \frac{-18.7}{8} = -2.337 d=N∑i=1Ndi=8−18.7=−2.337
The standard deviation sd s_d sd is calculated using the formula for sample standard deviation:
sd=∑(di−d‾)2n−1=1.68 s_d = \sqrt{\frac{\sum (d_i - \overline{d})^2}{n-1}} = 1.68 sd=n−1∑(di−d)2=1.68
We will test the hypothesis H0:μd=0 H_0: \mu_d = 0 H0:μd=0 against the alternative hypothesis H1:μd<0 H_1: \mu_d < 0 H1:μd<0 at the significance level α=0.05 \alpha = 0.05 α=0.05.
Calculate Standard Error (SE): SE=sdn=1.688≈0.594 SE = \frac{s_d}{\sqrt{n}} = \frac{1.68}{\sqrt{8}} \approx 0.594 SE=nsd=81.68≈0.594
Calculate Test Statistic: t=d‾−μ0SE=−2.337−00.594≈−3.9345 t = \frac{\overline{d} - \mu_0}{SE} = \frac{-2.337 - 0}{0.594} \approx -3.9345 t=SEd−μ0=0.594−2.337−0≈−3.9345
Calculate P-value: For a left-tailed test, the P-value is given by: P=T(z)≈0.0028 P = T(z) \approx 0.0028 P=T(z)≈0.0028
The calculated test statistic is approximately −3.9345-3.9345−3.9345 and the P-value is 0.00280.00280.0028. Since the P-value 0.00280.00280.0028 is less than the significance level 0.050.050.05, we reject the null hypothesis.
Thus, the answer is: H0:μd=0 \boxed{H_0: \mu_d = 0} H0:μd=0
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