Math  /  Data & Statistics

QuestionPart 4 of 8 HW Score:
Assume that the differences for the given data are normally distributed. Complete parts (a) through (d). Click here to view the sample data. Click here to view the table of critical t-values. Points: (a) Determine di=XiYid_{i}=X_{i}-Y_{i} for each pair of data. \begin{tabular}{|l|c|c|c|c|c|c|c|} \hline Observation & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline di\mathbf{d}_{\mathbf{i}} & -0.5 & 1.1 & -3.3 & -3.8 & 0.5 & -2.3 & -2.9 \\ \hline \end{tabular} (b) Compute d\overline{\mathrm{d}} and sd\mathrm{s}_{\mathrm{d}}. d=1.600\overline{\mathrm{d}}=-1.600 (Round to three decimal places as needed.) sd=1.950s_{d}=1.950 (Round to three decimal places as needed.) (c) Test if μd<0\mu_{\mathrm{d}}<0 at the α=0.05\alpha=0.05 level of significance.
Determine the null and alternative hypotheses. Choose the correct answer. A. H0:μd=0H_{0}: \mu_{d}=0 H1:μd<0\mathrm{H}_{1}: \mu_{\mathrm{d}}<0 B. H0:μd<0H_{0}: \mu_{d}<0 C. H0:μd>0H_{0}: \mu_{d}>0 H1:μd=0\mathrm{H}_{1}: \mu_{\mathrm{d}}=0 H1:μd<0H_{1}: \mu_{d}<0 D. H0:μd<0H_{0}: \mu_{d}<0 H1:μd>0\mathrm{H}_{1}: \mu_{\mathrm{d}}>0

Studdy Solution
We reject the null hypothesis (H0:μd0H_0: \mu_d \ge 0) in favor of the alternative hypothesis (H1:μd<0H_1: \mu_d < 0).
There is sufficient evidence at the α=0.05\alpha = 0.05 level of significance to conclude that μd<0\mu_d < 0.

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