Statistics Homework Solutions 147

# Statistics Homework Solutions 147 - 147 SECTION 6.8 7 071 2...

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Unformatted text preview: 147 SECTION 6.8 7 071. 2 643 Â¡ Â¤Â£ Â¨ Â¡ 3 015 Â¨ Â¡ Â¨ Â¡ Â¤ Â¡ Â¡ Â¡ 0 001. A computer package gives P 0 00040. Â¨ Â¤ Â¥ Â£ Â¨ From the t table, P 7 071 Â¡ âˆ†, the P-value is the sum of the areas to the right of t Since the alternate hypothesis is of the form ÂµD and to the left of t 7 071. Â¡ 7 0. Â¨ 1 6036 Â¡Â¥ 0 Â¨ 4 2857 Â¡ t 0 versus H1 : ÂµD Â¡ The null and alternate hypotheses are H0 : ÂµD Â¡ We can conclude that there is a difference in latency between motor point and nerve stimulation. 1 5.     1 Â¡ 10. There are 10 9 degrees of freedom. Â¡  Â¡  Â¡ 2 9907, n  Â¡ Â¨ 2 5, sD 15  Â¡ Â¨ Â¡ D 10734 Â¡ The differences are 4  3. 0. 2 643. âˆ†, the P-value is the sum of the areas to the right of t Â¨ Â¡ Â¨ Â¨ Â¤ Â¡ 0 05. A computer package gives P 0 027. Â¨ Â¡ Â¡ Â¡ P Â¡ Â¡ From the t table, 0 02 Â¨ Â¤ Â©Â¥ Â£ Since the alternate hypothesis is of the form ÂµD and to the left of t 2 643. Â¨ 10 Â¤Â£ 2 9907 Â¡Â¥ 0 Â¨ 25 Â¡ t 0 versus H1 : ÂµD Â¡ The null and alternate hypotheses are H0 : ÂµD Â¨ Â¡ We can conclude that the etch rates differ between center and edge. Â¡   Â¡   8. There are 8 20 1 13. 7 degrees of freedom. Â¡ 9   Â¡ Â¡ 9 3808, n 16 2 Â¡ Â¡ Â¨ 10, sD 21 4 Â¡ Â¡ Â¡ D 7  The differences are Â¡ 5. Â¡ 0. 3 015. âˆ†, the P-value is the sum of the areas to the right of t Â¤Â£ Â¨ Â¡ Â¨ Â¤ Â¡ Â¡ 0 02. A computer package gives P Â¨ Â¡ Â¡ Â¡ P 0 0195. Â¡ Â¡ From the t table, 0 01 Â¨ Â¤ Â¥ Â£ Since the alternate hypothesis is of the form ÂµD 3 015. and to the left of t Â¨ Â¡ We can conclude that the mean amount of corrosion differs between the two formulations. The differences are 35 57 14 29 31.  Â¨ 5. 4 degrees of freedom. 4 790. 0 005. A computer package gives P 0 0044. Â¡ Â¨ P 4 790. Â¨ Â¤ Â©Â¥ Â£ Â¡ Â¨ Â¤ Â¡ Â¡ Â¡ From the t table, 0 001 âˆ†, the P-value is the area to the right of t Â¡ Â¨ Since the alternate hypothesis is of the form ÂµD 0. Â¡ 5 Â¤Â£ 15 498 Â¡Â¥ 0 Â¨ 33 2 Â¨ t 0 versus H1 : ÂµD Â¨ Â¡ Â¡ The null and alternate hypotheses are H0 : ÂµD Â¡ Â¨ 1 Â¡ Â¡ There are 5  15 498, n Â¡ 33 2, sD  D  7. Â¨ We can conclude that the mean strength after 6 days is greater than the mean strength after 3 days. The differences are: 0 76 0 30 0 74 0 30 0 56 0 82.  Â¨  Â¨ 6. There are 6 1 Â¡ Â¨  Â¡ Â¨  Â¨ 0 233581, n Â¡ Â¨ Â¡ 0 58, sD  Â¨ Â¡ D Â¨ 5 degrees of freedom. Â¡ The null and alternate hypotheses are H0 : ÂµD 0 versus H1 : ÂµD Â¡ 9. 0. Â¨ 8 Â¡Â¥ 9 3808 Â¡ 0 Â¨ 10 Â¡ t 0 versus H1 : ÂµD Â¡ The null and alternate hypotheses are H0 : ÂµD ...
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