Statistics Homework Solutions 137

# Statistics Homework Solutions 137 - 137 SECTION 6.4 so P 0...

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Unformatted text preview: 137 SECTION 6.4 so P 0 1251. It cannot be concluded that the new process has a lower defect rate. Â¡ X 17, n Â¨ Â¡ 0 22667. 3 06. Â¡ Â¡ Â¡ Â£ Â©Â¥ Â¨ Â¤ Â¡ Â£Â¥ Â¨ Â¨ Â¡ Â¨ Â¤ Â¡ Â¡ 0 0011. It can be concluded that less than 40% of the fuses have burnout amperages greater than 15 A. Â¨ Â¡ so P 3 06, Â¡ p0 , the P-value is the area to the left of z Since the alternate hypothesis is of the form p Â¨ Â¡ 0 40 75 Â¡ 0 40 1 0 40. Â¨ 0 40 Â§ 0 22667 Â¨ Â¡ z 0 40 versus H1 : p Â¡ Â£ The null and alternate hypotheses are H0 : p Â¨ 17 75 Â¡ 75, p Â¨ 11. Â¡ 0 690. Â¡ 345 500 Â¨ Â£ Â¡ 13. (a) Sample p = p Â¡ 500. From part (a), p Â¡ Â¡ Â¨ Â¡ Â§ Â¨ 0 690. Â¨ Â¡ Â¡ p0 , the P-value is the area to the left of z 0 49. Â¨ Â¡ Â¡ Â£ Â©Â¥ Â¨ Â©Â£Â¥ Â¤Â¨ Â¡ Â¨ Â¨ Â¡ Â¤ Â¡ (c) Since the alternate hypothesis is of the form p Â¡ 0 3121. Â¨ Â¡ Thus P 0 7. n Â¨ (b) The null and alternate hypotheses are H0 : p 0 7 versus H1 : Âµ 0 690 0 700 0 7 1 0 7 500 0 49. z Section 6.4 Â¨ Â¡ Â¨ Â¡ 0 0264575 3 Â¤Â£ Â¨ Â¤ Â¥ Â£ 100 Â¡Â¥ Â¡ 100 01 100. 0 6547. Â¨ Â¨ Â¤ Â¡ t 2 degrees of freedom. 100 versus H1 : Âµ Â¡ Â¡ The null and alternate hypotheses are H0 : Âµ 1 Â¡ 3. There are 3 Â¡ 0 0264575, n Â¡ 100 01, s 1. (a) X Â¡ Âµ0 , the P-value is the sum of the areas to the right of t 0 6547 Â¨ Â¡ Since the alternate hypothesis is of the form Âµ and to the left of t 0 6547. Â¨ 0 80. A computer package gives P Â¡ Â¨ Â¡ P 0 580. Â¨ Â¡ Â¡ Â¡ From the t table, 0 50 Â¨ We conclude that the scale may well be calibrated correctly. (b) The t -test cannot be performed, because the sample standard deviation cannot be computed from a sample of size 1. 5 degrees of freedom. 2 449. Â¡ Âµ0 , the P-value is the area to the right of t 0 050. A computer package gives P Â¨ Â¡ P Â¡ Â¤ Â¥ Â£ Â¡ Â¤ Â¡ Since the alternate hypothesis is of the form Âµ Â¡ From the t table, 0 025 0 029. Â¨ 6 Â¤Â£ 4 Â¡Â¥ 35 Â¨ 39 35. Â¡ Â¡ t 35 versus H1 : Âµ 2 449. Â¨ Â¡ From part (a), the null and alternate hypotheses are H0 : Âµ Â¨ 1 Â¡ 6. There are 6 Â¡ 4, n Â¡ 39, s 35 Â¡ Â¨ (b) X 35 versus H1 : Âµ Â¡ 3. (a) H0 : Âµ Â¨ ...
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