solution-HW2c - Solution for ST516 HW2 2.11. The breaking...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution for ST516 HW2 2.11. The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is σ = 3 psi. A random sample of four specimens is tested. The results are y 1 =145, y 2 =153, y 3 =150 and y 4 =147. (a) State the hypotheses that you think should be tested in this experiment. H 0 : µ = 150 H 1 : > 150 (b) Test these hypotheses using α = 0.05. What are your conclusions? n = 4, = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.75 148.75 150 1.25 0.8333 3 3 2 4 o o y z n = = = = − Since z 0.05 = 1.645, do not reject. (c) Find the P -value for the test in part (b). From the z- table: ( )( ) 1 {1 0.7967 2 3 0.7995 0.7967 } 0.7986 P ≅− − + =   (d) Construct a 95 percent confidence interval on the mean breaking strength. The 95% confidence interval is ( ) ( ) ( ) ( ) 2 3 96 . 1 75 . 148 2 3 96 . 1 75 . 148 2 2 + + n z y n z y
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
145 81 151 69 . . µ 2.12. The viscosity of a liquid detergent is supposed to average 800 centistokes at 25 ° C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is σ = 25 centistokes. (a) State the hypotheses that should be tested. H 0 : = 800 H 1 : 800 (b) Test these hypotheses using α = 0.05. What are your conclusions? 812 800 12 1.92 25 25 4 16 o o y z n = = = = Since z /2 = z 0.025 = 1.96, do not reject. (c) What is the P -value for the test? P = = 2 00274 00549 ( . ) . (d) Find a 95 percent confidence interval on the mean. The 95% confidence interval is n z y n z y 2 2 + ( ) ( ) ( ) ( ) 25 824 75 799 25 12 812 25 12 812 4 25 96 1 812 4 25 96 1 812 . . . . . . + +
Background image of page 2
2.14. A normally distributed random variable has an unknown mean µ and a known variance σ 2 = 9. Find the sample size required to construct a 95 percent confidence interval on the mean, that has total length of 1.0. Since y N( µ ,9), a 95% two-sided confidence interval on is y z n y z n + α 2 2 y n y n + ( . ) ( . ) 196 3 3 If the total interval is to have width 1.0, then the half-interval is 0.5. Since z /2 = z 0.025 = 1.96, ( ) ( ) ( ) ( ) ( ) 139 30 138 76 11 76 11 5 0 3 96 1 5 0 3 96 1 2 = = = = = . . n . . . n . n .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.

Page1 / 12

solution-HW2c - Solution for ST516 HW2 2.11. The breaking...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online