Question

ework -/1 < Question 3 of 5 View Policies Current Attempt in Progress Two suppliers manufacture a plastic gear used in a laser printer. The impact strength of these gears, measured in foot-pounds, is an important characteristic. A random sample of 10 gears from supplier 1 results in x 1.289.30 and 51 - 22.5, and another random sample of 16 gears from the second supplier results in x 2 -322.10 and s2 - 21. Use only Table V of Appendix A (a) Is there evidence to support the claim that supplier 2 provides gears with higher mean impact strength? Use a -0.05, and assume that both populations are normally distributed but the variances are not equal. Round your answer to 4 decimal places < P-value 2 Is there evidence to support the claim? (b) Do the data support the claim that the mean impact strength of gears from supplier 2 is at least 25 foot-pounds higher than that of supplier 1? Find bounds on the P-value making the same assumptions as in part (a). Round your answer to 2 decimal places. < P-value . Is there evidence to support the claim? (c) Construct an appropriate 95% confidence interval on the difference in mean impact strength. Use only Table V of Appendix A. Round your answers to 3 decimal places Su 2-uis i

Answer 1

Anssy, 322.10 28.52 + 212 Pales) 1/9/2000 = 289.80, S, = 28.5, M1 =10 92 = 322.10, S2 = 21, N2 = 16 C=95) = 0.95 = 0.05 Ho: M1 = 42 Ha: MKM2 289.80 - är de Terk statistic sie SER ni + 16 N2 t= -32.8 32.8 50.6857 27.5625 578.1875 8.8484 t= -3.7094 df = A= (sa%i)? (58/m2) (22.5%) 2 (217129 Ni-1 N2-) 10-1 16-1 0.0050 p < 0.0010 -32.8 (28.52 Tot 16 mi temple + 6 P-1914e = 0.0008 P-value of 0.05 Reject to there is suffrent evidence to support the daim. Ho: M1-42=0 HQ : Al-M2C-25 test statistic ta ráe-(4-02) 289.80- 392.00-(-25) SIR 28.58 S22 ni f M2 10 TT6 t= - 328725 = -0.8881 8.8424

2 df = A= Minz 3 ) ni- 22.92 + al 16 18 (22.1971 109 101 16- 0.lo<p60.85 folue = 0.1947 P&=0.05 be fail to reject Ho Theace is not sufficerent widence to support the daien. = (1-9)/2 = 0.025, df =18 a +60125 = 2.101 Margin of Chora SIR N2 10 2.101 [28.52 812 & 18. 5778 Cu-ne) { = (289.30 – 348.10) - 18.9778 -32.8 -18.5778 -51.8778 -2) +(289.80- 392.16) + 18.5778 - 32.8t 18.5778 = -14.2882 - 51.378 S 4-MS -14.222