Question

To decide whether two different types of steel have the same true average fracture toughness values, n specimens of each type are tested, yielding the following results. Calculate the P-value for the appropriate two-sample z test, assuming that the data was based on n 100. (Round your answer to four decimal places.) Calculate the P-value for the appropriate two-sample z test, assuming that the data was based on n- 400. (Round your answer to four decimal places.) Is the small P-value for n 400 indicative of a difference that has practical significance? Would you have been satisfied with just a report of the P value? Comment briefly. You may need to use the appropriate table in the Appendix of Tables to answer this question. Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of braking systems. Use the two-sample t test at significance level 0.01 to test H0: μ 1-12-_10 versus Hai μι-H2 <-10 for the following data: m 6.. 114.8, si 5.01, n- 6,y-129.5, and s2 -5.36. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.) State the conclusion in the problem context O Reject Ho. The data suggests that the difference between mean stopping distances is less than-10. O Reject Ho. The data does not suggest that the difference between mean o Fail to reject Ho. The data suggests that the difference between mean O Fail to reject Ho. The data does not suggest that the difference between stopping distances is less than -10 stopping distances is less than -10. mean stopping distances is less than -10

Answer 1

t = (x1-x2) -d/sqrt(s1^2/n2+s2^2/n2)
= ((114.8 - 129.5) - (-10)/sqrt(5.01^2/6 + 5.36^2/6)
= -1.57

p value = .0886

Option D)
fail to reject H0.