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. Type Sample Average 60.6 60.4 Sample SD 1.0 1.0 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 = 500. (Round your answer to four decimal places.) Is the small p-value for n = 500 indicative of a difference that has practical significance? Would you have been satisfied with just a report of the P-value? Comment briefly.

Answer 1

Sol: Given that Type I Type I Sample average = 60-6 22 = 60.4 Sample Standard deviation - 1.0 Sg-1.0 Null hypothesis Hot Two different types of Steel have the some true average fracture Loughness values. Allemnative hypothesis 4 M + My (two tailed). + When = 100 g = 100 There The test Statistic - 1-12 606-604 - 02 SYSYV Foot too + - 0.2 = 1.414. P value based on Z scor is p > 0.1574 When = 500 = 500 Then -12 60-6-60.4 0.2 The test Statistic z. SV 500 +500 70.004 - 7. 0.66,- 3.165 P value based on 7 Score i p= 0.0016: yes the small p-value for n = 500 indicoltve ola difference that has practical significance. Yes, Satisfied with just a report 4 the p-value because in testing of hypothesis in p-value approach We compare p value with the level of significance & i Pd, we accept Ho, 4 peal, We reject- Ho