Question

Hypothesis Testing 02 (when o is unknow b) A research engineer for a tire manufacturer is investigating the tire life for a new rubber compound and has built 10 tires and tested them to end-of-life in a road test. The sample mean and standard deviation are: 60,139.7 km and s = 3645.94 km. Write down the null and alternative hypotheses to test if the mean tire life is less than 61,000 km b. What is the type of statistical test that that should be used? Explain Test the hypotheses at a 0.05, using the critical value approach. What is your decision on null hypothesis(H)? Interpret the test results. d. What is the P Value for the test? Test the hypotheses using the P Value. Do you get the same answer as part (c)? e. Construct a 95% confidence interval for the mean tire life. Test the hypotheses using the confidence interval approach. Do you get the same answer as parts (c) and (d)?

Answer 1

Date Page p.value = P Tn-lo a trall =PL T15,0.05 -0.944 from t-table IP-value = 0.1801 Here * He p-value ya 20.050 160 fail to reject Ho at 220.05 951. confidence interval is shot x = terit xs. from t-table, twittny & 415=241314 (60139.7 +2.1314x 3 645.94 ) 3) (60139.7 † 2:1314X 911.485) D C60189.7 + 19.4274) 658196.916 , 62082.684) (58196.916)< 4 < (62082 484) 7 Here Dat he 40 = 61000 € confidence interval fail to reject Ho at a. 0.05 -

Page -Ans) Giveno n-16 X = 60139.7 S = 3645.g4 ano.05 I To test hypothesis : Hoi 4 = 61000 vs Hiiy < 61000 here coe we t-test Because population std. den G tanol unknown & na 30 Test statistic 8- 3 60139.7 - 61000 36 45.94 NT6 tcala Tc-lo SIND -860.3 91.485 I tal = -0.9438 Decision Rule & Regeet Hoit teal a to at 2:0.0 Here tef=thold - 45,0.05 = -1.753 & Deation or tcal = -0.9438 Eult = -1.753 - he fail to reject Ho al d=0.05 conclusion - There is insufficient evidence to - reject to at a=0.0s. Of may not be conclude that the mean tire life is Tess than 61000 km