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Scan001 - A recent random sample of 15 pen barrels...

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Unformatted text preview: A recent random sample of 15 pen barrels yielded .379 .380 .378 .379 .381 .379 .380 .378, .379 .379 .381 .379 .380 .380 .380 Is it clear that, on the average, the outside diameter is less then .380 in? For Hazy < #01 Consider avhypothesis test. Step 1: State the Hypotheses . , We thus reject Ho 1f H0 I p = .380 , t < ’tn—lxx: ~ 110-: u < .380 » I where twm is the appropriate value from the t table in Step 2: State the Test Statistic ‘ the Appendix. t = or . p #0 s/fi “ Step 3: State the Critical or Rejection Region. The critical region depends upon (Ha. We thus reject H0 if t > twig»- Us y, textbook problems give a. 1 [‘ypical values for a are: I Step 5: ReachpConclusions and State in English ' I’lo . ' . Since t < -1.761, we have sufficient evidence to reject Ho. ' -05 (7.710“ P01711107”) We therefore have enough evidence to suggest that the - ° '01 ' true mean o.d. is less than .380. 11 our particular case, consider a = .05. ' A reasonable question: What are the “plausible” values 7 r Thus, we shall reject H0 if for the true mean o.d.. . . _ We can construct a 95% confidence interval for a by t < f’tn—lp ' ' _V I y 3*: tn—1.o/2s/\/fi t < “114,05 2 ‘ t < -1.761 .‘ ‘ ‘ tum atnvogs = 2.145- tep 4: Conduct Experiment and Calculate Test Statistic .0009 ‘ :21 ' i .3795 i 2.145 ~—-—— :~ , ‘ WE . y = .3795 _. ‘ :, s = .0009 .3795 i .0005 .. (.3790, .3800) y - M, .3795 — .380 ~ sfi " .0009/x/1‘5 = 42.152 140 2. Two-Tailed Tests An important characteristic of the grapes used to make fine wine is the sugar content. Basically, the wine maker can predict the final alcohol content of the wine by dividing the sugar content of the grapes by 2. A Napa Valley winery pays a premium to its wine grow- ers if they can deliver shipment with true mean alcohol contents of 26%. The winery tests grapes from five different, randomly selected locations in the shipment and determines the ‘ sugar content at each location. What is an appropriate method for determining if the wine growers deserves a premium? In our case, use a = .05. Thus; we shall reject the null hypothesis if » lti > t1£~1.0/2 ltl >.t4,.025 ltl > 2.777 Step 4: Conduct Experiment and Calculate Test Statistic. Suppose the next wine grower has y- =- 24.5 s 4—. 1L3 tgg’flo 8% 24.5 -— ‘26 1.3/\/5- ——2.580 M ii l5-l Step 1: State the Hypotheses. H0 : ,u = 26 V H, : p 75 26 Step 2: State the TestS'tatistic. t _~ 17 — Ho “SA/5' Step 3: State the critical region. For Ha 2p 75 #01, We thus reject H0 if Hi > tn—-I,a/2r where tn-1,a/2 is the appropriate value from the ttable in the Appendix. V Step 5: Reach Conclusion, State in English. Since It] <‘2.777. we fail to reject H0. Therefore we have insufficient evidence to show that the true sugar content is not 26%. ' Therefore, we should pay the grower the premium. Typically, we would want to check our assumptions. In this case, with n = 5, we cannot ~do very much. We must trust" that the data come from a very well behaved ‘(nearly normal) distribution. 155 ...
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