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Question;Question 4: (5 marks)In a study of distances travelled by buses before the first major engine failure, a sampling of 60 buses resulted in a mean of 152,600 km. This means that our sample had their first major engine failure at an average of 152,600 km. At the 0.05 level of significance, test the manufacturer?s claim that mean distance travelled before a major engine failure is more than 154,000 km. From previous research we know the population standard deviation is 6400 km.1.State 2.State the level of significance3.State the Decision Rule4. Compute the Test Statistic5. State your conclusionStep 1: Hypothesis statementsSteps 2 and 3: Level of Significance and the Decision RuleBell curve figure and tableUsing values you have obtained from Tables A.1 or A.2, enter the critical value z or t into the red boxes and the significant level into the yellow boxes. For one tail questions leave the appropriate boxes empty. Click on boxes to enter value.Step 4. Test statistic (formula).Step 5: Conclusion.Question 5: (5 marks)A night time cold medicine bears a label indicating the presence of 600 mg of acetaminophen in each fluid ounce of the drug. From previous studies the population standard deviation is known to be 21 mg. Suppose that Health Canada randomly selected 153 one-ounce samples and found that the mean acetaminophen content was 589 mg. Using, test the claim of the pharmaceutical company that the population mean is different from 600 mg. Would you buy this cold medicine? Give your reasons.1.State 2.State the level of significance3.State the Decision Rule4. Compute the Test Statistic5. State your conclusionStep 1: Hypothesis statementsSteps 2 and 3: Level of Significance and the Decision RuleBell curve figure and tableUsing values you have obtained from Tables A.1 or A.2, enter the critical value z or t into the red boxes and the significant level into the yellow boxes. For one tail questions leave the appropriate boxes empty. Click on boxes to enter value.Step 4. Test statistic (formula).Step 5: Conclusion.Question 6: (5 marks)The expense of moving the storage yard for the Consolidated Package Delivery Service (CPDS) is justified only if it can be shown that the daily mean travel distance will be significantly reduced from 342 km. In trial runs of 12 delivery trucks, the mean and standard deviation are found to be 318 km and 67 km, respectively. At the 0.01 level of significance, test the claim that the mean is less than 342 km. Do you think the storage yard be moved? Give your reasons.1.State 2.State the level of significance3.State the Decision Rule4. Compute the Test Statistic5. State your conclusionStep 1: Hypothesis statementsSteps 2 and 3: Level of Significance and the Decision RuleBell curve figure and tableUsing values you have obtained from Tables A.1 or A.2, enter the critical value z or t into the red boxes and the significant level into the yellow boxes. For one tail questions leave the appropriate boxes empty. Click on boxes to enter value.Step 4. Test statistic (formula).Step 5: Conclusion.

Paper#60493 | Written in 18-Jul-2015

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