This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: MGCR 271 Midterm Exam Summer 2010 MGCR 271: Business Statistics Midterm Exam 28 May 2010 12:00 - 2:00 pm Examiner: Ramnath Vaidyanathan Assoc Examiner: Student Name McGill ID INSTRUCTIONS 1. Please write your NAME and STUDENT NUMBER on the exam paper. 2. This examination is PRINTED ON BOTH SIDES of the paper. 3. This examination consists of a total of 5 QUESTIONS for 40 POINTS. 4. This examination consists of a total of 11 pages, including the cover page. 5. If you find any pages MISSING bring it to the attention of the invigilator immediately. 6. SPACE IS PROVIDED on the examination to answer all questions. 7. MARKS alloted to each question appear next to question numbers. 8. This is a CLOSED BOOK examination and counts for 20% of your final grade. 9. TABLES are handed out separately. 10. Your are permitted one single-sided CRIB SHEET printed or hand-written. 11. Show LOGIC since part marks will be given for method and partial results. 12. If you feel that a question is ambiguous, briefly explain your interpretation assumptions. 13. You are permitted TRANSLATION dictionaries ONLY. 14. STANDARD CALCULATOR permitted ONLY. 15. This examination and the tables provided MUST BE RETURNED. 16. GOOD LUCK! Page 1 of 11 MGCR 271 Midterm Exam Summer 2010 1. The federal government is putting pressure on hospitals to shorten the average length of stay of patients. A random sample of 41 hospitals in one state had a mean length of stay in 1995 of 3.9 days, with a standard deviation of 1.4 days. (a) (1 point) Use this information to construct a 85% confidence interval to estimate the mean length of stay of all hospitals in this particular state. The 85% confidence interval can be calculated to be x z * s n . From the table, we get z * = 1 . 44. Substituting for z * , we get the confidence interval to be (3 . 59 , 4 . 21). (b) (2 points) If you picked another random sample of 41 hospitals, would you expect that there is an 85% probability that the mean length of stay for this sample is within the interval calculated in (a)? Explain No. The Confidence Interval calculated in (a) is for the mean length of stay for all hospitals in the state and not for a sample. Hence, I would not expect 85% of the hospitals to have a length of stay within the interval calculated in (a). (c) (2 points) If the sample had included 93 hospitals (and all else turned out to be the same), how would the confidence interval have been affected? Explain and write down the new confidence interval? The sample size has increased from n = 41 to n = 93. Hence the width of the confidence interval would decrease. The new margin of error can be calculated to be moe = z * s n , which gives us moe = 0 . 21. Hence, the new confidence interval can be calculated as 3 . 9 . 21 which gives us (3 . 69 , 4 . 11)....
View Full Document
- Summer '10