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Unformatted text preview: STAT 409 Homework #8 Fall 2009 (due Friday, October 23, by 4:00 p.m.) 1. In a random sample of 100 Hawk & Hummingbird Airline ( HHA ) direct flights from New York to Boston, the average number of passengers was 56.3, with sample standard deviation 11.33. X = 56.3, s = 11.33, n = 100. ( n is large, can use z instead of t ) a) Construct a 95% confidence interval for the overall average number of passengers on this route. 56.3 2.22 or ( 54.08 , 58.52 ) ( z 0.025 = 1.960 ) b) To make a reasonable profit, HHA must average at least 58 passengers per flight on this route. The president of the airline is concerned that the average number of passengers is less than 58. Perform the appropriate test at = 0.05. H : 58 vs. H 1 : < 58. Test Statistic: Z =  1.50 . Rejection Region: Z <  z 0.05 =  1.645 . Do NOT Reject H . c) Suppose the actual overall average number of passengers on this flight is 57. Then in part (b) a ( Type I Error, Type II Error, correct decision ) was made. Type II Error . d) Find the pvalue of the test in part (b). pvalue = 0.0668 . e) Using the Pvalue obtained in part (d), state your decision ( Accept H or Reject H ) for = 0.10. Reject H . f) Describe in the context of the problem what happens if Type I Error occurs. The route is profitable, but the airline cancels it. g) Describe in the context of the problem what happens if Type II Error occurs. The route is not profitable, but the airline keeps it. 2. Hawk & Hummingbird Airline wants to determine the proportion of passengers that bring only carryon luggage to the flight from New York to Boston. n = 200, X = 44, p = 0.22 . a) In a random sample of 200 passengers, 44 passengers had only carryon luggage. Is there enough evidence to conclude that more than 20% of all passengers have only carryon luggage? Use = 0.10. H : p 0.20 vs. H 1 : p > 0.20. Test Statistic: Z = 0.71 . Rejection Region: Z > z 0.10 = 1.282 . Do NOT Reject H . b) Find the pvalue of the test in part (a). pvalue = 0.2389 . c) Find a 90% confidence interval for the overall proportion of passengers who have only carryon luggage. 0.22 0.048 or ( 0.172 , 0.268 ) ( z 0.05 = 1.645 ) d) Find the minimum sample size required in order to estimate the proportion of passengers who have only carryon luggage to within 2% with 90% confidence, if it is known that this proportion is at most 0.30....
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 Spring '11
 Stepanov
 Standard Deviation

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