Chandrasekaran,Ramasubramanian Third Exam

Chandrasekaran,Ramasubramanian Third Exam - Problem 1 2 3 4...

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Problem Value Earned 1 12 12 2 12 12 3 12 12 4 12 12 5 12 12 6 12 12 7 13 13 8 15 15 Total 100 100 1) In new housing developments, it is sometimes difficult to get utilities (water, electricity, natural gas) installed in a timely fashion. Your firm is considering investing in a new development which can finish housing construction in 90 days. A random sample of the times (in days) from the request for service connection to the time of service completion yielded the following data: 114 78 96 137 78 103 117 126 86 99 114 72 104 73 86 Is this data consistent with the hypothesis that on average it takes 90 days from request to connection of utility service? Use the 5% level of significance. Answer Mean is 98.87 sd 20.04 Ho: mu =90 Ha: mu !=90 n 15 Template for Confidence Interval Sample Sample Sample Mean SD n alpha Enter ===> 98.87 20.04 15 0.05 Confidence Interval is 87.77 to 109.96 Result: Since 90 falls into the interval we accept the null hypothesis

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And the average no of days is between 87.77 and 109.96
2) An auditor for a government agency is assigned the task of evaluating reimbursement for office visits to doctors paid by Medicare. The audit is to be conducted for all Medicare payments in a specific geographical area during a certain month. In that area and for that month there were a total of 25,056 visits. Since this is too large a number to process in a timely manner, the auditor decides to take a random sample from these visits. She wishes to estimate the average Medicare payment to within \$5.00 with a confidence of 99%. How large a sample should she take? In a preliminary sample of size 30, she found that the standard deviation of Medicare payments was \$35.42. Answer For a confidence interval of 99%, z = 2.58 Sample Size 30 Std deviation 35.42 Width 5 n >= 333 We can approximate it to 334 Result: To get a better result we would need 334 observations. Since we already took 30 observations we nee 334-30 =304 observations more.

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The marketing branch of the Mexican Tourist Office would like to increase the proportion of tourists who purchase silver jewelry while vactioning in Mexico from its present value of 40%. Toward this end, promotional literature describing both the beauty and value of the jewlery is prepared and distributed to all passengers on airplanes arriving at a certain seaside resort during a one week period. A random sample of 500 returning passengers was selected and it was found that 227 had purchased silver jewelry while in Mexico. Does this provide evidence that the promotional campaign was effective? You would like to be 95% sure of your conclusion. Answer Total number of passengers 500 No.of Passengers who purchased jewelry 227 Alpha 0.05 Hypothesis Definition: Ho : P = Po Ha : P != Po Where Po = 0.40 Template for Confidence Interval Sample Sample Successes Size x n Alpha Enter ==> 227 500 0.05 Sample Proportion = 0.45 Confidence Interval is 0.4104 to 0.4976 Result: Po= 0.40 does not lie within the confidence interval.
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This note was uploaded on 10/02/2011 for the course MIS 6312 taught by Professor Wiorkowski during the Spring '11 term at University of Texas at Dallas, Richardson.

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Chandrasekaran,Ramasubramanian Third Exam - Problem 1 2 3 4...

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