ENME 392 - Homework 9 Solutions - Sp12

ENME 392 - Homework 9 Solutions - Sp12 - ENME392 Spring...

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1 ENME392 Spring 2012 Homework 9 Solutions Total number of points: 100 Assignment: 1. The following data were collected for the number of M&Ms of each color in 17 small bags of candies. (20 pts) a) What is the probability that the red data could have come from a population with a mean value of 13% red? (3 pts) b) Construct a 95% confidence interval for the red population mean, based on the sample data. (3 pts) c) What is the probability, based on these samples, that red and blue M&Ms could have this difference in means if they came from parent populations of 13% red and 24% blue? (3 pts) d) What is the probability that red and blue M&Ms have the same average in the parent population, based on these samples? (3 pts) e) How many packets of M&Ms should be opened if we want to predict the population mean of the blue M&Ms in a bag with an error of < 1 candy with 95% confidence? (4 pts) f) Based on these data, give a tolerance interval for 99% of blue M&Ms with 95% certainty. (4 pts) Number of Candies in the Packages Number Red Blue Yellow Green Orange Brown 0 1 2 3 1 4 2
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2 5 3 1 1 1 6 4 1 3 7 4 2 3 1 1 2 8 2 5 1 4 9 2 1 2 4 2 2 10 4 4 4 3 11 3 1 2 3 12 1 2 2 3 13 2 1 1 1 14 1 3 15 1 2 16 1 1 17 1 18 19 20 21 Total 17 17 17 17 17 17 To compare to the population mean, we need to find the sample average, or x E( x ) xf ( x )  . This is obtained by taking each cell value, multiplying by the corresponding value for “Number”, and dividing by 17. x E( x ) xf ( x ) Expected Value Number Red Blue Yellow Green Orange Brown 0 0.000 0.000 0.000 0.000 0.000 0.000 1 0.000 0.000 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 0.000 0.000 0.000 3 0.000 0.000 0.176 0.000 0.000 0.000 4 0.471 0.000 0.000 0.000 0.000 0.000 5 0.882 0.294 0.294 0.000 0.000 0.294 6 1.412 0.000 0.000 0.353 0.000 1.059 7 1.647 0.824 1.235 0.412 0.412 0.824 8 0.941 0.000 2.353 0.471 0.000 1.882 9 1.059 0.529 1.059 2.118 1.059 1.059 10 0.000 2.353 2.353 2.353 1.765 0.000 11 0.000 1.941 0.647 1.294 1.941 0.000
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3 12 0.000 0.706 0.000 1.412 1.412 2.118 13 0.000 1.529 0.000 0.765 0.765 0.765 14 0.000 0.824 0.000 0.000 2.471 0.000 15 0.000 0.882 0.000 0.000 1.765 0.000 16 0.000 0.000 0.000 0.941 0.000 0.941 17 0.000 1.000 0.000 0.000 0.000 0.000 18 0.000 0.000 0.000 0.000 0.000 0.000 19 0.000 0.000 0.000 0.000 0.000 0.000 20 0.000 0.000 0.000 0.000 0.000 0.000 21 0.000 0.000 0.000 0.000 0.000 0.000 sum Total = Expected Value 6.412 10.882 8.118 10.118 11.588 8.941 56.06 Percent 11.4 19.4 14.5 18.0 20.7 15.9 % in Population 13 24 14 16 20 13 # in Population 7.28 13.44 7.84 8.96 11.2 7.28 56 We have added the “known” population values at the bottom of the table (in blue) from the M&M web site. We also need to find the variances. 22 x ( x ) f ( x )   Variance Number Red Blue Yellow Green Orange Brown 0 0.000 0.000 0.000 0.000 0.000 0.000 1 0.000 0.000 0.000 0.000 0.000 0.000 2 0.000 0.000 0.000 0.000 0.000 0.000 3 0.000 0.000 0.059 0.000 0.000 0.000 4 0.471 0.000 0.000 0.000 0.000 0.000 5 1.588 0.059 0.059 0.000 0.000 0.059 6 3.765 0.000 0.000 0.059 0.000 1.588 7 3.765 0.471 1.588 0.059 0.059 0.471 8 0.471 0.000 7.353 0.059 0.000 3.765 9 0.471 0.059 0.471 3.765 0.471 0.471 10 0.000 3.765 3.765 3.765 1.588 0.000 11 0.000 1.588 0.059 0.471 1.588 0.000 12 0.000 0.059 0.000 0.471 0.471 1.588 13 0.000 0.471 0.000 0.059 0.059 0.059 14 0.000 0.059 0.000 0.000 1.588 0.000 15 0.000 0.059 0.000 0.000 0.471 0.000
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4 16 0.000 0.000 0.000 0.059 0.000 0.059 17 0.000 0.059 0.000 0.000 0.000 0.000 18 0.000 0.000 0.000 0.000 0.000 0.000 19 0.000 0.000 0.000 0.000 0.000 0.000 20 0.000 0.000 0.000 0.000 0.000 0.000 21 0.000 0.000 0.000 0.000 0.000 0.000 Total = Variance 10.529 6.647 13.353 8.765 6.294 8.059 a) What is the probability that the red data could have come from a population with a mean value of 13% red? If the population is 13% red, then the population average would be 7.28. Since the population variance is unknown, we need to use the t-test. 6 412 7 28 0 868 0 868 0 340 10 53 4 123 2 554 10 53 17 X . . . .
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This note was uploaded on 04/05/2012 for the course ENME enme392 taught by Professor Cukier during the Spring '10 term at Maryland.

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ENME 392 - Homework 9 Solutions - Sp12 - ENME392 Spring...

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