Question

In: Statistics and Probability

Use the distribution below to test that Peanut M&Ms follow the stated distribution. Discuss your choice...

Use the distribution below to test that Peanut M&Ms follow the stated distribution. Discuss your choice of ?. Would a different ? have changed your conclusion? M&M states the following distribution for Peanut M&Ms: Red = 12%, Orange = 23%, Yellow = 15%, Green = 15%, Blue = 23%, Brown = 12%

Total Peanut M&Ms: 367

Red=44 Orange=81 Yellow=81 Green=56 Blue=61 Brown=44

Solutions

Expert Solution



Related Solutions

Use the distribution below to test that Peanut M&Ms follow the stated distribution. Discuss your choice...
Use the distribution below to test that Peanut M&Ms follow the stated distribution. Discuss your choice of ?. Would a different ? have changed your conclusion? M&M states the following distribution for Peanut M&Ms: Red = 12%, Orange = 23%, Yellow = 15%, Green = 15%, Blue = 23%, Brown = 12% Total number of plain M&M's: 665 Total number of red plain M&M's: 76 Total number of plain brown M&M's: 66 Total number of blue plain M&M's: 179 Total...
Use the distribution below to test that Plain M&Ms follow the stated distribution. Discuss your choice...
Use the distribution below to test that Plain M&Ms follow the stated distribution. Discuss your choice of ?. Would a different ? have changed your conclusion? M&M states the following distribution for Plain M&Ms: Red = 13%, Orange = 20%, Yellow = 13%, Green = 20%, Blue = 20%, Brown = 14% Total number of plain M&M's: 665 Total number of red plain M&M's: 76 Total number of plain brown M&M's: 66 Total number of blue plain M&M's: 179 Total...
The hypothesis being tested is: H0: Plain M&Ms follow the stated distribution Ha: Plain M&Ms does...
The hypothesis being tested is: H0: Plain M&Ms follow the stated distribution Ha: Plain M&Ms does not follow the stated distribution observed expected O - E (O - E)² / E Red 63 89.440 -26.440 7.816 Orange 155 137.600 17.400 2.200 Yellow 90 89.440 0.560 0.004 Green 101 137.600 -36.600 9.735 Blue 181 137.600 43.400 13.689 Brown 98 96.320 1.680 0.029 688 688.000 0.000 33.473 33.47 chi-square 5 df 3.03E-06 p-value The p-value is 0.0000. Since the p-value (0.0000) is...
Q.N. 7) According to the manufacturer of M& Ms, 12% of the peanut M&Ms in a...
Q.N. 7) According to the manufacturer of M& Ms, 12% of the peanut M&Ms in a bag should be brown, 15% yellow, 12% red, 23% blue, 23% orange, and 15% green. A student randomly selected a bag of peanut M&Ms. He counted the number of M&Ms that were each color and obtained the results shown in the table below. Test whether peanut M&Ms follow the distribution stated by M&M/Mars at the α = 0.05 level of significance. Color Frequency Brown...
According to Masterfoods, the company that manufactures M&Ms, 12% of peanut M&Ms are brown, 15% are...
According to Masterfoods, the company that manufactures M&Ms, 12% of peanut M&Ms are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. Assume that selecting multiple M&Ms are independent events. (Round your answers to three decimal places, for example: 0.123) Compute the probability that a randomly selected peanut M&M is not yellow. A: Compute the probability that four randomly selected peanut M&Ms are all yellow. A: If you randomly select five peanut...
Suppose that M&M claims that each bag of Peanut M&Ms should be 18 grams and Plain...
Suppose that M&M claims that each bag of Peanut M&Ms should be 18 grams and Plain M&Ms should be 13.5 grams. a. Test the claim that M&M is shorting its customers in bags of Plain M&Ms. b. Test the claim that M&M is overfilling Peanut bag bags of M&Ms. c. Discuss your choice of ?. i. Why did you choose the ? you did? ii. If you had chosen a different ?, would it have affected your conclusion? Total Plain...
The number of peanut M&Ms in a 2 ounce package is normally distributed with a mean...
The number of peanut M&Ms in a 2 ounce package is normally distributed with a mean of 28 and standard deviation 2; The number of Skittles in a 2 ounce package is normally distributed with a mean of 60 and standard deviation 4. Questions 1-3: Suppose that I purchase two 2-ounce packages of peanut M&Ms and one 2-ounce package of Skittles. 1. Let X= the total number of pieces of candy in all three bags combined. What is the distribution...
Use MS Excel to answer this. From the data given below test a hypothesis for variances...
Use MS Excel to answer this. From the data given below test a hypothesis for variances being equal. Use α = 0.05. Section A Section B 15 16 26 52 52 55 53 57.5 54 58 56.5 60 61 61 61.5 70 63 70 66 71 66 72 66.5 73 69 74.5 71 75 77 75.5 77 76 78 77 79 81 81 85 86 85.5 87 86 90 88 90 88.5 91 91 91 93 94 95 96 98...
Hypothesis test for M&Ms: This section will work through constructing and interpreting a hypothesis test for...
Hypothesis test for M&Ms: This section will work through constructing and interpreting a hypothesis test for proportions. You can find the information for constructing a hypothesis test in section 10.2. We will be using the p-value approach with a 5% significance level. 1. Treat your bag of M&M’s as a simple random sample. Choose your favorite color of M&M’s you will be working with for this project. State the color and give the counts below. Color of choice: Green The...
. It is known that scores on a certain IQ test follow a normal distribution with...
. It is known that scores on a certain IQ test follow a normal distribution with mean 100 and standard deviation 15. For the whole population of test-takers, what proportion of scores will be greater than 124.0? Also, the top 3% of test-takers will have scores greater than what value? Finally, consider a random group of 16 people who take the IQ test. For these 16 people, what is the probability that their average (mean) IQ score will be less...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT