Question

In: Statistics and Probability

In 1995 the Mars Company replaced tan M&Ms with blue M&Ms. A sample of 100 plain...

In 1995 the Mars Company replaced tan M&Ms with blue M&Ms. A sample of 100 plain M&Ms before the introduction of blue M&Ms had a mean weight 0.9160g with a standard deviation of 0.0433g. A sample of 100 plain M&Ms taken after 1995 had a mean weight of 0.9147g and a standard deviation of 0.0369g. (a) Construct the 98% confidence interval for the difference in the mean weight of plain M&Ms before and after the introduction of blue M&Ms. List: Parameters: u1 and u2 Null and Alt. Hypothesis p-value Initial conclusion Final conclusion (b) Is there a significant difference between mean weight of plain M&Ms before and after the introduction of blue? How do you know? List: t* ME Interval Interpret Interval (c) Does the confidence interval support the results of the hypothesis test? Why or why not?

Solutions

Expert Solution


Related Solutions

Q: The mars company claims that 13 percent of M&Ms plain candies distributed into bags are...
Q: The mars company claims that 13 percent of M&Ms plain candies distributed into bags are brown. Investigate this claim with an appropriate hypothesis test. Use a significance level of a= 0.05 Color Count Brown 33 Non-Brown 242 Total 275 1. The p-value for this test statistic is: _______________. 2. Null Hypothesis: 3. Alternative Hypothesis: 4. Conclusion: We REJECT/DO NOT REJECT the null hypothesis. (Circle the correct answer) State what this conclusion means in terms of the problem. 5. Would...
Mars, Inc. states that 24% of its M&M plain candies are blue and that 16% are...
Mars, Inc. states that 24% of its M&M plain candies are blue and that 16% are green. Assume that they are correct and a large random sample of plain M&Ms is randomly selected from the factory. A. What is the probability that if twenty M&Ms were eaten randomly from the sample, that exactly four of them would be green? B. Find the mean and standard deviation for the above binomial probability distribution in part c. Based on that, would it...
The Mars candy company claims that the percentage of blue M&Ms [2] candies is 24%. A...
The Mars candy company claims that the percentage of blue M&Ms [2] candies is 24%. A sample of 200 M&Ms candies was selected and 54 candies were blue. a) State the null hypothesis and the alternative hypothesis. b) Test the claim using a significance level of 0.05. c) State your conclusion. Should the Mars company take corrective action?
According to Mars, Inc., 20% of all M&Ms produced are blue. One bag of 50 M&Ms...
According to Mars, Inc., 20% of all M&Ms produced are blue. One bag of 50 M&Ms represents the sample for this problem. The sample data can be used to perform a two-sided hypothesis test to test whether 20% of all M&Ms are blue. In one bag of 50 M&Ms, there are 14 blue M&Ms. Use this data to test whether 20% of all M&Ms are blue.
A bowl of M&M's consists of 100 blue M&Ms and 90 gold M&Ms. Assume that the...
A bowl of M&M's consists of 100 blue M&Ms and 90 gold M&Ms. Assume that the M&Ms are all mixed up in the bowl, and that drawing out M&Ms is like simple random sampling. a) Suppose you draw out 10 M&Ms. Write a math expression for the chance that you get more blue M&Ms than gold. Use the cell below to find the numerical value of the answer, and write the output on your paper along with the math expression....
Assume that 15% of all M&Ms are blue in color. If 100 M&Ms are (randomly) poured...
Assume that 15% of all M&Ms are blue in color. If 100 M&Ms are (randomly) poured into a bowl, use the normal approximation to find the probability that more than 20 are blue. Round answer to four decimal places.
According to the manufacturer of M&Ms, 13% of the plain M&Ms in a bag should be...
According to the manufacturer of M&Ms, 13% of the plain M&Ms in a bag should be brown, 14% should be yellow, 13% should be red, 24% should be blue, 20% should be orange, and 16% should be green. A student randomly selected a bag of plain M&Ms. he counted the number of M&Ms that were each color and obtained the results shown in the table. Test whether plain M&Ms follow the distribution stated by M&M at the level of significance=0.05....
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...
Mars Inc., makers of M & M's candies, claims that they produce plain M & M's...
Mars Inc., makers of M & M's candies, claims that they produce plain M & M's with the following distribution Brown: 30% Red: 20% Yellow: 20% Orange: 10% Green: 1096 Blue: 10% A bag of plain M & M's was randomly selected trom the grocery store shelf, and the color counted were as follows Brown: 16 Red: 11Yellow: 19 Orange: 5 Green: 7 Blue: 3 How many total M & Ms were in this sample (This is your n)? What...
M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the...
M&M plain candies come in various colors. According to the M&M/Mars Department of Consumer Affairs, the distribution of colors for plain M&M candies is as follows. Suppose you have a bag of M&M’s and you choose one candy at random. Color Purple Yellow Red Orange Green Blue Brown Percentage 21% 17% 19% 9% 8% 6% 20% a.) Find the P (purple or green candy). b.) Find the P (not red). c.) Find the P (brown).
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT