Question

In: Statistics and Probability

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 less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that Plain M&Ms does not follow the stated distribution.

A different significance level would not have changed the conclusion.

3. Use the distribution below to test that Halloween Skittles follow the stated distribution. Discuss your choice of significance level. Would a different significance level have changed your conclusion?
Skittles states the following distribution:
Red = 18%, Orange = 22%, Green = 17%, Blue = 20%, Purple = 22%, Other (Zombie color) = 1%

Solutions

Expert Solution

observed frequencey, O expected proportion expected frequency,E (O-E) (O-E)²/E
63 0.180 123.84 -60.84 29.889
155 0.220 151.36 3.64 0.088
90 0.170 116.96 -26.96 6.214
101 0.200 137.60 -36.60 9.735
181 0.220 151.36 29.64 5.804
98 0.010 6.880 91.12 1206.810

chi square test statistic,X² = Σ(O-E)²/E =   1258.541              
                  
level of significance, α=   0.01              
Degree of freedom=k-1=   6   -   1   =   5
                  
P value =   0.0000   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value < α, Reject Ho                  

Therefore, we can conclude that Plain M&Ms does not follow the stated distribution.

A different significance level would not have changed the conclusion.

orchestra answered 3 years ago
Next > < Previous

Related Solutions

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...
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
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...
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....
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:...
Consider the following hypothesis test: H0: ? = 16 Ha: ? ? 16 A sample of...
Consider the following hypothesis test: H0: ? = 16 Ha: ? ? 16 A sample of 40 provided a sample mean of 14.17. The population standard deviation is 6. a. Compute the value of the test statistic (to 2 decimals). b. What is the p-value (to 4 decimals)? c. Using ? = .05, can it be concluded that the population mean is not equal to 16? Answer the next three questions using the critical value approach. d. Using ? =...
1. Suppose that M&M claims that their Plain M&Ms have an equal proportion of red and...
1. Suppose that M&M claims that their Plain M&Ms have an equal proportion of red and brown. a. Test the claim that the proportion of red M&Ms is greater than the proportion of brown M&Ms. b. 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?
Consider the following hypothesis test. H0: μ ≥ 70 Ha: μ < 70
   You may need to use the appropriate technology to answer this question. Consider the following hypothesis test. H0: μ ≥ 70 Ha: μ < 70 A sample of 100 is used and the population standard deviation is 12. Compute the p-value and state your conclusion for each of the following sample results. Use α = 0.01. (a) x = 68.5 Find the value of the test statistic. Find the p-value. (Round your answer to four decimal places.) p-value = State...
To test the hypothesis H0 : µ = 5 vs. Ha : µ 6= 5, a...
To test the hypothesis H0 : µ = 5 vs. Ha : µ 6= 5, a random sample of 18 elements is selected which yielded a sample mean of x¯ = 4.6 and a sample standard deviation of s = 1.2. The value of the test statistic is about: (a) −2.121 (b) −1.923 (c) −1.414 (d) 0.345 (e) 1.455
For a hypothesis test of H0: p = 0.65 against the alternative Ha: p < 0.65,...
For a hypothesis test of H0: p = 0.65 against the alternative Ha: p < 0.65, the z test statistic is found to be -2.35. What can be said about his finding? The finding is significant at both α = 0.05 and α = 0.01. The finding is significant at α = 0.05 but not at α = 0.01. The finding is significant at α = 0.01 but not at α = 0.05. The finding is not significant at α...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT