The UMUC Daily News reported that the color distribution for plain M&M;\'s was:

Question

The UMUC Daily News reported that the color distribution for plain M&M;'s was: 40% brow 20% yellow, 20% orange, 10% green, and 10% tan. Each piece of candy in a random sample of 100 plain M&M;'s was classified according to color, and the results are listed below. Use a 0.05 significance level to test the claim that the published color distribution is correct. SHow i work and justify your answer. (a) Identify the null hypothesis and the alternative hypothesis. (b) Determine the teat statistic. Show all work; writing the correct test statistic, without supporting work will receive no credit. (c) Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit. (d) Is there sufficient evidence to support the claim that the published color distribution is correct? Justify your answer. 18. A random sample of 4 professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars). (a) Find an equation of the least squares regression line. Show all work; writing the correct equation, without supporting work, will receive no credit. (b) Based on the equation from part (a), what is the predicted value of Y if X = 3? Show all work and justify your answer.

Explanation / Answer

17.

Let p1, p2, p3, p4, p5 = the probability of an M&M being brown, yellow, orange, green, and tan

a)

Ho: p1 = 0.40, p2 = 0.20, p3 = 0.20, p4 = 0.10, p5 = 0.10

Ha: The distribution of M&Ms is not p1 = 0.40, p2 = 0.20, p3 = 0.20, p4 = 0.10, p5 = 0.10.

******************

b)

Doing an observed/expected value table,          
O   E   (O - E)^2/E  
42   40   0.1  
21   20   0.05  
12   20   3.2  
7   10   0.9  
18   10   6.4  
          
Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    10.65   [ANSWER]

*********************

c)  
          
As df = a - 1,           
          
a =    5      
df = a - 1 =    4      
          
Then, the p value is, using technology,
          
p =    0.03079232   [ANSWER]

*******************

d)  
As P < 0.05, then there is significant evidence that the published color distribution is NOT CORRECT at 0.05 level.  
          

*******************************************

Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.