The UMUC Daily News reported that the color distribution for plain M&M’s was: 40
ID: 2927013 • Letter: T
Question
The UMUC Daily News reported that the color distribution for plain M&M’s was: 40% brown, 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 all work and justify your answer.
Color Brown Yellow Orange Green Tan
Number 42 21 12 7 18
(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test 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.
Explanation / Answer
Given table data is as below
set up null vs alternative as
null, Ho:true proportion of all 5 categories equal to their ex— pected value
alternative, H1: atleast one proportion of all 5 categories is diffrent from its ex
pected value.
level of significance, = 0.05
from standard normal table, chi square value at right tailed, ^2 /2 =9.488
since our test is right tailed,reject Ho when ^2 o > 9.488
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 10.65
critical value
the value of |^2 | at los 0.05 with d.f n-1 = 4 is 9.488
we got | ^2| =10.65 & | ^2 | =9.488
make decision
hence value of | ^2 o | < | ^2 | and here wereject Ho
^2 p_value =0.0308
there is evidence that atleast one proportion of all 5 categories is diffrent from its ex
pected value.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.