A manufacturer of colored candies states that 13% of the candies in a bag should
ID: 3133554 • Letter: A
Question
A manufacturer of colored candies states that 13% of the candies in a bag should be brown, 14% yellow, 13% red, 24% blue, 20% orange, and 16% green. A student randomly selected a bag of colored candies. He counted the number of candies of each color and obtained the results shown in the table. Test whether the bag of colored candies follows the distribution stated above at the = 0.05 level of significance Click the icon to view the table A-H,: The distribution of colors is the same as stated by the manufacturer. H: The distribution of colors is not the same as stated by the manufacturer. OB. H: The distribution of colors is not the same as stated by the manufacturer. H: The distribution of colors is the same as stated by the manufacturer. OC. None of these Observed Distribution of Colors Compute the expected counts for each color. ColorObserved Coun Expected Count Colored Candies in a bag BrownYellow RedBlue Orange Green Brown Yellow Red Blue Orange Green 59 63 52 60 Color Frequenc Claimed Proportion 59 63 52 60 67 0.130.140.130.24 0.20 0.16 Print Done 67 (Round to two decimal places as needed.)Explanation / Answer
There are 59+63+52+60+88+67 = 389 candies here.
We simply multiply the percentages by 389 to get the expected frequencies.
Doing an observed/expected value table,
[ANSWER, EXPECTED VALUES IN "E" COLUMN]
***************************************************
Using chi^2 = Sum[(O - E)^2/E],
chi^2 = 16.40661459 [ANSWER, TEST STATISTIC]
As df = a - 1,
a = 6
df = a - 1 = 5
Then, the critical chi^2 value is
significance level = 0.05
chi^2(crit) = 11.07049769
Also, the p value is
p = 0.005774149 [ANSWER, P VALUE]
Thus, comparing chi^2 and chi^2(crit) [or, p and significance level], we REJECT THE NULL HYPOTHESIS.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.