The manufacturer of a certain candy states that 15% of the candies are brown, 15% yellow, 25% red, 20% blue, 10% orange, 15% green. A student randomly selected a giant bag of candy and counted the number of each color. The results are in the table below. Does the candy follow the stated distribution?(a) Complete the table. Report all answers accurate to 2 decimal places.ColorBrownYellowRedBlueOrangeGreenFrequency273939242343Expected Frequency (b) Select the null and alternative hypotheses.H0H0: PBr=PY=PR=PBl=PO=PG=16PBr=PY=PR=PBl=PO=PG=16HaHa: at least one proportion is differentH0H0: PBrPBr=0.15, PYPY=0.15, PRPR=0.25, PBlPBl=0.20,POPO=0.1,PGPG=0.15,HaHa: at least one proportion is differentH0H0: The colors are independentHaHa: The colors are dependent(c) What is the chi-square test-statistic for this data? Round your answer to 3 decimals.χ2=χ2= (d) What is the P-Value? Round your answer to 4 decimals.P-Value = (e) For significance level alpha 0.05, what would be the decision of this hypothesis test?Reject the Null HypothesisFail to reject the Null Hypothesis(f) Select the conclusion.There is not sufficient evidence to support that the claim that at least one proportion is different. Therefore, the candy follows the stated distributionThere is sufficient evidence to support the claim that at least one proportion is different. Therefore, the candy does not follow the stated distribution.

Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question andmention the sub-parts to be solved. We are asked to find the chi square statistic, the hypothesis and expected frequencies. The chi square goodness of fit test is used to check the significance of and discrepancy between observed frequencies and expected frequencies. It enables us to check if the deviation of the experiment from theory is by chance or any inadequacy in fitting the data. a) The expected frequency of any observation is equal to the average of observed frequencies. n = total no of observations =6total observed frequencies = 195expected frequencies:brown =15100×195 =29.25yellow = 15100×195= 29.25red =25100×195=48.75blue =20100×195 =39.00orange =10100×195=19.50green=15100×195= 29.25 color observed expected brown 27 29.25 yellow 39 29.25 red 39 48.75 blue 24 39 orange 23 19.5 green 43 29.25 195 195 b) The hypothesis is given by: H0 : P(Br)= 0.15, P(Y)=0.25,P(Bl)=0.20, P(O)=0.1,P(G)=0.15 Ha : atleast one proportion is different c) The chi square goodness of fit statistic is given by:χ2 =∑i=1n (fi-ei)2eiwhere fi is the observed frequencyei is the expected frequencyχ2 is chi square statistic with (n-1) df. colour observed expected obs -exp (obs-exp)^2 (obs-exp)^2/ exp brown 27 29.25 -2.25 5.0625 0.173076923 yellow 39 29.25 9.75 95.0625 3.25 red 39 48.75 -9.75 95.0625 1.95 blue 24 39 -15 225 5.769230769 orange 23 19.5 3.5 12.25 0.628205128 green 43 29.25 13.75 189.0625 6.463675214 195 195 621.5 18.23418803 The chi square statistic is 18.234

Answered: The manufacturer of a certain candy…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The manufacturer of a certain candy states that 15% of the candies are brown, 15% yellow, 25% red, 20% blue, 10% orange, 15% green. A student randomly selected a giant bag of candy and counted the number of each color. The results are in the table below. Does the candy follow the stated distribution?

(a) Complete the table. Report all answers accurate to 2 decimal places.

Color	Brown	Yellow	Red	Blue	Orange	Green
Frequency	27	39	39	24	23	43
Expected Frequency

(b) Select the null and alternative hypotheses.

H0H0: PBr=PY=PR=PBl=PO=PG=16PBr=PY=PR=PBl=PO=PG=16
HaHa: at least one proportion is different
H0H0: PBrPBr=0.15, PYPY=0.15, PRPR=0.25, PBlPBl=0.20,POPO=0.1,PGPG=0.15,
HaHa: at least one proportion is different
H0H0: The colors are independent
HaHa: The colors are dependent

(c) What is the chi-square test-statistic for this data? Round your answer to 3 decimals.
χ2=χ2=

(d) What is the P-Value? Round your answer to 4 decimals.
P-Value =

(e) For significance level alpha 0.05, what would be the decision of this hypothesis test?

Reject the Null Hypothesis
Fail to reject the Null Hypothesis

(f) Select the conclusion.

There is not sufficient evidence to support that the claim that at least one proportion is different. Therefore, the candy follows the stated distribution
There is sufficient evidence to support the claim that at least one proportion is different. Therefore, the candy does not follow the stated distribution.

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