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 obtain the results shown in the table test whether the bag of colored candies follows the distribution stated above at the a=.05 level of signifcance

Color Frequency Claimed Proportion Colored Candies in a bag Brown Yellow Red Blue 63 64 52 62 0.13 0.14 0.13 0.24 Orange Green 97 66 0.20 0.16

Answer 1

Answer Date: 30/07/2020 The null and alternative hypothesis is, The null hypothesis is that the distribution of colors is same as stated by manufacturer. The alternative hypothesis is that the distribution of colors is not same as stated by manufacturer. The correct option is B The chi-square test statisticis, First, compute frequency expected then find chi-square test statistics. The expected frequency for brown is, fe = Npi = 404 x 13% = 52.52

The expected frequency for yellow is, fe= Npi = 404 x 14% = 56.56 The expected frequency for red is, fe = Npi = 404 x 13% = 52.52 The expected frequency for blue is, fe= Npi = 404 x 24% = 96.96

The expected frequency for orange is, fe = Npi = 404 x 20% = 80.80 The expected frequency for green is, fe= Npi = 404 x 16% = 64.64 63 Color Brown Yellow Red Blue Orange Green Total Frequency Proportion Expected count 0.13 52.52 64 0.14 56.56 52 0.13 52.52 62 0.24 96.96 0.20 80.80 66 0.16 64.64 404 1.00 404.00 97

The chi-square test statistic is, x?=(90-92) + fe (63-52.52) (64 - 56.56) (52 - 52.52) + 52.52 56.56 52.52 (62–96.96)2 (97–80.80)2 (66-64.64)2 + + 96.96 80.80 64.64 = 18.957 The chi-square test statistic is 18.957 The p-value is, By using Excel software, find p-value value with help of following steps: 1) Import the data. 2) Select CHITEST from Functions. Choose Actual_range and Expected_range. 4) Click OK. 3)

CHITEST Actual_range B2:37 Expected_range D2:07 {63;64;52;62;97;66) = {52.52;56.56;52.52;96.96;80.8;64.64 = 0.00195801 From Excel output, the p-value is 0.002 Decision The conclusion is that the significant p-value is less than 0.05 which is 0.002, so the null hypothesis is rejected at 5% level of significance. It can infer that the distribution of colors is not same as stated by manufacturer. The result is statistically significant. The correct option is A