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 α=0.05
level of significance.


Color | Frequency | Claimed Proportion

Brown | 60 | 0.13

Yellow | 65 | 0.14

Red | 56 | 0.13

Blue | 63 | 0.24

Orange | 82 | 0.20

Green | 63 | 0.16

Answer 1

Answer:

The null and alternative hypotheses are:

$H_{0}:$ The bag of colored candies follows the distribution stated

$H_{a}:$ The bag of colored candies does not follow the distribution stated

Under the null hypothesis, the test statistic is:

$\chi^{2} =\sum \frac{(O-E)^{2}}{O}$

Where:

$O$ is the observed frequencies

$E$ is the expected frequencies

The calculation are done in the attachment

$\therefore \chi^{2} =14.490$

Now, we have to find the chi-square critical value at 0.05 significance level for df = n - 1 = 6 - 1 = 5. Using the chi-square table, we have:

$\chi^{2}_{critical} = 11.070$

Since, the chi-square test statistic is greater than chi-square critical value. Therefore, we reject the null hypothesis and conclude that the bag of colored candies does not follow the distribution stated.