Amanufacturer 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

The null and alternative hypotheses are:

The bag of colored candies follows the distribution stated

The bag of colored candies does not follow the distribution stated

Under the null hypothesis, the test statistic is:

Where:

is the observed frequencies

is the expected frequencies

The calculation are done in the attachment

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:

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.

yo answer should be 6 if im not mistaken

step-by-step explanation:

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.