Question

calculate the test statistic $x^{2}$, and state your decision regarding the null hypothesis.
it has been rumored that the color distribution of m&ms is 30% brown, 20% yellow, 20% red, 10% orange, 10% blue, and 10% green. you suspect this rumor is
incorrect and decide to count how many of each color are contained in a randomly chosen bag in order to perform a chi-square goodness - of - fit test at a 5% significance
level.
provided the assumptions of the test are satisfied, calculate the test statistic $x^{2}$.
multiple choice 4 points
20.43, fail to reject ho
16.83, reject ho
4.15, fail to reject ho
4.15, reject ho

Explanation:

Step1: Calculate expected values

The total number of M&M's is \(n = 100\).
For brown: \(E_{brown}=0.3\times100 = 30\)
For yellow: \(E_{yellow}=0.2\times100=20\)
For red: \(E_{red}=0.2\times100 = 20\)
For orange: \(E_{orange}=0.1\times100=10\)
For blue: \(E_{blue}=0.1\times100 = 10\)
For green: \(E_{green}=0.1\times100=10\)

Step2: Calculate \(\chi^{2}\) statistic

The formula for the chi - square statistic is \(\chi^{2}=\sum\frac{(O - E)^{2}}{E}\)
For brown: \(\frac{(27 - 30)^{2}}{30}=\frac{(- 3)^{2}}{30}=\frac{9}{30}=0.3\)
For yellow: \(\frac{(16 - 20)^{2}}{20}=\frac{(-4)^{2}}{20}=\frac{16}{20}=0.8\)
For red: \(\frac{(21 - 20)^{2}}{20}=\frac{(1)^{2}}{20}=0.05\)
For orange: \(\frac{(12 - 10)^{2}}{10}=\frac{(2)^{2}}{10}=0.4\)
For blue: \(\frac{(9 - 10)^{2}}{10}=\frac{(-1)^{2}}{10}=0.1\)
For green: \(\frac{(15 - 10)^{2}}{10}=\frac{(5)^{2}}{10}=2.5\)

Step3: Sum up the values

\(\chi^{2}=0.3 + 0.8+0.05 + 0.4+0.1+2.5=4.15\)

Step4: Determine the critical value and make a decision

The degrees of freedom \(df=k - 1\), where \(k = 6\) (number of color categories). So \(df=6 - 1=5\)
For a significance level of \(\alpha = 0.05\), the critical value \(\chi_{0.05,5}^{2}=11.070\)
Since \(\chi^{2}=4.15<\chi_{0.05,5}^{2}=11.070\)

Answer:

The test statistic \(\chi^{2}=4.15\). Fail to Reject \(H_{0}\)