Answer:
The claim on the M&M’s website is not true.
Step-by-step explanation:
A Chi-square test for goodness of fit will be used in this case.
The hypothesis can be defined as:
H₀: The observed frequencies are same as the expected frequencies.
Hₐ: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
[tex]\chi^{2}=\sum\limits^{n}_{i=1}{\frac{(O_{i}-E_{i})^{2}}{E_{i}}}[/tex]
Here,
[tex]O_{i}[/tex] = Observed frequencies
[tex]E_{i}=N\times p_{i}[/tex] = Expected frequency.
The chi-square test statistic value is, 14.433.
The degrees of freedom is:
df = k - 1 = 6 - 1 = 5
Compute the p-value as follows:
[tex]p-value=P(\chi^{2}_{k-1} >14.433) =P(\chi^{2}_{5} >14.433) =0.013[/tex]
*Use a Chi-square table.
The significance level is, α = 0.05.
p-value = 0.013 < α = 0.05.
So, the null hypothesis will be rejected at 5% significance level.
Thus, concluding that the claim on the M&M’s website is not true.
