ASAPPPPPP!! PLEASE help me!!!!!!!!!!
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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:
<em>H</em>₀: The observed frequencies are same as the expected frequencies.
<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.
The test statistic is given as follows:
Here,
= Observed frequencies
= Expected frequency.
The chi-square test statistic value is, 14.433.
The degrees of freedom is:
df = <em>k</em> - 1 = 6 - 1 = 5
Compute the <em>p</em>-value as follows:
*Use a Chi-square table.
The significance level is, <em>α</em> = 0.05.
p-value = 0.013 < <em>α</em> = 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.
