According to the manufacturer of M&Ms, 13 % of the plain...

According to the manufacturer of M\&Ms, $13 \%$ of the plain M\&Ms in a bag should be brown, $14 \%$ yellow, $13 \%$ red, $24 \%$ blue $, 20 \%$ orange, and $16 \%$ green. A student randomly selected a bag of plain M\&Ms. He counted the number of $\mathrm{M} \& \mathrm{Ms}$ that were each color and obtained the results shown in the table. Test whether plain M\&Ms follow the distribution stated by M\&M/Mars at the $\alpha=0.05$ level of significance. $$ \begin{array}{lc} \text { Color } & \text { Frequency } \\ \hline \text { Brown } & 57 \\ \hline \text { Yellow } & 64 \\ \hline \text { Red } & 54 \\ \hline \text { Blue } & 75 \\ \hline \text { Orange } & 86 \\ \hline \text { Green } & 64\\ \hline \end{array} $$

According to the manufacturer of M\&Ms, $13 \%$ of the plain M\&Ms in a bag should be brown, $14 \%$ yellow, $13 \%$ red, $24 \%$ blue $, 20 \%$ orange, and $16 \%$ green. A student randomly selected a bag of plain M\&Ms. He counted the number of $\mathrm{M} \& \mathrm{Ms}$ that were each color and obtained the results shown in the table. Test whether plain M\&Ms follow the distribution stated by M\&M/Mars at the $\alpha=0.05$ level of significance.
$$
\begin{array}{lc}
\text { Color } & \text { Frequency } \\
\hline \text { Brown } & 57 \\
\hline \text { Yellow } & 64 \\
\hline \text { Red } & 54 \\
\hline \text { Blue } & 75 \\
\hline \text { Orange } & 86 \\
\hline \text { Green } & 64\\
\hline
\end{array}
$$

Key Concepts

Hypothesis Testing Framework

This concept involves setting up two competing hypotheses: the null hypothesis, which posits that the observed data follow the specified distribution, and the alternative hypothesis, which contends that the data do not follow the distribution. The framework includes computing a test statistic, comparing it to a critical value or using a p-value, and then making a decision regarding the null hypothesis.

Significance Level

The significance level (alpha) is the threshold for rejecting the null hypothesis. It represents the probability of making a Type I error, i.e., rejecting the null hypothesis when it is actually true. A common significance level such as 0.05 helps in determining whether the magnitude of the test statistic indicates a statistically significant deviation from the expected distribution.

Degrees of Freedom

The degrees of freedom in a chi-squared goodness-of-fit test are generally equal to the number of categories minus one. They dictate the shape of the chi-squared distribution used to evaluate the test statistic. Correctly calculating degrees of freedom is essential for accurately determining p-values and critical values.

Chi-Squared Goodness-of-Fit Test

This statistical test assesses whether the observed frequencies of categorical data match the expected frequencies based on a given distribution. It does so by calculating a test statistic which quantifies the discrepancy between observed and expected counts. The test statistic is compared to a chi-squared distribution to decide whether to reject the null hypothesis that the data follows the specified distribution.

Expected Frequency Calculation

Expected frequencies are computed by multiplying the total sample size by the probability (or proportion) assigned to each category under the null hypothesis. This is crucial for determining the divergences between observed frequencies and theoretical expectations, which form the basis of the chi-squared test statistic.

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