Question

In each of Exercises $12.24-12.33$, apply the chi-square goodness-of-fit test, using either the critical-value approach or the$P-$value approach, to perform the required hypothesis test.
Observing that the proportion of blue M&Ms in his bowl of candy appeared to be less than that of the other colors, R. Fricker, Jr., decided to compare the color distribution in randomly chosen bags of M&Ms to the theoretical distribution reported by M&M/MARS consumer affairs. Fricker published his findings in "The Mysterious Case of the Blue M&Ms" (Chance, Vol. 9(4), pp. 19-22). The following table gives the theoretical distribution.
For his study, Fricker bought three bags of M&Ms from local stores and counted the number of each color. The average number of each color in the three bags was distributed as follows.

Do the data provide sufficient evidence to conclude that the color distribution of M&Ms differs from that reported by M&M/MARS consumer affairs? Use $\alpha =0.05$.

Short answer

The data doesn't provide enough evidence to indicate the color distribution of M&Ms differ from that M&Ms/MARS consumer affairs.

Step-by-step solution

Step 1. Given Information

Color distribution table:

Hypothesis

The null and alternative hypotheses are as follows:

$H_0:$The data doesn't provide enough evidence to indicate that the color distribution of M&Ms differs from that of M&Ms/MARS consumer affairs.
$H_a:$The data doesn't provide enough evidence to indicate that the color distribution of M&Ms differs from that of M&Ms/MARS consumer affairs.

Step 3. Calculating the Goodness-of-fit

Calculating the Goodness-of-fit:

${\chi }^2=\sum \frac{(Obs-Exp{)}^2}{Exp}$
$=4.091$
The degrees of freedom the given data has:
$df=k-1$
$$\begin{gathered} =6-1 \\ =5 \end{gathered}$$
The critical value of ${\chi }^2$at $0.05$significance level with $5df$is$11.070$.
The value of the test statistics is ${\chi }^2=4.091$.${\chi }^2=4.091<{\chi }_{0.05}^2=11.070$ as it does not fall in the rejection region, the investigator failed to reject the null hypothesis$H_0$.