Question

sally was interested in whether the number of m&ms was uniform over all 6 colors (the same number of m&ms for each color). the table categorizes the number of m&ms of each color found in an 1-pound bag.
frequency of each color

	color
number	70	85	81	79	93	96

state the appropriate null hypothesis for a chi-square goodness-of-fit test.
○ $h_0:$ the proportion of each color is 1/6
○ $h_0:$ the number of m&ms is independent of color
○ $h_0:$ the proportion for at least one of the colors is not 1/6
○ $h_0: \mu = \frac{1}{6}$

Explanation:

To determine the null hypothesis for a chi - square goodness - of - fit test in this context, we consider the research question: whether the number of M&Ms is uniform over all 6 colors. A uniform distribution over 6 colors means that the proportion of each color should be \( \frac{1}{6} \).

• Option 2: The statement "The number of m&m's is independent of color" is related to a chi - square test of independence, not goodness - of - fit. So this option is incorrect.
• Option 3: "The proportion for at least one of the colors is not \( \frac{1}{6} \)" is more like an alternative hypothesis (or a statement that would be part of testing against the null of uniformity), not the null hypothesis itself.
• Option 4: The symbol \( \mu \) (mean) is not the appropriate parameter here. We are dealing with proportions of each color, not a mean. So this option is incorrect.

• Option 1: Since we want to test if the distribution of M&M colors is uniform (same number of each color, so same proportion \( \frac{1}{6} \) for each color), the null hypothesis \( H_0 \): The proportion of each color is \( \frac{1}{6} \) is correct.

Answer:

A. \( H_0 \): The proportion of each color is \( 1/6 \)