Question

determine the expected count under the null hypothesis. 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 | | brown | red | yellow | blue | orange | green | number | 70 | 85 | 81 | 79 | 93 | 96 | under the chi-square goodness-of-fit null hypothesis, what would be the expected number of each color for this bag of m&ms? 75 504 84 100

Explanation:

Step1: Calculate total number of M&Ms

Sum the numbers of each color: \( 70 + 85 + 81 + 79 + 93 + 96 \).
\( 70+85=155 \), \( 155+81=236 \), \( 236+79=315 \), \( 315+93=408 \), \( 408+96=504 \). So total \( n = 504 \).

Step2: Find expected count per color

Under null hypothesis (uniform distribution), expected count for each color is total divided by number of colors (6).
\( \text{Expected} = \frac{504}{6} = 84 \).

Answer:

84