Question

state your conclusion to the hypothesis test.
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	color	color	color	color	color
number	70	85	81	79	93	96

perform a chi-square goodness-of-fit test at the 5% significance level and state your conclusion.

hint: the test-statistics is \\(\chi^2 = 5.43\\)

\\(\bigcirc\\) there is not sufficient sample evidence to suggest that the distribution of colors is not uniform.
\\(\bigcirc\\) there is not sufficient sample evidence to suggest that the distribution of colors is uniform.
\\(\bigcirc\\) there is sufficient sample evidence to suggest that the distribution of colors is not uniform.
\\(\bigcirc\\) there is sufficient sample evidence to suggest that the distribution of colors is uniform.

Explanation:

Step1: Determine Degrees of Freedom

For a chi - square goodness - of - fit test with \(k\) categories, the degrees of freedom \(df=k - 1\). Here, there are \(k = 6\) colors, so \(df=6 - 1=5\).

Step2: Find Critical Value

Using a chi - square distribution table with \(\alpha=0.05\) and \(df = 5\), the critical value \(\chi^{2}_{0.05,5}=11.070\) (from chi - square distribution tables or statistical software).

Step3: Compare Test Statistic and Critical Value

The test statistic is given as \(\chi^{2}=5.43\). Since \(5.43<11.070\) (the critical value), we fail to reject the null hypothesis. The null hypothesis for a chi - square goodness - of - fit test for uniformity is that the distribution is uniform (\(H_0\): the number of M&Ms of each color is the same, i.e., the distribution is uniform; \(H_1\): the distribution is not uniform). Failing to reject \(H_0\) means that there is not sufficient sample evidence to suggest that the distribution of colors is not uniform.

Answer:

There is not sufficient sample evidence to suggest that the distribution of colors is not uniform.