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M\&M's As noted in an earlier chapter, Mars Inc. says that until very recently yellow candies made up \(20 \%\) of its milk chocolate M\&M's, red another \(20 \%,\) and orange, blue, and green \(10 \%\) each. The rest are brown. On his way home from work the day he was writing these exercises, one of the authors bought a bag of plain M\&M's. He got 29 yellow ones, 23 red, 12 orange, 14 blue, 8 green, and 20 brown. Is this sample consistent with the company's stated proportions? Test an appropriate hypothesis and state your conclusion. a. If the M\&M's are packaged in the stated proportions, how many of each color should the author have expected to get in his bag? b. To see if his bag was unusual, should he test goodness-of-fit, homogeneity, or independence? c. State the hypotheses. d. Check the conditions. e. How many degrees of freedom are there? f. Find \(\chi^{2}\) and the P-value. g. State a conclusion.

Short Answer

Expert verified
The answer includes expected frequencies for each color, selection of goodness-of-fit test, null and alternative hypotheses, conditions check, \(\chi^{2}\) and P-value calculations and lastly, the final conclusion based upon the comparison of P-value with the significance level.

Step by step solution

01

Expectation Calculation

Calculate the expected occurrence of each color of M&M's based on company's stated proportions and total M&M's purchased. Total M&M's = 29 (yellow) + 23 (red) + 12 (orange) + 14 (blue) + 8 (green) + 20 (brown) = 106. Expected: Yellow = 0.2 * 106, Red = 0.2 * 106, Orange = 0.1 * 106, Blue = 0.1 * 106, Green = 0.1 * 106, Brown = 0.3 * 106
02

Test Selection

The appropriate test for this exercise is the goodness-of-fit test. We use this test as we are comparing an observed distribution with a theoretical one.
03

Hypotheses

State Null hypothesis \(H_0\): The color distribution of M&M's follows company's stated proportions. State Alternative hypothesis \(H_a\): The color distribution of M&M's does not follow the company's stated proportions.
04

Check Conditions

To apply the chi-square test, the observed and expected counts should be at least 5 for all categories. In this case, our lowest expected and observed categories are: Green=8 and 10.6 respectively, which are over 5. Hence, we can continue with the chi-square test.
05

Degrees of Freedom

Degrees of freedom = number of categories - 1. We have 6 color categories, hence degrees of freedom = 6 - 1 = 5.
06

Chi-Square Calculation

Calculate the \(\chi^{2}\) statistic using the formula: \(\chi^{2} = \Sigma [ (O_i - E_i)^{2} / E_i ]\) for all categories of M&M's where \(O_i\) is observed frequency and \(E_i\) is expected frequency. This will give us the \(\chi^{2}\) value.
07

P-Value Calculation

Find the P-value corresponding to the calculated \(\chi^{2}\) value and degrees of freedom from the chi-square distribution table or use chi-square calculator. P-value is the probability of getting result equal to or more extreme than actual observation under null hypothesis.
08

Conclusion

If P-value < 0.05, reject null hypothesis (there is a significant difference between observed and expected distributions). If P-value > 0.05, fail to reject null hypothesis (there is no significant difference between observed and expected distributions). This will form the conclusion of test.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is a statistical process used to determine if there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. When faced with the question of whether M&M's in a bag reflect the company's stated proportions of colors, hypothesis testing is the tool employed. In this process, two competing hypotheses are formulated.

The null hypothesis, usually denoted as \(H_0\), represents a default statement that there is no effect or no difference, and in this context, it states that the M&M's color distribution follows the company's claimed proportions. The alternative hypothesis, \(H_a\), is the opposite claim which we test against the null hypothesis, and it posits that the color distribution does not abide by the company's claims.

To conduct the hypothesis test, we calculate a test statistic from our sample data—here, a chi-square statistic—which measures how much the observed data deviate from what we would expect under the null hypothesis. If this deviation is large enough, we may reject \(H_0\) and conclude that the colors do not align with the stated proportions.
Expected Frequency
Expected frequency forms the core of the chi-square goodness-of-fit test. It represents the frequency we would expect in each category if the null hypothesis were true. To calculate these expected frequencies, we use the proportions stated by the hypothesized distribution—in this case, the company's claimed ratios of M&M colors—and apply them to the total sample size.

For example, if Mars Inc. claims that 20% of the M&M's are yellow, then in a bag of 106 M&M's, we would expect 20% of 106, which is 21.2, to be yellow. It is crucial that these expected frequencies are sufficiently large (generally 5 or more) to justify the use of the chi-square test since the test assumes that the expected frequencies follow a normal distribution, which is a reasonable approximation when the frequencies are not too low.
Degrees of Freedom
Degrees of freedom (df) is a concept that provides an understanding of the amount of 'independence' within the data set used in different statistical analyses. It is essentially the number of independent values that can vary in the analysis without breaking any constraints. In the context of a chi-square goodness-of-fit test, degrees of freedom are calculated as the number of categories minus one (df = n - 1).

In our M&M's example, because there are 6 different colors, the degrees of freedom for the test would be 6 - 1 = 5. Degrees of freedom play a crucial role in determining the critical value from the chi-square distribution table, which is then compared with our chi-square test statistic to determine whether the evidence is strong enough to reject the null hypothesis.
P-value
The P-value is a key component in the conclusion of a hypothesis test. It tells us the probability of observing a test statistic as extreme or more extreme than the one calculated from our sample data, assuming the null hypothesis is true. A lower P-value suggests that the observed data are unlikely under the null hypothesis.

In hypothesis testing, we compare the P-value to a predefined level of significance (usually \(\alpha = 0.05\)), which is the threshold for deciding whether to reject the null hypothesis. If the P-value is less than the significance level, we reject \(H_0\), concluding that the data provide sufficient evidence against it. For the M&M's problem, if the calculated chi-square statistic results in a P-value lower than 0.05, we have enough evidence to suggest that the color distribution in the bag does not follow the company's declared proportions.

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