Problem 29 - Basic Statistics for Business and Economics with Student

Question

Did you ever purchase a bag of M\&M's candies and wonder about the distribution of colors? You can go to the website www.baking.m-ms.com and click the United States on the map, then click About M\&M's, then History of M\&M's Brand, Product Information, and Peanut and find the percentage breakdown according to the manufacturer, as well as a brief history of the product. Did you know in the beginning they were all brown? For peanut M\&M's 12 percent are brown, 15 percent yellow, 12 percent red, 23 percent blue, 23 percent orange, and 15 percent green. A 6 -oz. bag purchased at the Book Store at Coastal Carolina University on November \(1,2005,\) had 12 blue, 14 brown, 13 yellow, 14 red, 7 orange, and 12 green. Is it reasonable to conclude that the actual distribution agrees with the expected distribution? Use the .05 significance level. Conduct your own trial. Be sure to share with your instructor.

Step-by-Step Solution

Answer

The observed distribution does not agree with the expected distribution.

Step 1: State the Hypotheses

To determine if the observed distribution of M&M’s matches the expected distribution, we will perform a chi-square test for goodness of fit. First, we state our hypotheses: - Null Hypothesis (\(H_0\)): The observed distribution of colors matches the expected distribution. - Alternative Hypothesis (\(H_a\)): The observed distribution of colors does not match the expected distribution.

Step 2: Calculate Expected Counts

Calculate the expected number of candies of each color based on the total number of candies and the expected percentage distribution. Total number of candies is \(12 + 14 + 13 + 14 + 7 + 12 = 72\). - Brown: \(72 \times 0.12 = 8.64\) - Yellow: \(72 \times 0.15 = 10.8\) - Red: \(72 \times 0.12 = 8.64\) - Blue: \(72 \times 0.23 = 16.56\) - Orange: \(72 \times 0.23 = 16.56\) - Green: \(72 \times 0.15 = 10.8\)

Step 3: Calculate the Chi-Square Test Statistic

For each color, use the formula \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. - Brown: \(\frac{(14 - 8.64)^2}{8.64} = 3.395\) - Yellow: \(\frac{(13 - 10.8)^2}{10.8} = 0.473\) - Red: \(\frac{(14 - 8.64)^2}{8.64} = 3.395\) - Blue: \(\frac{(12 - 16.56)^2}{16.56} = 1.256\) - Orange: \(\frac{(7 - 16.56)^2}{16.56} = 5.515\) - Green: \(\frac{(12 - 10.8)^2}{10.8} = 0.133\) - Sum: \(\chi^2 = 3.395 + 0.473 + 3.395 + 1.256 + 5.515 + 0.133 = 14.167\)

Step 4: Determine the Critical Value and Decision

The degrees of freedom for the test are \(6 - 1 = 5\) (number of categories minus 1). At a significance level of \(\alpha = 0.05\), the critical chi-square value from the chi-square distribution table is approximately 11.070. - If \(\chi^2 > \text{critical value}\), reject \(H_0\). Here, \(14.167 > 11.070\).

Step 5: Conclude

Since the calculated \(\chi^2\) statistic of 14.167 is greater than the critical value of 11.070, we reject the null hypothesis. This suggests that the observed distribution does not agree with the expected distribution at the 0.05 significance level.

Key Concepts

Hypothesis Testing

Hypothesis testing is a fundamental statistical method used to make decisions about a population based on sample data. It's like a structured way of asking and answering questions.

In the chi-square test for goodness of fit, we begin by setting up two opposing statements, known as the null hypothesis and the alternative hypothesis.

• The **Null Hypothesis** (\(H_0\)) states that there is no difference between the observed distribution of data and the expected distribution. For instance, in our M&M example, \(H_0\) claims that the color distribution in the bought bag aligns with the manufacturer's stated distribution.
• The **Alternative Hypothesis** (\(H_a\)) suggests that there's a significant difference between the observed data and what's expected. This might mean that the colors aren't as evenly distributed as the manufacturer claims.

After stating the hypotheses, the next step is to use the sample data to test them. We decide to "reject \(H_0\)" if the evidence strongly indicates the observed data doesn't fit the expected distribution. If there's no strong evidence, we "fail to reject \(H_0\)", accepting it as a plausible explanation.

Expected Distribution

The expected distribution is a crucial element in hypothesis testing, especially in the chi-square test for goodness of fit. It provides a benchmark to compare with the observed data.

In our M&M scenario, we determine the expected number of each candy color based on the total count of candies and their respective percentages. This step is essential because we're comparing what we see in reality (observed data) with what we were told to expect by the manufacturer.

• To calculate the expected count of each color, we multiply the total number of candies, which is 72 in this case, by the percentage expected for each color.
• For example, if 12% of the candies should be brown, the expected count is \(72 \times 0.12 = 8.64\).

This calculation is repeated for all colors. Having these expected values allows us to perform the chi-square test, determining if what we see matches what was expected. The smaller the difference between observed and expected counts, the higher the likelihood the null hypothesis is true.

Significance Level

The significance level, often represented by \(\alpha\), plays a key role in hypothesis testing. It helps in deciding whether to reject or fail to reject the null hypothesis.

In simple terms, the significance level is a threshold for making this decision. It defines how much evidence we need before we decide there's a real effect or difference present.

• In many tests, including our M&M example, a common choice for \(\alpha\) is 0.05. This indicates a 5% risk we're willing to take to incorrectly reject the null hypothesis – called a "Type I error".
• If the probability (p-value) that the observed data would differ from the expected data due to random chance is less than \(\alpha\), we reject \(H_0\).

Choosing \(\alpha = 0.05\) is like saying, "We're okay with a 5% chance of being wrong in our decision." It creates a balance between being too strict or too lenient in hypothesis testing.

Degrees of Freedom

Degrees of freedom are a concept in statistics that determine the number of values in a calculation that are free to vary. It impacts the critical values used in statistical tests.

In the chi-square test for goodness of fit, degrees of freedom help in finding the appropriate threshold for comparison, which tells us whether the observed differences are significant.

• The degrees of freedom are calculated based on the number of categories minus one. So in our M&M example, with six color categories, the degrees of freedom are \(6 - 1 = 5\).
• This number is critical when looking up the chi-square distribution table to find the critical value at a chosen significance level.

Understanding degrees of freedom is important because it affects how strictly we view our chi-square test statistic. The more categories we have, the higher the degrees of freedom, and the more wiggle room there is in the comparison.