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According to the manufacturer of M\&Ms, \(13 \%\) of the plain M\&Ms in a bag should be brown, \(14 \%\) yellow, \(13 \%\) red, \(24 \%\) blue \(, 20 \%\) orange, and \(16 \%\) green. A student randomly selected a bag of plain M\&Ms. He counted the number of \(\mathrm{M} \& \mathrm{Ms}\) that were each color and obtained the results shown in the table. Test whether plain M\&Ms follow the distribution stated by M\&M/Mars at the \(\alpha=0.05\) level of significance. $$ \begin{array}{lc} \text { Color } & \text { Frequency } \\ \hline \text { Brown } & 57 \\ \hline \text { Yellow } & 64 \\ \hline \text { Red } & 54 \\ \hline \text { Blue } & 75 \\ \hline \text { Orange } & 86 \\ \hline \text { Green } & 64\\\ \hline \end{array} $$

Short Answer

Expert verified
Fail to reject the null hypothesis. The distribution matches the manufacturer's stated distribution.

Step by step solution

01

- State the null and alternative hypotheses

The null hypothesis (H_0) states that the distribution of M&Ms' colors follows the manufacturer's distribution. The alternative hypothesis (H_a) states that the distribution of M&Ms' colors does not follow the manufacturer's distribution.
02

- Calculate the expected frequencies

First, calculate the total number of M&Ms: 57 (Brown) + 64 (Yellow) + 54 (Red) + 75 (Blue) + 86 (Orange) + 64 (Green) = 400. Then find the expected frequency for each color by multiplying the total number by the given percentages: Expected Brown = 0.13 * 400 = 52, Expected Yellow = 0.14 * 400 = 56, Expected Red = 0.13 * 400 = 52, Expected Blue = 0.24 * 400 = 96, Expected Orange = 0.20 * 400 = 80, Expected Green = 0.16 * 400 = 64.
03

- Use the chi-square formula

The chi-square formula is shown below for each color component, summarize them as follows: Chi-Square = (57 − 52)2 52 + (64 − 56)2 56 + (54–52)2 52 +(75-96)2/96 + (86-80)*80 + (64-64)2/64 Chi-square is calculated by the sum of these individual values
04

- Determine the critical value and compare

First, calculate the degrees of freedom as the number of categories minus 1: 6 − 1 = 5. With \( \alpha = 0.05 \) and 5 degrees of freedom, use a chi-square table to find the critical value, which is approximately 11.07. Compare the calculated chi-square value to this critical value.
05

- Make a decision

Since the calculated chi-square value (3.8) is less than the critical value (11.07), we fail to reject the null hypothesis.
06

- Conclusion

There is not enough evidence to conclude that the distribution of M&Ms' colors differs from the manufacturer's stated distribution at the \( \alpha = 0.05 \) significance level.

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Key Concepts

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

Expected Frequency
In a chi-square test, the concept of **expected frequency** is crucial. Expected frequency is the number of occurrences that we expect to find in each category, assuming the null hypothesis is true. We use this to compare against our observed data.
For our M&Ms example, if the manufacturer claims that 13% of the M&Ms should be brown, and we have 400 M&Ms in total, the expected number of brown M&Ms is calculated by: \ \(Expected~Brown = 0.13 * 400 = 52\).
We repeat this for each color: \ - Expected Yellow = 0.14 * 400 = 56 \ - Expected Red = 0.13 * 400 = 52 \-Expected Blue = 0.24 * 400 = 96 \ - Expected Orange = 0.20 * 400 = 80 \ - Expected Green = 0.16 * 400 = 64.
These expected numbers help us to understand if the observed frequencies deviate significantly from what we would anticipate.
Null Hypothesis
The **null hypothesis** (often written as \(H_0\)) in statistical testing is our starting assumption. It usually states that there is no effect or no difference. For the M&Ms example, the null hypothesis is that the color distribution of the M&Ms in the bag follows the manufacturer's stated percentages.
This means that we assume the proportions they provided (13% brown, 14% yellow, etc.) are accurate, and any deviations in our sample are just due to random chance.
If our data strongly contradicts the null hypothesis, we have grounds to reject it, concluding that the actual distribution may be different from what the manufacturer claims.
Degrees of Freedom
In a chi-square test, **degrees of freedom** (df) are important because they help us determine the critical value from the chi-square distribution table.
The degrees of freedom in our test are calculated by taking the number of categories and subtracting one. For our M&Ms data, we have 6 colors, so df = 6 - 1 = 5.
The degrees of freedom reflect the number of values that are free to vary, considering we are using the data to estimate parameters. A higher number of degrees of freedom typically indicates a larger dataset and usually leads to more reliable results.
Significance Level
The **significance level** (often denoted by \(\alpha\)) is the threshold at which we decide whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true (a type I error).
Common significance levels are 0.05, 0.01, or 0.10. For the M&Ms example, we use \(\alpha = 0.05\), meaning we are willing to accept a 5% chance of wrongly rejecting the null hypothesis.
We compare our chi-square test statistic to the critical value corresponding to our significance level and degrees of freedom from the chi-square table. If our test statistic is greater than the critical value, we reject the null hypothesis.
Statistical Hypothesis Testing
The chi-square test is a form of **statistical hypothesis testing** used to see if there's a significant difference between the observed frequencies and the expected frequencies. This kind of testing involves the following steps:
1. State the null and alternative hypotheses. \ - \(H_0\): There is no significant difference between the observed and expected frequencies. \-\(H_a\): There is a significant difference between the observed and expected frequencies.
2. Calculate the expected frequencies based on the null hypothesis.
3. Use the chi-square formula \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) is observed frequency and \(E_i\) is expected frequency.
4. Determine the degrees of freedom and find the critical value from the chi-square table.
5. Compare the calculated chi-square value to the critical value and make a decision.
6. Conclude whether to reject or fail to reject the null hypothesis based on this comparison.
This structured approach allows us to make objective decisions about the data.

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Most popular questions from this chapter

Determine the expected counts for each outcome. $$ \begin{array}{lllll} \hline \boldsymbol{n}=\mathbf{5 0 0} & & & & \\ \hline p_{i} & 0.2 & 0.1 & 0.45 & 0.25 \\ \hline \text { Expected counts } & & & & \\ \hline \end{array} $$

The following table contains the number of successes and failures for three categories of a variable. $$ \begin{array}{lccc} & \text { Category } 1 & \text { Category } 2 & \text { Category } 3 \\ \hline \text { Success } & 76 & 84 & 69 \\ \hline \text { Failure } & 44 & 41 & 49 \\ \hline \end{array} $$ Test whether the proportions are equal for each category at the \(\alpha=0.01\) level of significance.

Determine \((a)\) the \(\chi^{2}\) test statistic, \((b)\) the degrees of freedom, (c) the critical value using \(\alpha=0.05,\) and (d) test the hypothesis at the \(\alpha=0.05\) level of significance. \(H_{0}: p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=\frac{1}{4}\) \(H_{1}\) : At least one of the proportions is different from the others. $$ \begin{array}{lcccc} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \text { Observed } & 30 & 20 & 28 & 22 \\ \hline \text { Expected } & 25 & 25 & 25 & 25 \\ \hline \end{array} $$

In a famous study by the Physicians Health Study Group from Harvard University from the late 1980 s, 22,000 healthy male physicians were randomly divided into two groups; half the physicians took aspirin every other day, and the others were given a placebo. Of the physicians in the aspirin group, 104 heart attacks occurred; of the physicians in the placebo group, 189 heart attacks occurred. The results were statistically significant, which led to the advice that males should take an aspirin every other day in the interest of reducing the chance of having a heart attack. Does the same advice apply to women? In a randomized, placebo-controlled study, 39,876 healthy women 45 years of age or older were randomly divided into two groups. The women in group 1 received \(100 \mathrm{mg}\) of aspirin every other day; the women in group 2 received a placebo every other day. The women were monitored for 10 years to determine if they experienced a cardiovascular event (such as heart attack or stroke). Of the 19,934 in the aspirin group, 477 experienced a heart attack. Of the 19,942 women in the placebo group, 522 experienced a heart attack. Source: Paul M. Ridker et al. "A Randomized Trial of Low-Dose Aspirin in the Primary Prevention of Cardiovascular Disease in Women." New England Journal of Medicine \(352: 1293-1304 .\) (a) What is the population being studied? What is the sample? (b) What is the response variable? Is it qualitative or quantitative? (c) What are the treatments? (d) What type of experimental design is this? (e) How does randomization deal with the explanatory variables that were not controlled in the study? (f) Determine whether the proportion of cardiovascular events in each treatment group is different using a two-sample \(Z\) -test for comparing two proportions. Use the \(\alpha=0.05\) level of significance. What is the test statistic? (g) Determine whether the proportion of cardiovascular events in each treatment group is different using a chi-square test for homogeneity of proportions. Use the \(\alpha=0.05\) level of significance. What is the test statistic? (h) Square the test statistic from part (f) and compare it to the test statistic from part \((\mathrm{g}) .\) What do you conclude?

In Section 10.2, we tested hypotheses regarding a population proportion using a z-test. However, we can also use the chi-square goodness-of-fit test to test hypotheses with \(k=2\) possible outcomes. In Problems 25 and \(26,\) we test hypotheses with the use of both methods. According to the U.S. Census Bureau, \(7.1 \%\) of all babies born are of low birth weight \((<5 \mathrm{lb}, 8 \mathrm{oz})\) An obstetrician wanted to know whether mothers between the ages of 35 and 39 years give birth to a higher percentage of low-birth-weight babies. She randomly selected 240 births for which the mother was 35 to 39 years old and found 22 low-birth-weight babies. (a) If the proportion of low-birth-weight babies for mothers in this age group is \(0.071,\) compute the expected number of low-birth-weight births to 35 - to 39 -year-old mothers. What is the expected number of births to mothers 35 to 39 years old that are not low birth weight? (b) Answer the obstetrician's question at the \(\alpha=0.05\) level of significance using the chi-square goodness-of-fit test. (c) Answer the question by using the approach presented in Section 10.2

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