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Chapter 11: Problem 12

Suppose we have a multinomial population with four categories: \(A, B, C,\) and \(D .\) The null hypothesis is that the proportion of items is the same in every category. The null hypothesis is \\[ H_{0}: p_{\mathrm{A}}=p_{\mathrm{B}}=p_{\mathrm{C}}=p_{\mathrm{D}}=.25 \\] A sample of size 300 yielded the following results. \\[ \begin{array}{llll} A: 85 & B: 95 & C: 50 & D: 70 \end{array} \\] Use \(\alpha=.05\) to determine whether \(H_{0}\) should be rejected. What is the \(p\) -value?

Short Answer

Expert verified
Reject the null hypothesis; p-value is approximately 0.0015.

Step by step solution

01

Calculate Expected Frequencies

The total sample size is 300, and under the null hypothesis, each category has an equal probability of 0.25. Therefore, the expected frequency for each category is given by \( E_i = 0.25 \times 300 = 75 \).
02

Compute the Chi-Square Statistic

The formula for the chi-square statistic is \( \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \), where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency. For A, B, C, and D, compute:\[ \chi^2 = \frac{(85-75)^2}{75} + \frac{(95-75)^2}{75} + \frac{(50-75)^2}{75} + \frac{(70-75)^2}{75} \] Simplifying, \( \chi^2 = \frac{100}{75} + \frac{400}{75} + \frac{625}{75} + \frac{25}{75} = 1.33 + 5.33 + 8.33 + 0.33 = 15.33 \).
03

Determine the Degrees of Freedom

The degrees of freedom for a chi-square test in a multinomial distribution is given by \( df = k - 1 \), where \( k \) is the number of categories. Here, \( df = 4 - 1 = 3 \).
04

Find the Critical Value and Compare

Using a chi-square distribution table, find the critical value for \( \alpha = 0.05 \) with 3 degrees of freedom, which is approximately 7.815. Since the calculated chi-square value (15.33) is greater than the critical value (7.815), we reject the null hypothesis.
05

Calculate the p-value

Use a chi-square distribution table or a calculator to find the p-value corresponding to \( \chi^2 = 15.33 \) with 3 degrees of freedom. The p-value is the probability of observing a chi-square statistic as extreme as, or more extreme than, 15.33. This p-value is approximately 0.0015.

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Key Concepts

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

Multinomial Distribution
A multinomial distribution is a generalization of the binomial distribution. It's used to model probabilities of outcomes in experiments where there are more than two possible results. Think of it as a setup where you categorize items into multiple groups or categories.
For example, if we perform a survey and classify responses into categories such as category A, B, C, and D, we might want to know how likely we are to get the counts we observe under certain assumptions. This is where the multinomial distribution comes into play.
  • Each category has its probability, called a multinomial probability.
  • The sum of all these probabilities equals one, as they represent all possible outcomes in an experiment.
In the original problem, the assumption is that the probabilities of choosing a particular category (A, B, C, or D) are equal, each being 0.25. This assumption is critical to test using a chi-square statistic.
Null Hypothesis
A null hypothesis is a foundational concept in statistical hypothesis testing. It represents a default or baseline assumption that there is no effect or difference in the context of an experiment. In scientific research, the null hypothesis is often stated with an equality sign.
In the exercise provided, the null hypothesis ( H_0 ) states that the proportion of items in each category is identical, namely 25% for A, B, C, and D. This means, under this hypothesis, there should be no significant difference in the data we collect, and any deviation is attributed to random chance.
  • The null hypothesis is a crucial part of testing where you decide to reject it or not based on the data.
  • Failing to reject the null hypothesis suggests that there is no statistical evidence to support a significant difference.
When the calculated test statistic is way higher than a critical threshold or the p-value is really low, it suggests that the observed data is unlikely under the null hypothesis, leading us to reject it.
Degrees of Freedom
Degrees of freedom refer to the number of independent values or quantities that can be assigned to a statistical distribution. They are a way of expressing how many values in a calculation are free to vary.
In the context of a chi-square test dealing with a multinomial distribution, the degrees of freedom are calculated using the formula: df = k - 1 , where k represents the number of categories.
  • This formula accounts for the fact that all category probabilities must sum to one, leaving one less than the total number of categories free to vary.
  • In the problem we are solving, with four categories (A, B, C, and D), the degrees of freedom amount to 3 since it's calculated as 4 - 1 = 3 .
Understanding degrees of freedom is essential because it influences the critical value of the chi-square distribution and thus affects decisions about the null hypothesis.

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Most popular questions from this chapter

An American Automobile Association (AAA) study investigated the question of whether a man or a woman was more likely to stop and ask for directions (AAA, January 2006). The situation referred to in the study stated the following: "If you and your spouse are driving together and become lost, would you stop and ask for directions?" A sample representative of the data used by AAA showed 300 of 811 women said that they would stop and ask for directions, while 255 of 750 men said that they would stop and ask for directions. a. The AAA research hypothesis was that women would be more likely to say that they would stop and ask for directions. Formulate the null and alternative hypotheses for this study. b. What is the percentage of women who indicated that they would stop and ask for directions? c. What is the percentage of men who indicated that they would stop and ask for directions? d. \(\quad\) At \(\alpha=.05,\) test the hypothesis. What is the \(p\) -value, and what conclusion would you expect AAA to draw from this study?

M\&M/MARS, makers of M\&M \(^{\oplus}\) Chocolate Candies, conducted a national poll in which more than 10 million people indicated their preference for a new color. The tally of this poll resulted in the replacement of tan-colored M\&Ms with a new blue color. In the brochure "Colors," made available by M\&M/MARS Consumer Affairs, the distribution of colors for the plain candies is as follows: \(\begin{array}{ccccc}\text { Brown } & \text { Yellow } & \text { Red } & \text { Orange } & \text { Green } & \text { Blue } \\ 30 \% & 20 \% & 20 \% & 10 \% & 10 \% & 10 \%\end{array}\) In a follow-up study, samples of 1-pound bags were used to determine whether the reported percentages were indeed valid. The following results were obtained for one sample of 506 plain candies. \(\begin{array}{ccccc}\text { Brown } & \text { Yellow } & \text { Red } & \text { Orange } & \text { Green } & \text { Blue } \\ 177 & 135 & 79 & 41 & 36 & 38\end{array}\) Use \(\alpha=.05\) to determine whether these data support the percentages reported by the company.

Medical tests were conducted to learn about drug-resistant tuberculosis. Of 142 cases tested in New Jersey, 9 were found to be drug-resistant. Of 268 cases tested in Texas, 5 were found to be drug-resistant. Do these data suggest a statistically significant difference between the proportions of drug-resistant cases in the two states? Use a .02 level of significance. What is the \(p\) -value, and what is your conclusion?

Where do women most often buy casual clothing? Data from the U.S. Shopper Database provided the following percentages for women shopping at each of the various outlets (The Wall Street Journal, January 28,2004 ). \(\begin{array}{lccc}\text { Outlet } & \text { Percentage } & \text { Outlet } & \text { Percentage } \\ \text { Wal-Mart } & 24 & \text { Kohl's } & 8 \\\ \text { Traditional department stores } & 11 & \text { Mail order } & 12 \\\ \text { JC Penney } & 8 & \text { Other } & 37\end{array}\) The other category included outlets such as Target, Kmart, and Sears as well as numerous smaller specialty outlets. No individual outlet in this group accounted for more than \(5 \%\) of the women shoppers. A recent survey using a sample of 140 women shoppers in Atlanta, Georgia, found 42 Wal-Mart, 20 traditional department store, 8 JC Penney, 10 Kohl's, 21 mail order, and 39 other outlet shoppers. Does this sample suggest that women shoppers in Atlanta differ from the shopping preferences expressed in the U.S. Shopper Database? What is the \(p\) -value? Use \(\alpha=.05 .\) What is your conclusion?

In July 2001 , the Harris Ad Track Research Service conducted a survey to evaluate the effectiveness of a major advertising campaign for Kodak cameras (USA Today, August 27, 2001 ). In a sample of 430 respondents, 163 thought the ads were very effective. In another sample of 285 respondents to other ad campaigns, 66 thought the ads were very effective. a. Estimate the proportion of respondents who thought the Kodak ads were very effective and the proportion of respondents who felt the other ads were very effective. b. Provide a \(95 \%\) confidence interval for the difference in proportions. c. \(\quad\) On the basis of your results in part (b), do you believe the Kodak advertising campaign is more effective than most advertising campaigns?
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