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Chapter 12: Problem 5

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

Short Answer

Expert verified
100, 50, 225, 125

Step by step solution

01

- Understand the Problem

You are given a total number of trials, represented by the parameter n, which in this case is 500. You are also provided with probabilities for each outcome, denoted as \( p_i \) for the various categories. Your task is to determine the expected count for each outcome.
02

- Formula for Expected Count

The formula to calculate the expected count for each outcome is given by \( E_i = n \times p_i \), where \( E_i \) is the expected count, \( n \) is the total number of trials, and \( p_i \) is the probability of the outcome.
03

- Calculate the Expected Counts

Using the formula from Step 2, you can calculate the expected counts for each outcome: For the first outcome: \( E_1 = 500 \times 0.2 = 100 \)For the second outcome: \( E_2 = 500 \times 0.1 = 50 \)For the third outcome: \( E_3 = 500 \times 0.45 = 225 \)For the fourth outcome: \( E_4 = 500 \times 0.25 = 125 \)

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Key Concepts

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

probability distribution
A probability distribution is a mathematical function that gives the probabilities of occurrence of different possible outcomes. For example, in our exercise, we have four possible outcomes with given probabilities: 0.2, 0.1, 0.45, and 0.25. Each number represents the chance of that particular outcome happening. The sum of all probabilities in a distribution always equals 1, showing that one of the possible outcomes will definitely occur.
expected count formula
The expected count formula helps to determine the average outcome expected over a large number of trials. The formula is: \( E_i = n \times p_i \), where:
  • \( E_i \) is the expected count for a specific outcome.
  • \( n \) is the total number of trials.
  • \( p_i \) is the probability of that outcome.
outcome analysis
Outcome analysis involves understanding and predicting outcomes based on given probabilities. From our exercise, using 500 trials and the given probabilities, we have calculated the expected counts for four different outcomes. For the first outcome, with a probability of 0.2, the expected count is 100. This analytic step is crucial in many fields, like quality control and market research.
multinomial distribution
The multinomial distribution is a generalization of the binomial distribution. It models the probabilities of obtaining a particular combination of counts for each category. In our case, we consider the probabilities for four different outcomes. Each outcome follows the multinomial distribution, represented by its own probability. This distribution helps to calculate the expected counts precisely and predict various scenarios in fields like genetics or polling data.
statistical calculation
Statistical calculation is the backbone of analyzing data. Using the steps outlined for our problem, we derive the expected counts by multiplying the total number of trials (500) by each outcome's probability. Performing these calculations:
  • For the first outcome: \( 500 \times 0.2 = 100 \)
  • For the second outcome: \( 500 \times 0.1 = 50 \)
  • For the third outcome: \( 500 \times 0.45 = 225 \)
  • For the fourth outcome: \( 500 \times 0.25 = 125 \)
Each step demonstrates the utility of statistical calculations in predicting and analyzing outcomes accurately.

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Most popular questions from this chapter

Explain why chi-square goodness-of-fit tests are always right tailed.

Determine the expected counts for each outcome. $$ \begin{array}{lllll} \hline \boldsymbol{n}=\mathbf{7 0 0} & & & & \\ \hline p_{i} & 0.15 & 0.3 & 0.35 & 0.20 \\ \hline{\text {Expected counts }} & & & & \\ \hline \end{array} $$

Family Structure and Sexual Activity A sociologist wants to discover whether the sexual activity of females between the ages of 15 and 19 years and family structure are associated. She randomly selects 380 females between the ages of 15 and 19 years and asks each to disclose her family structure at age 14 and whether she has had sexual intercourse. The results are shown in the table. Data are based on information obtained from the National Center for Health Statistics. $$ \begin{array}{lcccc} &&{\text { Family Structure }} \\ \hline & \text { Both Biological } & & & \\ \text { Had Sexual } & \text { or Adoptive } & \text { Single } & \text { Parent and } & \text { Nonparental } \\ \text { Intercourse } & \text { Parents } & \text { Parent } & \text { Stepparent } & \text { Guardian } \\ \hline \text { Yes } & 64 & 59 & 44 & 32 \\ \hline \text { No } & 86 & 41 & 36 & 18 \\ \hline \end{array} $$ (a) Compute the expected values of each cell under the assumption of independence. (b) Verify that the requirements for performing a chi-square test of independence are satisfied. (c) Compute the chi-square test statistic. (d) Test whether family structure and sexual activity of 15 - to 19-year-old females are independent at the \(\alpha=0.05\) level of significance. (e) Compare the observed frequencies with the expected frequencies. Which cell contributed most to the test statistic? Was the expected frequency greater than or less than the observed frequency? What does this information tell you? (f) Construct a conditional distribution by family structure and draw a bar graph. Does this evidence support your conclusion in part (d)?

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. $$ \begin{array}{lc} \text { Color } & \text { Frequency } \\ \hline \text { Brown } & 57 \\ \hline \text { Yellow } & 64 \\ \hline \text { Red } & 54 \\ \hline \text { Blue } & 75 \\ \hline \text { Orange } & 86 \\ \hline \text { Green } & 64\\\ \hline \end{array} $$

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. \(H_{0}:\) The random variable \(X\) is binomial with \(n=4, p=0.3\) \(H_{1}:\) The random variable \(X\) is not binomial with \(n=4\) \(p=0.3\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 260 & 400 & 280 & 50 & 10 \\ \hline \text { Expected } & 240.1 & 411.6 & 264.6 & 75.6 & 8.1 \\ \hline \end{array} $$
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