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A sociologist is studying various aspects of the personal lives of preeminent nineteenth-century scholars. A total of one hundred twenty subjects in her sample had families consisting of two children. The distribution of the number of boys in those families is summarized in the following table. Can it be concluded that the number of boys in two-child families of preeminent scholars is binomially distributed? Let \(\alpha=0.05\). $$ \begin{array}{lrrr} \hline \text { Number of boys } & 0 & 1 & 2 \\ \text { Number of families } & 24 & 64 & 32 \\ \hline \end{array} $$

Short Answer

Expert verified
Yes, it can be concluded that the number of boys in two-child families of preeminent scholars follows a binomial distribution since the test statistic is less than the critical value at \(\alpha\)=0.05, implying that we fail to reject the null hypothesis of a binomial distribution.

Step by step solution

01

Calculate Binomial Probabilities

Firstly, we calculate the theoretical binomial probabilities using the formula mentioned above. Here, n = 2 (children) and p = 0.5 (probability of a child being a boy). Thus, the expected probabilities for 0, 1 and 2 boys respectively will be:\nFor 0 boys: \(P(X=0) = C(2,0) \cdot (0.5^0) \cdot ((1-0.5)^2) = 0.25\)\nFor 1 boy: \(P(X = 1) = C(2,1) \cdot (0.5^1) \cdot ((1-0.5)^(2-1)) = 0.5\)\nAnd for 2 boys \(P(X = 2) = C(2,2) \cdot (0.5^2) \cdot ((1-0.5)^(2-2)) = 0.25\)
02

Calculate Expected Frequencies

Next, we calculate the expected frequencies by multiplying the theoretical probabilities with the total number of families. The total number of families = 120.\nTherefore, expected frequencies for 0 boys = \(.25 \cdot 120 = 30\) families\nExpected frequencies for 1 boy = \(.5 \cdot 120 = 60\) families\nAnd lastly, expected frequencies for 2 boys = \(.25 \cdot 120 = 30\) families
03

Using Chi-Square Goodness of Fit Test

The Chi-Square goodness of fit test checks if observed outcomes fit with expected outcomes. The test statistic is calculated as: \(\chi^2 = \Sigma [(O_i - E_i)^2/E_i]\), where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency.\nWe calculate:\nFor 0 boys: \((24-30)^2/30 = 1.2\)\nFor 1 boy: \((64-60)^2/60 = 0.27\)\nFor 2 boys: \((32-30)^2/30 = 0.13\)\nThen we sum up these values to find \(\chi^2 = 1.2 + 0.27 + 0.13 = 1.6\) which is the chi-square statistic.
04

Test Statistic against Critical Value

Our degree of freedom will be = number of categories - 1 = 3 - 1 = 2. From chi-square distribution table (df=2), the critical value at \(\alpha = 0.05\) is approximately 5.991. Our test statistic (1.6) is less than the critical value (5.991), so we fail to reject the null hypothesis.

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Key Concepts

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

Chi-Square Goodness of Fit Test
The Chi-Square Goodness of Fit Test is a robust statistical tool used to determine if observed data fits a specific probability distribution. When working with categorical data, such as the number of boys in families, we may have an expected distribution based on theory or prior knowledge—the binomial distribution in this case.

The Chi-Square test is particularly valuable for hypothesis testing, where we compare the observed frequencies (actual count of outcomes) to expected frequencies (theoretical count based on our presumed distribution). The formula used in the test is: \(\chi^2 = \Sigma [(O_i - E_i)^2/E_i]\), where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency for each category.

If the calculated Chi-Square statistic is less than the critical value from the Chi-Square distribution table, we maintain our initial assumption (the null hypothesis) that the observed distribution conforms to the expected distribution. Otherwise, we reject the null hypothesis, suggesting that the distribution of the data is significantly different from what we expected.
Probability Theory
Probability theory forms the foundation of statistical analysis, underpinning various distributions, including the binomial distribution. It deals with the likelihood of an event's occurrence and quantifies uncertainty within mathematically defined frameworks. The binomial distribution, in particular, represents the probability of a fixed number of successes in a specific number of independent Bernoulli trials, each with two possible outcomes (success or failure).

In this example, if we presume the birth of a boy or girl is equally likely (a common assumption for human births), we can describe the distribution using binomial probabilities where \( p = 0.5 \) for the success in each trial (any child being a boy). The probability of having 0, 1, or 2 boys in a two-child family can thus be computed using the formula: \(P(X = k) = C(n,k) \cdot p^k \cdot (1-p)^{n-k}\), where \(n\) is the number of trials (children), \(k\) is the number of successes (boys), and \(p\) is the probability of success on each trial.
Statistical Hypothesis Testing
Statistical hypothesis testing is a methodological framework used to make inferences about populations based on sample data. A researcher may propose a null hypothesis (\(H_0\)), a statement of no effect or no difference, and an alternative hypothesis (\(H_a\)) that contradicts the null hypothesis.

In the context of the Chi-Square Goodness of Fit Test, the null hypothesis typically asserts that the observed frequencies match the expected distribution. Conversely, the alternative hypothesis suggests a deviation from this expected pattern. The decision to reject or fail to reject the null hypothesis is made by comparing the test statistic to a critical value corresponding to a pre-determined significance level (\(\alpha\)).

The significance level, such as \(\alpha = 0.05\), represents the threshold for the probability of committing a Type I error—inadvertently rejecting a true null hypothesis. If the test statistic exceeds the critical value, it suggests sufficient evidence to reject the null hypothesis in favor of the alternative, indicating that the observed data likely come from a different distribution than expected.

The steps followed in hypothesis testing—from formulating hypotheses and calculating the test statistic to making a conclusion—provide a standardized approach to making inferences and drawing conclusions from dataset comparisons.

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Most popular questions from this chapter

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