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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. $$ \begin{array}{lcccc} \hline\text { Outcome } & \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \\\ \hline \text { Observed } & 30 & 20 & 28 & 22 \\ \hline \text { Expected } & 25 & 25 & 25 & 25 \\ \hline \end{array} $$

Short Answer

Expert verified
Test statistic: 2.72, degrees of freedom: 3, critical value: 7.815. We fail to reject the null hypothesis.

Step by step solution

01

Calculate the Chi-Square Test Statistic

Use the formula for the chi-square test statistic: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]Where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. For each category A, B, C, and D, calculate the terms:\[\frac{(30 - 25)^2}{25} = 1,\frac{(20 - 25)^2}{25} = 1,\frac{(28 - 25)^2}{25} = 0.36,\frac{(22 - 25)^2}{25} = 0.36\]Sum these up to get the \(\chi^2\) test statistic:\[\chi^2 = 1 + 1 + 0.36 + 0.36 = 2.72\]
02

Determine Degrees of Freedom

The degrees of freedom (df) for a chi-square test are calculated with the formula: \[df = (k - 1)\]Where \(k\) is the number of categories. Here, there are 4 categories (A, B, C, D), so the degrees of freedom are:\[df = 4 - 1 = 3\]
03

Find the Critical Value

Using the chi-square distribution table, find the critical value for \(\alpha = 0.05\) and \(df = 3\). The critical value is 7.815.
04

Compare Test Statistic with Critical Value

Compare the calculated \(\chi^2\) test statistic (2.72) with the critical value (7.815). Since 2.72 < 7.815, we fail to reject the null hypothesis.
05

Conclusion

At the \(\alpha = 0.05\) level of significance, the test does not provide sufficient evidence to reject the null hypothesis. Therefore, we conclude that there is no significant difference between the observed and expected frequencies.

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Key Concepts

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

degrees of freedom
Degrees of freedom (df) are an essential concept in statistics, particularly in hypothesis testing. They refer to the number of independent values or quantities that can vary in a statistical calculation. In the context of a chi-square test, degrees of freedom help determine the shape of the chi-square distribution, which in turn helps find the critical value needed for decision-making.

The formula for calculating degrees of freedom in a chi-square test is straightforward:\[df = (k - 1)\]where \(k\) is the number of categories being compared. For example, if you have 4 categories (A, B, C, D), the degrees of freedom would be: \[df = 4 - 1 = 3\].
Degrees of freedom are crucial because they influence the critical value that you will look up in chi-square distribution tables. The higher the degrees of freedom, the more the chi-square distribution shifts and expands.
critical value
The critical value is a threshold used in statistical tests to decide whether to accept or reject the null hypothesis. It is based on the desired significance level (\(\alpha\)), which is usually set at 0.05 for a 95% confidence level. For a chi-square test, the critical value is found in the chi-square distribution table and is determined by the degrees of freedom and \(\alpha\).

For instance, let's say you have 3 degrees of freedom and an \(\alpha\) of 0.05. To find the critical value, you would look at the chi-square distribution table under these parameters. In our example, the critical value is 7.815.

Comparing the calculated chi-square test statistic (2.72) to this critical value (7.815) can help you make a conclusion. If your test statistic is less than the critical value, you fail to reject the null hypothesis; if it is greater, you reject the null hypothesis.
null hypothesis
The null hypothesis (\(H_0\)) is a fundamental aspect of statistical testing. It represents a statement of no effect or no difference. For a chi-square test, the null hypothesis often states that the observed frequencies are equal to the expected frequencies, meaning that any differences are due to random chance.

In our example, the null hypothesis is: \[ H_0: p_A = p_B = p_C = p_D = \frac{1}{4} \].
This means we assume that the proportions across categories A, B, C, and D are equal unless proven otherwise.
Alternatively, the alternative hypothesis (\(H_1\)) states that at least one of these proportions is different. After calculating the chi-square test statistic and the critical value, comparing them helps decide whether to reject the null hypothesis. If the test statistic is less than the critical value, it means we don't have enough evidence to reject \(H_0\), so we assume that the observed frequencies match the expected frequencies.

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Most popular questions from this chapter

Explain the differences between the chi-square test for independence and the chi-square test for homogeneity. What are the similarities?

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.

If the expected count of a category is less than 1 , what can be done to the categories so that a goodness-of-fit test can still be performed?

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 \(n=4\) \(p=0.3\) $$ \begin{array}{llllll} \boldsymbol{X} & \mathbf{0} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \text { Observed } & 260 & 400 & 280 & 50 & 10 \\ \hline \text { Expected } & 240.1 & 411.6 & 264.6 & 75.6 & 8.1 \\ \hline \end{array} $$

The following table contains observed values and expected values in parentheses for two categorical variables, \(X\) and \(Y\), where variable \(X\) has three categories and variable \(Y\) has two categories: $$ \begin{array}{cccc} & \boldsymbol{X}_{\mathbf{1}} & \boldsymbol{X}_{\mathbf{2}} & \boldsymbol{X}_{3} \\ \hline \boldsymbol{Y}_{\mathbf{1}} & 34(36.26) & 43(44.63) & 52(48.11) \\ \hline \boldsymbol{Y}_{\mathbf{2}} & 18(15.74) & 21(19.37) & 17(20.89) \\ \hline \end{array} $$ (a) Compute the value of the chi-square test statistic. (b) Test the hypothesis that \(X\) and \(Y\) are independent at the \(\alpha=0.05\) level of significance.

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