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Chapter 12: Problem 7

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?

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)?

Does the location of your seat in a classroom play a role in attendance or grade? To answer this question, professors randomly assigned 400 students a general education physics course to one of four groups. The 100 students in group 1 sat 0 to 4 meters from the front of the class, the 100 students in group 2 sat 4 to 6.5 meters from the front, the 100 students in group 3 sat 6.5 to 9 meters from the front, and the 100 students in group 4 sat 9 to 12 meters from the front. (a) For the first half of the semester, the attendance for the whole class averaged \(83 \% .\) So, if there is no effect due to seat location, we would expect \(83 \%\) of students in each group to attend. The data show the attendance history for each group. How many students in each group attended, on average? Is there a significant difference among the groups in attendance patterns? Use the \(\alpha=0.05\) level of significance. (b) For the second half of the semester, the groups were rotated so that group 1 students moved to the back of class and group 4 students moved to the front. The same switch took place between groups 2 and \(3 .\) The attendance for the second half of the semester averaged \(80 \% .\) The data show the attendance records for the original groups (group 1 is now in back, group 2 is 6.5 to 9 meters from the front, and so on ). How many students in each group attended, on average? Is there a significant difference in attendance patterns? Use the \(\alpha=0.05\) level of significance. Do you find anything curious about these data? $$ \begin{array}{lllll} \hline \text { Group } & 1 & 2 & 3 & 4 \\ \hline \text { Attendance } & 0.84 & 0.81 & 0.78 & 0.76\\\ \hline \end{array} $$ (c) At the end of the semester, the proportion of students in the top \(20 \%\) of the class was determined. Of the students in group \(1,25 \%\) were in the top \(20 \%\); of the students in group \(2,21 \%\) were in the top \(20 \%\); of the students in group \(3,15 \%\) were in the top \(20 \%\); of the students in group \(4,19 \%\) were in the top \(20 \% .\) How many students would we expect to be in the top \(20 \%\) of the class if seat location plays no role in grades? Is there a significant difference in the number of students in the top \(20 \%\) of the class by group? (d) In earlier sections, we discussed results that were statistically significant, but did not have any practical significance. Discuss the practical significance of these results. In other words, given the choice, would you prefer sitting in the front or back?

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