Math Problem Statement

The student population at the state college consists of 30% freshmen, 25% sophomores, 25% juniors, and 20% seniors. The college theater department recently staged a production of a modern musical. A researcher recorded the class status of each student entering the theater and found a total of 20 freshmen, 23 sophomores, 22 juniors, and 15 seniors. Is the distribution of class status for theatergoers significantly different from the distribution for the general college? Test at the .05 level of significance.

χ² critical value = (to three decimal places) χ² = (to two decimal places) Conclusion:

Fail to reject the null hypothesis; the distribution for theater goers is not significantly different from the distribution for the population of students. Reject the null hypothesis; the distribution for theater goers is significantly different from the distribution for the population of students. Fail to reject the null hypothesis; the distribution for theater goers is significantly different from the distribution for the population of students. Reject the null hypothesis; the distribution for theater goers is not significantly different from the distribution for the population of students.

Solution

To determine if the class distribution for theatergoers is significantly different from the general college population, we conduct a chi-square goodness-of-fit test.

Step 1: State the hypotheses

  • Null hypothesis (H₀): The distribution of theatergoers' class status matches the distribution of the general college population.
  • Alternative hypothesis (H₁): The distribution of theatergoers' class status is significantly different from the general college population.

Step 2: Calculate expected frequencies

The total number of theatergoers = 20+23+22+15=8020 + 23 + 22 + 15 = 80.
Use the college population proportions to compute the expected counts:

  • Freshmen: 0.30×80=240.30 \times 80 = 24
  • Sophomores: 0.25×80=200.25 \times 80 = 20
  • Juniors: 0.25×80=200.25 \times 80 = 20
  • Seniors: 0.20×80=160.20 \times 80 = 16

Step 3: Compute the chi-square statistic

The chi-square statistic is calculated using:

χ2=(OiEi)2Ei\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}

Where:

  • OiO_i = observed frequencies
  • EiE_i = expected frequencies
Class StatusObserved (OiO_i)Expected (EiE_i)(OiEi)2/Ei(O_i - E_i)^2 / E_i
Freshmen2024(2024)224=1624=0.667\frac{(20 - 24)^2}{24} = \frac{16}{24} = 0.667
Sophomores2320(2320)220=920=0.450\frac{(23 - 20)^2}{20} = \frac{9}{20} = 0.450
Juniors2220(2220)220=420=0.200\frac{(22 - 20)^2}{20} = \frac{4}{20} = 0.200
Seniors1516(1516)216=116=0.0625\frac{(15 - 16)^2}{16} = \frac{1}{16} = 0.0625

χ2=0.667+0.450+0.200+0.0625=1.3795\chi^2 = 0.667 + 0.450 + 0.200 + 0.0625 = 1.3795

Rounded to two decimal places: χ2=1.38\chi^2 = 1.38.

Step 4: Determine the critical value

The degrees of freedom (df) = number of categories1=41=3\text{number of categories} - 1 = 4 - 1 = 3.
At the α=0.05\alpha = 0.05 significance level, the critical value from the chi-square table is:

χcritical2=7.815\chi^2_{\text{critical}} = 7.815

Step 5: Compare and conclude

  • χ2=1.38\chi^2 = 1.38
  • χcritical2=7.815\chi^2_{\text{critical}} = 7.815

Since χ2=1.38<7.815\chi^2 = 1.38 < 7.815, we fail to reject the null hypothesis.

Conclusion

The distribution for theatergoers is not significantly different from the distribution for the general college population.


Summary

  • χcritical2=7.815\chi^2_{\text{critical}} = 7.815
  • χ2=1.38\chi^2 = 1.38
  • Fail to reject the null hypothesis. The distribution for theatergoers is not significantly different from the general student population.

Do you want further clarification or detailed breakdowns of any step?
Here are 5 related questions:

  1. How is the chi-square critical value determined?
  2. What are the assumptions of the chi-square goodness-of-fit test?
  3. How does changing the significance level affect the conclusion?
  4. How would the result change if the observed frequencies were more extreme?
  5. How is the degrees of freedom calculated?

Tip: Always compare χ2\chi^2 with the critical value to decide on the hypothesis.

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Math Problem Analysis

Mathematical Concepts

Chi-Square Goodness-of-Fit Test
Probability
Hypothesis Testing

Formulas

Chi-Square Formula: χ² = Σ((O_i - E_i)² / E_i)
Expected Frequency Formula: E_i = (proportion × total)

Theorems

Chi-Square Distribution

Suitable Grade Level

College Level