Worked solution P70-Easy-1. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-2. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-3. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-4. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-5. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-6. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-7. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-8. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-9. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-10. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-11. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-12. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-13. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-14. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-15. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-16. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Easy-17. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-1. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-2. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-3. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-4. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-5. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-6. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-7. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-8. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-9. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-10. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-11. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-12. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-13. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-14. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-15. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-16. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Tough-17. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-1. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-2. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-3. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-4. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-5. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-6. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-7. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-8. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-9. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-10. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-11. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-12. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-13. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-14. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=6.960, df=3, p=0.0732. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=6.960,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-15. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=8.480, df=3, p=0.0371. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=8.480,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-16. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=10.160, df=3, p=0.0173. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=10.160,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.

Worked solution P70-Toughest-17. Expected counts are [25.0, 25.0, 25.0, 25.0]; chi-square=5.600, df=3, p=0.1328. $${\chi }^2=\sum \frac{(O-E{)}^2}{E}=5.600,df=4-1=3.$$ Interpretation: A small p-value indicates evidence against the null distributional claim, but the design controls whether the conclusion concerns association, homogeneity, or treatment effects. Validity: Use a random sample and expected counts large enough for the chi-square approximation. This procedure is removed from revised AP Statistics core. Error to reject: Do not present goodness-of-fit as a required May 2027 procedure.