- Count how many categorical variables are recorded for each individual. - Decide whether the question concerns association between those variables or fit to one claimed distribution. - Match the structure of the data to the chi-square procedure.

1. The data contain two categorical variables measured on one random sample. 2. The question asks whether the variables are associated. 3. The appropriate procedure is a chi-square test of independence for a two-way table.

A chi-square test of independence for a two-way table.

- Identify the number of rows and columns that contain categories, not totals. - Degrees of freedom depend on the table dimensions. - Reduce each dimension by one before multiplying.

1. For a two-way table, \(df=(r-1)(c-1)\). 2. Here \(r=2\) and \(c=3\), so \(df=(2-1)(3-1)=2\).

\(2\)

- Think about what knowing a student's grade would tell you if the variables were independent. - Compare the whole pattern of choices, not just the most popular class.

1. Under the null, music-class preference does not depend on grade level. 2. The percentages choosing band, choir, and music production should each be the same in ninth and tenth grade in the population.

\(H_0\): The percentages preferring band, choir, and music production are the same among ninth and tenth graders in the population.

- Separate the count recorded in the survey from the count predicted by the model. - Which count belongs in the denominator of a chi-square contribution?

1. The survey gives the observed count \(O=32\); the independence model gives \(E=25\). 2. The cell's contribution is \(\frac{(32-25)^2}{25}\).

\(O=32\), \(E=25\), contribution \(\frac{(32-25)^2}{25}\).

- Compare the two observed-minus-expected differences. - Think about the effect of squaring a negative difference. - Check whether the denominators are also the same.

1. The departures are \(14-10=4\) and \(6-10=-4\). 2. Their squares are both \(16\), and each contribution is \(\frac{16}{10}=1.6\). 3. The contributions are equal because chi-square squares the departures.

They contribute equally: \(1.6\) each, because the positive and negative departures have equal squares and equal expected counts.

- Use the p-value and the stated cutoff to make the statistical decision. - Write the conclusion about what this sample provides evidence for, not what it proves.

1. Since \(0.62>0.05\), fail to reject the null hypothesis. 2. The sample does not provide significant evidence of an association between grade level and preferred lunch place. It does not establish independence.

Fail to reject \(H_0\). The sample does not give statistically significant evidence that grade and preferred lunch place are associated; it does not prove independence.

- How many student populations are being compared? - Notice whether the data came from one sample or a separate sample at each school.

1. Separate samples were drawn from two school populations. 2. The question compares their categorical lunch-choice distributions, so the appropriate procedure is a chi-square test of homogeneity.

Chi-square test of homogeneity, because there is a separate sample from each school and the question compares the schools' lunch-choice distributions.

- Focus on distributions of the response categories across the groups being compared. - The null should describe equality of those categorical distributions. - The alternative does not need to claim that every store differs from every other store.

1. The null hypothesis states that the categorical response distribution is the same across the three stores. 2. The alternative states that the response distributions are not all the same across the stores.

\(H_0\): The satisfaction-response distribution is the same at all three stores. \(H_a\): The satisfaction-response distributions are not all the same at the three stores.

- Check the sampling method separately from the numerical cell-count condition. - Compare the sample size with one tenth of the population size. - For the expected-count check, the smallest expectation is enough to decide whether all cells clear the threshold.

1. The random-sample condition is met because the problem states that a simple random sample was used. 2. The \(10\%\) condition is met because \(150\) is at most \(10\%\) of a population larger than \(1500\). 3. The expected-count condition is met because the smallest expected count is \(6.2\), so every expected count is at least \(5\).

All three stated conditions are met.

- A chi-square contribution uses both the size of the observed-minus-expected difference and the expected count. - The departure is squared before it is scaled. - Make sure the denominator is the expected count, not the observed count.

1. A cell contributes \(\frac{(O-E)^2}{E}\) to \(\chi^2\). 2. Here the contribution is \(\frac{(18-12)^2}{12}=\frac{36}{12}=3\).

\(3\)

- A larger raw difference does not automatically mean a larger chi-square contribution. - Scale each squared departure by its own expected count. - Compare the resulting contributions, not just \(|O-E|\).

1. The contributions are \(\frac{(22-20)^2}{20}=0.2\), \(\frac{(12-8)^2}{8}=2\), and \(\frac{(30-25)^2}{25}=1\). 2. The second cell has the largest contribution.

The cell with \((O,E)=(12,8)\), with contribution \(2\).

- Use the degrees-of-freedom relationship as an equation rather than only as a forward formula. - Substitute the known number of row categories and leave the column count unknown. - Remember that the formula reduces each table dimension by one.

1. Let the number of column categories be \(c\). For a two-way table, \(df=(r-1)(c-1)\). 2. With \(r=3\), \(6=(3-1)(c-1)=2(c-1)\). 3. Therefore \(c-1=3\), so \(c=4\).

\(4\) column categories.

- Compare the p-value with the significance level to choose a test decision. - Say which two preferences the conclusion concerns. - Ask whether a survey that observes choices can establish cause and effect.

1. Since \(0.015<0.05\), reject the null hypothesis of independence. 2. The survey provides statistically significant evidence that study location and preferred note-taking method are associated. 3. A survey can show association, but it does not establish that one of the variables causes the other.

Reject \(H_0\). There is statistically significant evidence that study location and note-taking preference are associated. No causal effect is established by the survey.

- Distinguish “fail to reject” from “accept” or “prove.” - Compare the p-value with the significance level first. - State what the data fail to show rather than claiming certainty about the null model.

1. Since \(0.28>0.05\), fail to reject the null hypothesis of independence. 2. A large p-value does not prove the null hypothesis; it means the sample does not provide sufficiently strong evidence against it. 3. The appropriate conclusion is that the data do not provide statistically significant evidence of an association between the variables.

Omar's statement is incorrect. Fail to reject \(H_0\); the data do not provide statistically significant evidence of an association, but independence has not been proved.

- Identify exactly which counts the chi-square approximation condition refers to. - The observed table and the null-model expected table play different roles. - Check the smallest expected count against the threshold.

1. Narin's check uses the wrong quantities. 2. The usual high-school chi-square condition is stated in terms of expected cell counts, not observed cell counts. 3. Because one expected count is \(3.8<5\), that condition is not met, even though all observed counts exceed \(5\). 4. The chi-square approximation should not be used as stated for this table.

The reasoning is incorrect. The relevant condition concerns expected counts, and the expected count \(3.8\) violates the stated minimum of \(5\).

- Find the boundary where the shaded right tail begins. - Compare the reported test statistic with that boundary rather than estimating a distance on the graph. - For the shaded area's meaning, think about repeated samples if the null model were correct.

1. The graph marks the right-tail cutoff at \(5.99\). 2. Since \(6.40>5.99\), the statistic lies in the shaded rejection region. Reject the null hypothesis at \(\alpha=0.05\). 3. If the null hypothesis is true, the probability of falling in the shaded right tail is \(0.05\).

a) Reject \(H_0\): \(T=6.40\) is beyond the cutoff \(5.99\), in the rejection region. b) The shaded tail contains the largest \(5\%\) of chi-square values under the null model.

- Divide each laptop count by the total number in its own grade. - Compare percentages rather than raw counts because the grade totals differ. - What could a chi-square test tell you that the two sample percentages alone cannot?

1. Ninth graders total \(40\), so the laptop percentage is \(\frac{24}{40}=60\%\). 2. Tenth graders total \(60\), so the laptop percentage is \(\frac{15}{60}=25\%\). 3. The laptop percentage is \(35\) percentage points higher among ninth graders in this sample. The observed distribution differs between the grades. 4. These descriptive percentages alone do not establish statistical significance or causation.

Ninth grade: \(60\%\) laptop. Tenth grade: \(25\%\) laptop. The sampled ninth graders are more likely to prefer laptops; the percentages differ by \(35\) percentage points.

- Treat the junior total and the chess-choice total as table margins. - Compare the actual junior–chess count with the independence prediction. - Use the sign of \(O-E\) to describe the direction of the difference.

1. The expected count under independence is \(E=\frac{40\cdot36}{120}=12\). 2. The observed count is \(O=20\), which is \(20-12=8\) above expectation. 3. The sample has more junior chess-club choices than the independence model predicts for that cell.

Expected count: \(12\). The observed count is \(8\) above expectation.

- Separate the expected-count condition from how students entered the sample. - Think about who is unlikely to attend the sports fair. - Ask whether the sample represents the population named in the conclusion.

1. Expected counts of at least \(10\) satisfy the usual expected-count condition. 2. But students who visit a sports fair are not a random sample of all students; they may have different sports interests. 3. Large expected counts do not fix this sampling bias, so a school-wide population conclusion is not justified by the described design.

No. The expected-count condition holds, but sampling only sports-fair visitors can bias the results. The findings should not be generalized to all students without appropriate sampling.

- Start by saying what the null hypothesis claims. - A p-value is calculated under an assumption. What is the assumption in this test? - Does the test assign a probability to the hypothesis itself or to results from samples?

1. A p-value is calculated assuming the null hypothesis of independence is true; it is not the probability that independence is true. 2. A p-value of \(0.03\) means that if the variables were independent, the probability of a chi-square statistic at least as large as the observed one would be \(3\%\).

The statement is incorrect. Assuming grade level and lunch choice are independent, the chance of obtaining a chi-square statistic at least as large as the observed one is \(3\%\). The p-value is not a \(3\%\) chance that independence is true.

- Find the two grade totals and the two study-location totals. - Check whether symmetry helps with the expected counts. - Include each of the four cells in the chi-square sum.

1. Both row totals and column totals are \(50\), with grand total \(100\). Every expected count is \(25\). 2. Each cell contributes \(\frac{(5)^2}{25}=1\). 3. The statistic is \(\chi^2=1+1+1+1=4\), with \(df=(2-1)(2-1)=1\).

\(\chi^2=4\) and \(df=1\).

- Use the margins to predict the count for each grade-program combination. - Look for repeated discrepancies to simplify the arithmetic. - Sum all six contributions and determine degrees of freedom from the table dimensions.

1. Each row has total \(60\), each column has total \(40\), and the grand total is \(120\). Each expected count is \(20\). 2. The two Sports cells contribute \(\frac{10^2}{20}=5\) each. Each of the four other cells contributes \(\frac{5^2}{20}=1.25\). 3. Hence \(\chi^2=2\cdot5+4\cdot1.25=15\), with \(df=(2-1)(3-1)=2\).

\(\chi^2=15\) and \(df=2\).

- Pay attention to whether the data come from one sample with two variables or from separate samples of several populations. - Focus on what the null hypothesis says about categorical distributions. - Separate the sampling interpretation from the mechanics of the two-way-table calculation.

1. Because separate samples are taken from three populations and their categorical distributions are being compared, the natural framing is a chi-square test of homogeneity. 2. The data are still arranged in a two-way table, and the null model says the activity distribution is the same across schools. 3. Expected counts, chi-square contributions, the test statistic, and degrees of freedom are computed with the same two-way-table formulas used for a test of independence.

a) Chi-square test of homogeneity. b) The null model still produces expected counts from the two-way table margins, so the same chi-square statistic and degrees-of-freedom calculations apply.

- First find the junior and senior totals and the totals for each club preference. - Work out the expected counts before computing contributions to the statistic. - After comparing the reported p-value with the significance level, name the two variables in your conclusion.

1. The row totals are \(40\) and \(40\); the column totals are \(36\) and \(44\); the grand total is \(80\). 2. Each row has expected counts \((18,22)\). 3. The contributions are \(\frac{36}{18}=2\), \(\frac{36}{22}\approx1.636\), \(2\), and \(\frac{36}{22}\approx1.636\). 4. Thus \(\chi^2\approx7.273\), with \(df=(2-1)(2-1)=1\). 5. Since \(0.0070<0.05\), reject \(H_0\). The sample gives statistically significant evidence that grade level and preferred club type are associated.

a) Expected rows: \((18,22)\) and \((18,22)\). b) \(\chi^2\approx7.273\), \(df=1\). c) Reject \(H_0\). There is statistically significant evidence of an association between grade level and preferred club type.

- Check whether the two grade totals are equal. What does that suggest about each row's share of an expected column total? - A program can have many students but still fit the independence prediction closely. - Would the program with the highest observed bar necessarily produce the largest chi-square contribution?

1. For Sports, Music, and Art, the observed rows are juniors \((30,20,10)\) and seniors \((15,25,20)\). The row totals are both \(60\); the column totals are \((45,45,30)\) and the grand total is \(120\). 2. Each row's expected counts are \((22.5,22.5,15)\). 3. Sports contributes \(2.5+2.5=5\), Music contributes \(\frac{6.25}{22.5}+\frac{6.25}{22.5}\approx0.556\), and Art contributes \(\frac{25}{15}+\frac{25}{15}\approx3.333\). 4. Sports contributes the most, and \(\chi^2\approx8.889\).

a) Both expected rows are \((22.5,22.5,15)\). b) Contributions by Sports, Music, and Art are approximately \(5\), \(0.556\), and \(3.333\); Sports contributes the most. c) \(\chi^2\approx8.889\).

- First predict each bus and walking count from the row and column totals. - Compare what happens when signed residuals are added with what the chi-square formula adds. - Check all four cells before deciding whether the statistic is zero.

1. The row totals are \(30\) and \(70\), column totals \(40\) and \(60\), and grand total \(100\). Expected counts are \((12,18)\) and \((28,42)\). 2. The residuals are \((4,-4)\) and \((-4,4)\), whose signed sum is \(0\). 3. The chi-square statistic uses squared differences divided by expected counts, so negative and positive residuals cannot cancel. 4. Thus \(\chi^2=\frac{16}{12}+\frac{16}{18}+\frac{16}{28}+\frac{16}{42}\approx3.175\).

The signed residuals cancel, but the squared chi-square contributions do not. The correct statistic is \(\chi^2\approx3.175\).

- For each plan, compare renewals with the number of people sampled from that plan. - What does a small p-value say about the evidence against equal population rates? - Would the same numerical difference matter equally to every business decision?

1. The renewal rates in the samples are \(\frac{2500}{5000}=50\%\) for Basic and \(\frac{2700}{5000}=54\%\) for Premium. 2. The difference is \(54\%-50\%=4\) percentage points. 3. With random samples and the stated large populations, the small p-value gives strong evidence that the two plans' population renewal rates differ. 4. It does not show that a \(4\)-point difference is important to the fitness center. That depends on how a difference of this size affects its decisions and costs.

a) Basic: \(50\%\). Premium: \(54\%\). Difference: \(4\) percentage points. b) There is strong evidence that the two plans' population renewal rates differ. c) No. Statistical significance does not tell us whether a \(4\)-point difference is important for the fitness center's decisions.

- Imagine carrying the marked value of \(T\) horizontally from one panel to the other. - Ask whether an unchanged statistic must occupy the same place relative to both shaded tails. - What feature of the two tests determines which chi-square reference curve they use?

1. In panel a), the critical value is \(5.99\). Since \(7.50>5.99\), reject the null hypothesis. 2. In panel b), the critical value is \(12.59\). Since \(7.50<12.59\), do not reject the null hypothesis. 3. The reference chi-square distribution changes with the degrees of freedom, so the critical value at the same significance level can change.

a) Reject in panel a) with \(df=2\); do not reject in panel b) with \(df=6\). b) Different degrees of freedom use different chi-square reference distributions and therefore different critical values.

- Find the boundary between the shaded and unshaded parts of each curve. - Decide where \(T\) falls relative to each boundary. - How does the set of values that lead to rejection change when the shaded tail grows?

1. In panel a), the critical value is \(7.78\). Since \(8.00>7.78\), reject the null hypothesis at \(\alpha=0.10\). 2. In panel b), the critical value is \(9.49\). Since \(8.00<9.49\), do not reject the null hypothesis at \(\alpha=0.05\). 3. A larger significance level creates a larger rejection region and a smaller critical value, so the same statistic can cross one cutoff but not the other.

a) At \(\alpha=0.10\), reject the null hypothesis. At \(\alpha=0.05\), do not reject it. b) The significance level determines the size of the rejection region and therefore the critical value.

- Each bar gives a percentage of its own grade, not a percentage of all students. - Use each grade's sample size to find its library count. - For the independence prediction, first find the library share among all surveyed students.

1. The charts show \(60\%\) of juniors and \(40\%\) of seniors chose the library. These give \(0.60\cdot40=24\) juniors and \(0.40\cdot60=24\) seniors. 2. The counts match, but the percentages do not. Grade groups of different sizes can have equal counts and different preference rates. 3. There are \(24+24=48\) library choices among \(100\) students. Independence predicts a \(48\%\) library rate in each grade. 4. The expected counts are \(0.48\cdot40=19.2\) juniors and \(0.48\cdot60=28.8\) seniors.

a) \(24\) juniors and \(24\) seniors. b) Juniors have a \(60\%\) library rate, but seniors have a \(40\%\) rate. Equal counts do not mean equal preferences when group sizes differ. c) Expected counts: \(19.2\) juniors and \(28.8\) seniors.

- Each row is a different home area; make sure the percentages use the people in that row. - Would the overall bus share be the same in each area if home area and travel method were independent? - Keep the observed commuter pattern separate from what the test says about the wider population.

1. Downtown, \(\frac{35}{50}=70\%\) take the bus. In the suburbs, \(\frac{20}{50}=40\%\) take the bus. 2. In all, \(55\) of \(100\) commuters take the bus. Under independence, the expected downtown–bus count is \(\frac{50\cdot55}{100}=27.5\). 3. The observed downtown–bus count is \(35\), which is \(7.5\) above the independence prediction. 4. Since \(0.0026<0.05\), reject independence. There is statistically significant evidence that home area and main travel method are associated among bus-or-train commuters living downtown or in the suburbs in this city. The observed bus percentages and downtown excess indicate the direction of this association, not its cause.

a) Downtown: \(70\%\). Suburbs: \(40\%\). b) Expected downtown–bus count: \(27.5\). The observed count of \(35\) is \(7.5\) higher. c) Reject independence. There is statistically significant evidence of an association between home area and main travel method among the city's downtown and suburban bus-or-train commuters. The downtown bus percentage and count exceed the independence predictions; the test does not show why.

- Try a small table with its margins already fixed. - How many cells in one row can you choose before the row total fixes the last one? - What does the last row depend on when the column totals are known?

1. Choose counts in the first \(r-1\) rows and the first \(c-1\) columns. 2. The last cell in each of those rows is then fixed by its row total. The last row is fixed by the column totals. 3. There are \((r-1)(c-1)\) freely chosen cells. For a \(3\times4\) table, this gives \(2\cdot3=6\) degrees of freedom.

Once the counts in the first \(r-1\) rows and first \(c-1\) columns are chosen, their row totals fix the rest of those rows, and the column totals fix the last row. So there are \((r-1)(c-1)\) free cells. A \(3\times4\) table has \(2\cdot3=6\) degrees of freedom.

- Treat the chi-square contribution formula as an equation with the observed count unknown. - Undo the division by the expected count before taking a square root. - Use the stated direction of the departure to choose between the two possible signs.

1. Use \(\frac{(O-20)^2}{20}=1.8\). 2. Then \((O-20)^2=36\), so \(|O-20|=6\). 3. Because the observed count is greater than expected, \(O-20=6\), giving \(O=26\).

\(26\)

- Equal grade totals make it possible to check your expected table by symmetry. - A large difference between two observed bars is not always the largest difference from independence. - Keep your description of the biggest contribution separate from the test's overall conclusion.

1. For Art, Tech, and Sports, the observed rows are \((50,30,20)\) for ninth grade and \((30,45,25)\) for tenth grade. Both row totals are \(100\); the column totals are \((80,75,45)\) and the grand total is \(200\). 2. The expected counts in each row are \((40,37.5,22.5)\). 3. The combined contributions by club are \(5.000\), \(3.000\), and about \(0.556\). Hence \(\chi^2\approx8.556\); Art contributes the most. 4. Because \(0.0139<0.05\), reject the null hypothesis. There is significant evidence that grade and club preference are associated.

a) Observed rows: ninth grade \((50,30,20)\), tenth grade \((30,45,25)\). Both expected rows are \((40,37.5,22.5)\). b) \(\chi^2\approx8.556\); Art contributes the most. c) Reject \(H_0\); there is significant evidence of an association between grade level and club preference.

- Use program totals and resource totals to work out expectations. - Keep the nine chi-square contributions separate so you can compare their sizes. - After deciding from the p-value, use the category names to explain what the strongest departure means. - Ask whether students were randomly assigned to programs before making a causal claim.

1. \(H_0\): Academic program and campus resource preference are independent. \(H_a\): They are associated. 2. Each row total is \(60\); the column totals are \(45\), \(60\), and \(75\). Each row has expected counts \((15,20,25)\), so all expected counts are at least \(5\). 3. The cell contributions are \(5.4\), \(0.2\), \(1.96\), \(0.6\), \(1.25\), \(0.16\), \(2.4\), \(0.45\), and \(3.24\). Their sum is \(\chi^2=15.66\). 4. The degrees of freedom are \(df=(3-1)(3-1)=4\). 5. Since \(0.0035<0.05\), reject \(H_0\). There is statistically significant evidence of an association between program and preferred campus resource. 6. The Health sciences–Library cell contributes most: \(\frac{(24-15)^2}{15}=5.4\). More health sciences students prefer the library than expected under independence. 7. The test identifies association, not cause. No randomized assignment to academic programs is described.

a) \(H_0\): Academic program and preferred campus resource are independent. \(H_a\): They are associated. b) Every expected row is \((15,20,25)\); all are at least \(5\); \(\chi^2=15.66\), \(df=4\). c) Reject \(H_0\); there is statistically significant evidence of an association. d) Health sciences–Library contributes most, \(5.4\). The observed count \(24\) is above its expected count \(15\). e) No. Association alone does not show causation.

- The campus sample sizes differ. Compare a bar with its own predicted count, not the other campus's bar. - Try comparing the absolute residuals for Band and Art before comparing contributions. - If two contributions start with the same squared departure, what else could make them unequal?

1. For Band, Choir, and Art, East has observed counts \((30,15,5)\) and West has \((50,30,20)\). The row totals are \(50\) and \(100\); column totals are \(80,45,25\); the grand total is \(150\). 2. Expected counts are \((\frac{80}{3},15,\frac{25}{3})\) at East and \((\frac{160}{3},30,\frac{50}{3})\) at West, or approximately \((26.67,15,8.33)\) and \((53.33,30,16.67)\). 3. Band contributes \(\frac{5}{12}+\frac{5}{24}=0.625\) across the two cells; Art contributes \(\frac{4}{3}+\frac{2}{3}=2.000\). 4. Although both electives have raw departures of size \(\frac{10}{3}\) in each row, Art has smaller expected counts, making its standardized chi-square contribution greater.

a) Observed counts: East \((30,15,5)\), West \((50,30,20)\). Expected counts are approximately East \((26.67,15,8.33)\), West \((53.33,30,16.67)\). b) Band contributes \(0.625\); Art contributes \(2.000\). c) Art has smaller expected counts, so equal-sized residuals yield larger standardized contributions.

- Decide whether this is one sample with two variables or separate samples whose response distributions are being compared. - Check the sampling conditions before doing any chi-square arithmetic. - Build row totals, column totals, and the grand total before calculating expected counts. - Keep the individual cell contributions visible if you need to diagnose which cell drives the statistic most strongly. - Phrase a failure-to-reject conclusion as a statement about evidence, not certainty.

1. The null hypothesis is that the class-format distribution is the same at all three campuses; the alternative is that the distributions are not all the same. 2. The samples are stated to be separate simple random samples. The sample sizes are \(80\), \(100\), and \(120\), each less than \(10\%\) of a campus enrollment of at least \(2000\), so the sampling-independence condition is met. 3. The column totals are \(146\), \(92\), and \(62\), with grand total \(300\). The expected rows are approximately \((38.93,24.53,16.53)\), \((48.67,30.67,20.67)\), and \((58.40,36.80,24.80)\). Every expected count exceeds \(5\). 4. Summing all nine contributions gives \(\chi^2\approx7.942\). 5. The degrees of freedom are \((3-1)(3-1)=4\). 6. Since \(0.0937>0.05\), fail to reject the null hypothesis. The samples do not provide statistically significant evidence that class-format preferences have different distributions across the three campuses. 7. The largest contribution is North–In person, approximately \(2.111\), because \(48\) students were observed there versus about \(38.93\) expected under the common-distribution model. 8. Failing to reject means the observed differences are not strong enough for rejection at the chosen significance level; it does not establish that the population distributions are exactly identical.

a) \(H_0\): The class-format distribution is the same at the North, Central, and South campuses. \(H_a\): The distributions are not all the same. b) The stated separate simple random samples meet the random-sample condition; each sample is less than \(10\%\) of its campus population; all expected counts exceed \(5\). c) Expected rows are approximately \((38.93,24.53,16.53)\), \((48.67,30.67,20.67)\), and \((58.40,36.80,24.80)\); \(\chi^2\approx7.942\); \(df=4\). d) Fail to reject \(H_0\). There is not statistically significant evidence at the \(0.05\) level that the class-format distributions differ among the campuses. e) North–In person contributes the most, approximately \(2.111\); the observed count is above expectation. f) Failure to reject indicates insufficient evidence against the common-distribution model, not proof that the three population distributions are exactly equal.

- Focus first on how the samples were obtained, not on the fact that the displayed table is the same. - Compare what the null hypothesis says about one population in Study I with what it says about two sampled populations in Study II. - Once the observed margins are fixed, ask what part of the chi-square calculation can and cannot depend on the sampling design. - Keep the numerical test result separate from the population statement it supports.

1. Study I uses a chi-square test of independence. \(H_0\): school setting and preferred class format are independent in the county student population. \(H_a\): they are associated. 2. Study II uses a chi-square test of homogeneity. \(H_0\): the preferred-format distribution is the same in the urban-school and suburban-school populations. \(H_a\): the distributions differ. 3. The row totals are \(90\) and \(90\), the column totals are \(75\), \(60\), and \(45\), and the grand total is \(180\). Each expected row is \((37.5,30,22.5)\). 4. The six contributions sum to \(\chi^2=8.00\), and \(df=(2-1)(3-1)=2\). The arithmetic is the same because both tests compare the same observed table with the same margin-based expected table. 5. Since \(0.0183<0.05\), Study I rejects independence and provides evidence of an association between school setting and preferred class format among county students. 6. Study II rejects homogeneity and provides evidence that preferred-format distributions differ between the county's urban-school and suburban-school populations. 7. The calculations coincide, but the sampling designs determine different population questions. One asks about association between two variables in one population; the other compares the distribution of one response across separately sampled populations.

a) Study I: chi-square test of independence; \(H_0\): school setting and preferred class format are independent; \(H_a\): they are associated. Study II: chi-square test of homogeneity; \(H_0\): the preferred-format distributions are the same for urban-school and suburban-school populations; \(H_a\): they differ. b) Expected rows: \((37.5,30,22.5)\) and \((37.5,30,22.5)\); \(\chi^2=8.00\); \(df=2\). The numerical calculation is identical because the observed table and its margins are identical. c) Study I: reject independence; there is evidence of an association. Study II: reject homogeneity; there is evidence that the two population distributions differ. d) The sampling design determines the inferential question and population interpretation even when the chi-square arithmetic is identical.