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- 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 \(df=(2-1)(3-1)=2\).

\(2\)

- 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.

- 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 \(5\).

- Compare the horizontal positions of the observed-statistic line and the critical-value line. - A test statistic inside the shaded rejection region leads to rejection of the null hypothesis. - Think about why unusually large chi-square values count as evidence against the null model.

1. The graph labels the critical value as \(c=5.99\) and the observed statistic as \(T=6.40\). 2. The observed statistic lies to the right of the critical value, inside the shaded rejection region. 3. Therefore, reject the null hypothesis at \(\alpha=0.05\). 4. The shaded region consists of chi-square values at least as extreme as the critical cutoff under the null model; its total area is \(0.05\).

a) Critical value: \(5.99\); observed statistic: \(6.40\). b) Reject the null hypothesis. c) The shaded right tail is the \(5\%\) rejection region under the null distribution.

- Start by adding the observed counts to obtain all margins. - The balanced margins make the expected table especially simple. - Add all four cell contributions, then determine degrees of freedom from the table dimensions.

1. Both row totals and both column totals are \(50\), with grand total \(100\), so every expected count is \(25\). 2. Each cell contributes \(\frac{(5)^2}{25}=1\) to \(\chi^2\). 3. Therefore \(\chi^2=1+1+1+1=4\). 4. The degrees of freedom are \((2-1)(2-1)=1\).

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

- Find the row and column totals before calculating expected counts. - Look for repeated expected counts or repeated departures to reduce unnecessary arithmetic. - The test statistic is the sum of all six cell contributions.

1. Each row total is \(60\), each column total is \(40\), and the grand total is \(120\), so every expected count is \(20\). 2. The first and fourth cells contribute \(\frac{10^2}{20}=5\) each. 3. Each of the other four cells differs from expectation by \(5\), so each contributes \(\frac{5^2}{20}=1.25\). 4. Thus \(\chi^2=5+1.25+1.25+5+1.25+1.25=15\). 5. The degrees of freedom are \((2-1)(3-1)=2\).

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

- The p-value measures how surprising a test statistic at least this large would be under the null model. - Compare the p-value with the stated significance level before writing the conclusion. - Phrase the conclusion as evidence about association, not as proof of causation.

1. Compare the p-value with the significance level: \(0.015<0.05\). 2. Reject the null hypothesis of independence. 3. The data provide statistically significant evidence of an association between study setting and preferred note-taking method.

Reject \(H_0\). There is statistically significant evidence at the \(0.05\) level that study setting and preferred note-taking method are associated.

- 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.

- Begin by finding all row totals, column totals, and the grand total. - Compute expected counts before any chi-square contributions. - Keep the test-statistic calculation separate from the p-value decision.

1. The row totals are \(40\) and \(40\), the column totals are \(36\) and \(44\), and the grand total is \(80\). 2. The expected counts are \((18,22)\) in each row. 3. The four 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 \(df=1\). 5. Since \(0.0070<0.05\), reject the null hypothesis. The data provide statistically significant evidence of an association between group and outcome.

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.

- Read the six counts from the bar heights and use them to build the table margins. - Because the two row totals are equal, compare what that implies for their expected counts. - Group the two cell contributions by category only after computing them with their matching expected counts.

1. The observed rows are \((30,20,10)\) and \((15,25,20)\). Both row totals are \(60\); the column totals are \((45,45,30)\); the grand total is \(120\). 2. The expected counts in each row are \((22.5,22.5,15)\). 3. Category 1 contributes \(2.5+2.5=5\). Category 2 contributes \(\frac{6.25}{22.5}+\frac{6.25}{22.5}\approx0.556\). Category 3 contributes \(\frac{25}{15}+\frac{25}{15}\approx3.333\). 4. Category 1 contributes the most, and \(\chi^2\approx8.889\).

a) Expected rows: \((22.5,22.5,15)\) and \((22.5,22.5,15)\). b) Category totals are approximately \(5\), \(0.556\), and \(3.333\); Category 1 contributes the most. c) \(\chi^2\approx8.889\).

- First calculate the expected table from the margins. - Compare the formula for a raw residual with the formula for a chi-square contribution. - Consider what squaring does to positive and negative departures.

1. The row totals are \(30\) and \(70\), the column totals are \(40\) and \(60\), and the grand total is \(100\). The expected counts are \((12,18)\) and \((28,42)\). 2. The residuals are \((4,-4)\) and \((-4,4)\), so their signed sum is indeed \(0\). 3. A chi-square statistic does not add raw residuals. It adds \(\frac{(O-E)^2}{E}\), so opposite signs cannot cancel. 4. The contributions are \(\frac{16}{12}\), \(\frac{16}{18}\), \(\frac{16}{28}\), and \(\frac{16}{42}\), which sum to \(\chi^2\approx3.175\).

The raw residuals cancel because the margins are preserved, but chi-square uses squared, scaled residuals. The correct statistic is \(\chi^2\approx3.175\).

- Treat the two panels as separate hypothesis tests even though they have the same observed statistic. - In each panel, compare \(T\) with the labeled critical boundary. - The degrees of freedom determine which chi-square reference distribution is used.

1. In panel a), the critical value is \(5.99\). Since \(7.50>5.99\), the statistic is in the rejection region, so the null hypothesis is rejected. 2. In panel b), the critical value is \(12.59\). Since \(7.50<12.59\), the statistic is not in the rejection region, so the null hypothesis is not rejected. 3. The reference chi-square distribution depends on the degrees of freedom, so the cutoff for the same significance level changes with \(df\).

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

- Compare \(T\) with the critical line separately in each panel. - A larger significance level places more null-distribution area in the rejection region. - The observed statistic does not change when only the decision threshold changes.

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.

- 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 \(O=26\).

\(26\)

- Compare the conditional renewal percentages, not only the p-value. - Separate evidence that a difference exists from the magnitude of that difference. - Large samples can make relatively small departures from a null model statistically detectable.

1. The renewal rates are \(\frac{2500}{5000}=50\%\) for Group A and \(\frac{2700}{5000}=54\%\) for Group B. 2. The small p-value provides strong statistical evidence that membership group and renewal status are associated in the population represented by the study. 3. Statistical significance does not by itself show that the association is large or practically important. 4. Here the observed difference in renewal rates is \(4\) percentage points, so practical importance should be judged using context in addition to the p-value.

The result gives strong evidence of an association, but the p-value does not measure the association's practical size. The observed renewal-rate difference is \(4\) percentage points, so practical importance requires contextual judgment.

- Use the count axis and grid lines to read each bar height before building the two-way table. - Find the row totals, column totals, and grand total before calculating expected counts. - Track the two cell contributions within each category as you add the chi-square statistic.

1. The observed rows are \((50,30,20)\) and \((30,45,25)\). Each row total is \(100\); the column totals are \((80,75,45)\); the grand total is \(200\). 2. The expected counts in each row are \((40,37.5,22.5)\). 3. The combined contributions by category are \(5.000\), \(3.000\), and approximately \(0.556\), so \(\chi^2\approx8.556\). Category 1 contributes the most. 4. Because \(0.0139<0.05\), reject the null hypothesis of independence. The data provide statistically significant evidence of an association between group and category.

a) Observed rows: \((50,30,20)\) and \((30,45,25)\). Expected rows: \((40,37.5,22.5)\) and \((40,37.5,22.5)\). b) \(\chi^2\approx8.556\); Category 1 contributes the most. c) Reject \(H_0\); there is statistically significant evidence of an association.

- Build the margins first; the equal row totals create a useful expected-count pattern. - Keep a separate running list of cell contributions rather than jumping directly to the final statistic. - After the p-value decision, return to the table to interpret which observed-versus-expected departures matter most. - Distinguish the scope of a random-sample inference from the stronger requirements for a causal conclusion.

1. \(H_0\): academic program and preferred campus resource are independent. \(H_a\): they are associated. 2. Each row total is \(60\); the column totals are \(45\), \(60\), and \(75\). Therefore every row has expected counts \((15,20,25)\). 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 \((3-1)(3-1)=4\). 5. Since \(0.0035<0.05\), reject \(H_0\). There is statistically significant evidence of an association between academic program and preferred resource. 6. The largest contribution is Program A–Resource 1: \(\frac{(24-15)^2}{15}=5.4\). That cell has more observations than expected under independence. 7. The result establishes association, not causation; no causal assignment is described.

a) \(H_0\): program and resource preference are independent. \(H_a\): they are associated. b) Each expected row is \((15,20,25)\); \(\chi^2=15.66\), \(df=4\). c) Reject \(H_0\); there is statistically significant evidence of an association. d) Program A–Resource 1 contributes the most, \(5.4\), because \(24\) were observed versus \(15\) expected. e) No. The test supports association, not causation.

- Read each panel as one row of the observed two-way table. - Compare raw residuals first, then remember that chi-square contributions divide squared residuals by expected counts. - Equal raw departures need not have equal chi-square contributions when their expected counts differ.

1. The observed rows are Group A \((30,15,5)\) and Group B \((50,30,20)\). The row totals are \(50\) and \(100\), the column totals are \(80,45,25\), and the grand total is \(150\). 2. The expected rows are Group A \((26.67,15,8.33)\) and Group B \((53.33,30,16.67)\), rounded to two decimals. 3. For category 1, the two contributions are approximately \(0.417\) and \(0.208\), totaling \(0.625\). 4. For category 3, the two contributions are approximately \(1.333\) and \(0.667\), totaling \(2.000\). 5. The raw residual magnitudes tie, but category 3 has smaller expected counts. Dividing the same squared residual by smaller expected counts produces larger chi-square contributions.

a) Observed rows: \((30,15,5)\) and \((50,30,20)\). Expected rows: approximately \((26.67,15,8.33)\) and \((53.33,30,16.67)\). b) Category 1 contributes \(0.625\); category 3 contributes \(2.000\). c) Category 3 has smaller expected counts, so the same raw residual size is more substantial after standardization.

- 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.