- Translate “equally likely” into a probability for one category. - Expected counts come from sample size multiplied by the null-model probability. - Check that all expected counts together sum to the sample size.

1. Equal likelihood gives each category probability \(\frac{1}{4}\). 2. The expected count is \(100\cdot\frac{1}{4}=25\) for each category.

\(25\) in each category.

- Count the number of outcome categories in the null distribution. - One constraint comes from the fact that all category counts must sum to the sample size. - Reduce the category count by one.

1. For a goodness-of-fit test with \(k\) categories and no parameters estimated from the data, \(df=k-1\). 2. Here \(df=5-1=4\).

\(4\)

- How many categorical variables are recorded per customer? - The sample is compared with a fixed claim, not with a second variable.

1. Only one categorical variable, cookie flavor, is recorded for each customer. 2. The observed flavor counts are compared with a fixed claimed distribution, so use a chi-square goodness-of-fit test.

A chi-square goodness-of-fit test.

- Separate the claim made before observing customers from the percentages found in the sample. - The null hypothesis is the distribution the data are being checked against.

1. The null hypothesis states the garage's claim about the population, not what happened in this sample. 2. Therefore \(H_0\) uses \((0.60,0.25,0.15)\). The sample frequencies are used to assess the claim.

\(H_0\) uses \((0.60,0.25,0.15)\), the claimed population probabilities. The sample frequencies are observations, not the null model.

- The sample provides the observed count. - Think about how many yogurt choices the claimed percentage predicts in a sample of this size.

1. The observed count is the number who actually chose yogurt: \(O=24\). 2. The model predicts \(E=80\cdot0.25=20\) yogurt choices.

\(O=24\), \(E=20\).

- First compare the two observed-minus-expected differences. - Does an equal difference guarantee an equal chi-square contribution? - After checking the raw differences, what else in the contribution formula could affect their sizes?

1. Both differences have magnitude \(4\), so both squared differences are \(16\). 2. The contributions are \(\frac{16}{10}=1.6\) for pretzels and \(\frac{16}{40}=0.4\) for popcorn. 3. Pretzels contribute more because the same difference is larger relative to the smaller expected count.

Pretzels contribute more: \(1.6\) versus \(0.4\) for popcorn, because the expected pretzel count is smaller.

- Compare the observations with the same expected count in each category. - How does squaring affect a larger departure?

1. Survey A's departures are \((1,0,-1)\), while Survey B's are \((3,0,-3)\). 2. All expected counts match across the two surveys, but Survey B's squared departures are larger. Survey B therefore has the larger chi-square statistic.

Survey B, because its counts differ more from the same expected counts.

- The null model is a claim about all three size probabilities, not just the most common size. - The alternative need not say in advance which size will differ. - Write hypotheses about the population's choices, not a particular sample's results.

1. The null hypothesis specifies probabilities \(0.50\), \(0.30\), and \(0.20\) for small, medium, and large boxes, respectively. 2. The alternative says the actual population distribution of box sizes differs from the claimed one for at least one size.

\(H_0\): The population probabilities for small, medium, and large boxes are \((0.50,0.30,0.20)\). \(H_a\): The population box-size distribution differs from \((0.50,0.30,0.20)\).

- Imagine a batch of \(100\) tickets first. How could that help with the given total? - How should a predicted count change if the number of tickets doubles but the proportion stays the same? - What total should all three predicted counts account for?

1. Multiply the total number of tickets by each claimed proportion. 2. The expected counts are \(200\cdot0.45=90\) Adult, \(200\cdot0.35=70\) Child, and \(200\cdot0.20=40\) Senior tickets.

Adult: \(90\); Child: \(70\); Senior: \(40\) tickets.

- Convert each null probability into an expected number of observations. - The condition is checked using expected counts rather than observed counts. - Inspect every category after calculating the expectations.

1. Multiply \(60\) by each null probability to get expected counts \(42\), \(12\), \(4.8\), and \(1.2\). 2. Two expected counts are below \(5\). 3. Therefore, the stated expected-count condition does not hold.

a) \((42,12,4.8,1.2)\) b) No.

- Compare the observed count with the count predicted by the null model. - Square the departure before scaling it. - Use the expected count as the denominator.

1. A category contributes \(\frac{(O-E)^2}{E}\). 2. Here the contribution is \(\frac{(28-20)^2}{20}=\frac{64}{20}=3.2\).

\(3.2\)

- Count how many categorical variables are measured on each sampled individual. - Decide whether the comparison is between two variables or between one observed distribution and a fixed model. - For a fully specified goodness-of-fit model, degrees of freedom depend on the number of categories.

1. There is one categorical variable whose observed distribution is being compared with a specified probability model, so a chi-square goodness-of-fit test is appropriate. 2. With \(6\) categories and a fully specified model, \(df=6-1=5\).

a) Chi-square goodness-of-fit test. b) \(df=5\).

- The claim concerns a pattern across three methods. What would it mean for that pattern to fit the sample? - How does the p-value compare with the chosen significance level? - Limit the conclusion to the kind of customers actually sampled.

1. Since \(0.030<0.05\), reject the null hypothesis that customers' most recent contact methods follow the claimed proportions. 2. The data provide statistically significant evidence that this distribution differs from the claim among customers who have contacted the store.

Reject \(H_0\). There is statistically significant evidence that the most recent contact methods of customers who have contacted the store differ in proportion from the bookstore's claim.

- Distinguish a failure to find evidence against a model from proof that the model is true. - Compare the p-value with the significance level first. - Phrase the conclusion in terms of what the sample does or does not provide evidence for.

1. Since \(0.41>0.05\), fail to reject the null hypothesis. 2. A large p-value does not prove that the claimed distribution is exactly correct. 3. It means the sample does not provide statistically significant evidence that the population distribution differs from the claim.

The conclusion is too strong. Fail to reject \(H_0\); the data do not provide statistically significant evidence against the claimed distribution, but the claim has not been proved.

- Use the labeled critical boundary rather than judging whether the statistic merely “looks large.” - Check whether \(T\) actually lies inside the shaded region. - A decision rule is tied to a specific degrees of freedom and significance level.

1. The graph shows a critical value of \(9.49\). 2. The observed statistic is \(T=7.20\), which lies to the left of \(9.49\) and outside the shaded rejection region. 3. Therefore, the null hypothesis should not be rejected at \(\alpha=0.05\). 4. Visual position alone is not enough; the statistic must be compared with the rejection boundary for the correct degrees of freedom and significance level.

Ingrid's claim is incorrect. Since \(7.20<9.49\), the statistic is outside the rejection region, so do not reject the null hypothesis at \(\alpha=0.05\).

- Use the model percentages, rather than the sample's observed proportions, to find expected counts. - Match each percentage to the same animal category in the observed list. - Once you have the model's cat prediction, compare it with the actual count.

1. Expected counts are \(80\cdot0.55=44\) dogs, \(80\cdot0.30=24\) cats, and \(80\cdot0.15=12\) small animals. 2. The observed cat count is \(18\), which is \(24-18=6\) below the expected count. 3. These expected counts are predictions under the claimed model, not a replacement for what was observed.

Expected counts: dogs \(44\), cats \(24\), small animals \(12\). Cats are \(6\) below expectation.

- Ask what assumption a p-value calculation starts with. - Compare the p-value with the significance level separately from interpreting its meaning. - Avoid describing nonrejection as proof of the model.

1. The p-value is not the probability that the null model is true. If the claimed distribution were true, \(0.08\) is the chance of a chi-square statistic at least as large as the observed one. 2. Since \(0.08>0.05\), fail to reject the null hypothesis. 3. The sample does not give statistically significant evidence that the ticket-type distribution differs from the claim.

The student is incorrect: \(p=0.08\) is a right-tail probability assuming the claimed model, not the probability that it is true. Fail to reject \(H_0\) at \(\alpha=0.05\).

- An ogive gives totals up to boundaries, not the counts in each separate interval. - Subtract the cumulative values at the two ends of the requested interval. - Use the final cumulative count to find the sample size for the model expectation.

1. The ogive's cumulative count at \(10\) is \(10\), and at \(20\) it is \(40\). The observed count in \([10,20)\) is \(40-10=30\). 2. The last cumulative count is \(80\), so the sample has \(80\) games. 3. Each of four equally likely intervals has expected count \(80\cdot\frac{1}{4}=20\).

a) \(30\) scores. b) \(20\) expected scores.

- Think about which plan is most likely to give a count close to the minimum allowed. - What would a model probability tell you about a sample of a particular size? - Consider whether a check on one plan can also rule out problems with the other plans.

1. The Family plan has the smallest claimed probability, \(0.04\), so it gives the smallest expected count. 2. With \(n=100\), the smallest expectation is \(100\cdot0.04=4\), below \(5\). 3. With \(n=150\), the smallest expectation is \(150\cdot0.04=6\), at least \(5\). All other categories have larger expected counts.

Sampling \(150\) customers meets the expected-count requirement; \(100\) does not.

- Start by translating the equal-probability model into expected counts. - Compute one contribution for each category before adding them. - Degrees of freedom depend on the number of categories in the fully specified null model.

1. Equal probabilities give expected counts of \(80\cdot\frac{1}{4}=20\) in each category. 2. The contributions are \(\frac{(30-20)^2}{20}=5\), \(0\), \(\frac{(18-20)^2}{20}=0.2\), and \(\frac{(12-20)^2}{20}=3.2\). 3. Therefore \(\chi^2=5+0+0.2+3.2=8.4\). 4. The degrees of freedom are \(4-1=3\).

Expected counts: \((20,20,20,20)\); \(\chi^2=8.4\); \(df=3\).

- Make sure Mystery's predicted count comes from the library's claim, not its sample frequency. - Try one genre at a time to avoid pairing an observed count with the wrong genre. - Do different genres with similar differences necessarily contribute equally to the statistic?

1. The expected counts for Mystery, Science Fiction, History, and Biography are \((35,25,25,15)\). 2. Their contributions are \(\frac{121}{35}\approx3.457\), \(\frac{1}{25}=0.04\), \(\frac{49}{25}=1.96\), and \(\frac{9}{15}=0.6\). 3. Thus \(\chi^2\approx6.057\). 4. With four fully specified categories, \(df=4-1=3\).

Expected counts: \((35,25,25,15)\); \(\chi^2\approx6.057\); \(df=3\).

- Count the categorical variables recorded in each study. - Ask whether there is a fixed target distribution or a second categorical variable. - Do not choose a procedure merely because both calculations involve observed and expected counts.

1. Study 1 uses a chi-square goodness-of-fit test because one categorical variable is compared with a specified probability distribution. 2. Study 2 uses a chi-square test of independence because two categorical variables are recorded on each sampled individual and the question concerns association. 3. The distinction comes from the data structure and null model, not from the fact that both procedures use a chi-square statistic.

Study 1: chi-square goodness of fit. Study 2: chi-square test of independence. Goodness of fit compares one categorical distribution with a fixed model; independence tests association between two categorical variables.

- Compute contributions category by category rather than comparing raw differences alone. - The denominator changes with the expected count. - After identifying the largest contribution, return to the sign of \(O-E\) to describe its direction.

1. The contributions are \(\frac{(40-50)^2}{50}=2\), \(\frac{(36-30)^2}{30}=1.2\), and \(\frac{(24-20)^2}{20}=0.8\). 2. The first category contributes the most. 3. Its observed count \(40\) is below its expected count \(50\).

a) Contributions: \((2,1.2,0.8)\). b) The first category contributes the most, and it is below expectation.

- Separate conditions about how observations were collected from conditions about expected cell sizes. - Ask whether every member of a target population had a probability-based chance to enter the sample. - A large sample can still be systematically unrepresentative.

1. The expected-count condition is satisfied because all expected counts exceed \(20\). 2. The sampling condition is problematic because the responses come from a voluntary-response sample rather than a random sample from a defined population. 3. Large expected counts address the chi-square approximation, but they do not remove selection bias or justify population inference from a self-selected sample.

The sampling condition is the problem. A voluntary-response sample can be biased, and large expected counts do not make it representative of the target population.

- Reverse the relationship between expected count, sample size, and null probability for Category C. - All claimed probabilities must add to \(1\). - After completing the probability model, convert each probability back to an expected count.

1. Since Category C has expected count \(18\), its claimed probability is \(\frac{18}{120}=0.15\). 2. The remaining probability for Category D is \(1-0.40-0.25-0.15=0.20\). 3. Multiplying the four probabilities by \(120\) gives expected counts \((48,30,18,24)\).

a) \(0.15\) b) \(0.20\) c) \((48,30,18,24)\)

- Ask where the null probabilities came from and whether they existed before the sample was observed. - A useful test needs a genuine discrepancy that the data are allowed either to support or contradict. - Consider what happens to every \(O-E\) term when expected proportions are copied from the observed sample.

1. In a goodness-of-fit test of a specified claim, the null probabilities must come from the claimed population model, not be chosen to match the same sample being tested. 2. Setting the null probabilities equal to the sample proportions forces every expected count to equal its observed count. 3. That guarantees \(\chi^2=0\) by construction and therefore provides no test of an external claim. 4. Kabelo must state the null distribution independently of these observed sample proportions before carrying out the goodness-of-fit comparison.

The null model was fitted to reproduce the observed sample exactly, so \(\chi^2=0\) is guaranteed and does not test a previously stated claim. The null probabilities must be specified independently of the sample for this goodness-of-fit task.

- Ask whether stronger evidence against a goodness-of-fit model comes from smaller or larger chi-square statistics. - The p-value includes null-distribution outcomes at least as extreme as the observed statistic. - Compare the shaded side of \(T\) in the two panels.

1. A chi-square statistic measures the total standardized discrepancy between observed and expected counts. 2. Larger chi-square values indicate greater disagreement with the null model. 3. Therefore, the p-value is the area at or to the right of the observed statistic \(T=9.00\). 4. Panel b) shades that right-tail region. Panel a) shades values less extreme than the observed statistic and therefore does not represent the p-value.

Panel b). The chi-square p-value is the area to the right of the observed statistic because larger chi-square values are more inconsistent with the null model.

- Read each bar height from the frequency grid before using the probability model. - Convert each model probability to an expected count using the sample size. - Compare the completed test statistic with the stated critical value.

1. The observed counts are \((5,25,35,25,10)\) for the five intervals. 2. Multiplying the model probabilities by \(100\) gives expected counts \((10,20,40,20,10)\). 3. The contributions are \(2.5\), \(1.25\), \(0.625\), \(1.25\), and \(0\), so \(\chi^2=5.625\). 4. Since \(5.625<9.49\), do not reject the proposed delivery-time distribution at \(\alpha=0.05\).

a) Observed: \((5,25,35,25,10)\); expected: \((10,20,40,20,10)\). b) \(\chi^2=5.625\). c) Do not reject the proposed model at \(\alpha=0.05\).

- Use the series labels and bar widths to distinguish observed counts from expected counts. - Compute one contribution \(\frac{(O-E)^2}{E}\) for each die face, then add them. - To find the largest contribution, compare standardized discrepancies rather than only noting whether a count is above or below expected.

1. From the graph, the observed counts for faces \(1\) through \(6\) are \((8,11,9,10,16,6)\). The expected count is \(10\) for each face. 2. The chi-square statistic is \(\frac{(8-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(16-10)^2}{10}+\frac{(6-10)^2}{10}=5.80\). 3. Since \(5.80<11.07\), do not reject the fair-die model at \(\alpha=0.05\). 4. Face \(5\) contributes \(\frac{36}{10}=3.60\), the largest single contribution.

a) Observed counts: \((8,11,9,10,16,6)\); \(\chi^2=5.80\). b) Do not reject the fair-die model at \(\alpha=0.05\). c) Face \(5\).

- Read each observed frequency from the bar height and the count grid. - Convert the four model probabilities into expected counts using the total of \(56\). - Compare the resulting statistic with the stated critical value after summing all four contributions.

1. The bar chart gives observed counts \((6,24,18,8)\). 2. Multiplying \(56\) by the stated probabilities gives expected counts \((7,21,21,7)\). 3. The chi-square statistic is \(\frac{1}{7}+\frac{9}{21}+\frac{9}{21}+\frac{1}{7}=\frac{8}{7}\approx1.14\). 4. Since \(1.14<7.81\), do not reject the proposed distribution at \(\alpha=0.05\).

a) Observed counts: \((6,24,18,8)\). b) Expected counts: \((7,21,21,7)\); \(\chi^2\approx1.14\). c) Do not reject the proposed model at \(\alpha=0.05\).

- Chi-square p-values are right-tail probabilities, so compare the test statistic with the listed upper-tail cutoffs. - Locate the two consecutive critical values between which the observed statistic falls. - Use the resulting p-value range, rather than the statistic alone, for the significance decision.

1. The observed statistic \(10.60\) is greater than the \(0.05\) upper-tail critical value \(9.488\) but less than the \(0.025\) critical value \(11.143\). 2. Therefore the right-tail probability satisfies \(0.025<p<0.05\). 3. Because the p-value is below \(0.05\), reject the null hypothesis at \(\alpha=0.05\). 4. The p-value range means that, if the claimed category distribution were correct, the probability of obtaining a chi-square statistic at least as large as \(10.60\) would be between \(2.5\%\) and \(5\%\).

a) \(0.025<p<0.05\). b) Reject \(H_0\) at \(\alpha=0.05\). c) Under the claimed distribution, the probability of a chi-square statistic at least as large as \(10.60\) is between \(2.5\%\) and \(5\%\).

- Use the frequency-axis grid to read the observed count for each category. - Expected counts come from the sample size and the stated null probabilities, not from the bar heights. - Compare the category contributions before using the supplied p-value for the decision.

1. From the chart, the observed counts are \((50,20,20,10)\). The expected counts are \((40,30,20,10)\). 2. The contributions are \(\frac{100}{40}=2.5\), \(\frac{100}{30}\approx3.333\), \(0\), and \(0\). 3. Thus \(\chi^2\approx5.833\), and Category 2 contributes the most. 4. Since \(0.120>0.05\), fail to reject the null hypothesis. The data do not provide statistically significant evidence that the category distribution differs from the null model.

a) Observed counts: \((50,20,20,10)\); expected counts: \((40,30,20,10)\). b) \(\chi^2\approx5.833\); Category 2 contributes the most. c) Fail to reject \(H_0\); there is not statistically significant evidence of a different distribution.

- Is the Park count a missing part of the observed ride total or a missing model probability? - Check that the predicted counts still sum to the same sample size. - A single location's departure and the whole test's result answer different questions.

1. The observed start-location counts must sum to \(100\), so \(x=100-25-20-15=40\) at Park. 2. The expected counts are \((32,28,25,15)\). 3. The contributions are \(\frac{64}{32}=2\), \(\frac{9}{28}\approx0.321\), \(\frac{25}{25}=1\), and \(0\), giving \(\chi^2\approx3.321\). 4. With four fully specified categories, \(df=3\). 5. Since \(0.345>0.05\), do not reject the null hypothesis. The sample does not provide significant evidence that the start-location proportions differ from the service's claim.

a) \(x=40\) rides starting at Park. b) \(\chi^2\approx3.321\), \(df=3\) c) Do not reject \(H_0\); the data do not provide statistically significant evidence that the distribution of ride-start locations differs from the claim.

- Read each observed count from the bar height using the count-axis grid. - Equal likelihood means all four expected counts are the same. - Compare the completed chi-square statistic directly with the stated critical value.

1. From the chart, the observed counts are \((20,25,35,20)\). Equal probabilities give expected counts \((25,25,25,25)\). 2. The contributions are \(\frac{25}{25}=1\), \(0\), \(\frac{100}{25}=4\), and \(\frac{25}{25}=1\). 3. Thus \(\chi^2=6\), and Category 3 has the largest contribution. 4. Because \(6<7.815\), fail to reject the null hypothesis at \(\alpha=0.05\). 5. The data do not provide statistically significant evidence that the four category probabilities differ from equality.

a) Observed counts: \((20,25,35,20)\); \(\chi^2=6\); Category 3 contributes the most. b) Fail to reject \(H_0\); there is not statistically significant evidence that the categories are not equally likely.

- Match the model probabilities to the named club categories before finding expected counts. - For part d), raw differences and contributions rank cells differently in some tables. What are you comparing in each case? - For part e), separate what the sample tells you from what the testing rule asks you to choose.

1. \(H_0\): The population club-choice probabilities are \((0.40,0.25,0.20,0.10,0.05)\). \(H_a\): The distribution differs from this claim. 2. The expected counts are \((80,50,40,20,10)\), all at least \(5\), so the expected-count condition holds. 3. The contributions are \(\frac{196}{80}=2.45\), \(\frac{64}{50}=1.28\), \(\frac{36}{40}=0.90\), \(\frac{49}{20}=2.45\), and \(\frac{49}{10}=4.90\), giving \(\chi^2=11.98\). 4. With five fully specified categories, \(df=5-1=4\). 5. At \(\alpha=0.05\), reject \(H_0\); at \(\alpha=0.01\), fail to reject \(H_0\), since \(0.01<0.0175<0.05\). 6. Other contributes most, \(4.90\). Its expected count is much smaller than Sports', so a smaller raw departure can give a larger scaled contribution. 7. The sample and p-value do not change; the chosen cutoff for rejecting the null changes.

a) \(H_0\): Club-choice probabilities \((0.40,0.25,0.20,0.10,0.05)\). \(H_a\): The distribution differs. b) Expected counts \((80,50,40,20,10)\), all at least \(5\); \(\chi^2=11.98\), \(df=4\). c) Reject at \(\alpha=0.05\); fail to reject at \(\alpha=0.01\). d) Other contributes most, \(4.90\), because its smaller expected count makes the standardized discrepancy larger despite a smaller raw difference. e) The same p-value crosses one significance cutoff but not the other.

- An ogive gives cumulative counts, so individual interval counts come from successive differences. - Read each cumulative point from the frequency grid before taking differences. - After reconstructing the frequency table, use the usual goodness-of-fit contributions.

1. The cumulative counts at the upper interval boundaries are \(10,30,65,90,100\). 2. Taking successive differences gives observed interval counts \((10,20,35,25,10)\). 3. Equal likelihood gives expected counts of \(20\) in every interval. 4. The chi-square statistic is \(\frac{100}{20}+0+\frac{225}{20}+\frac{25}{20}+\frac{100}{20}=22.50\). 5. Since \(22.50>9.49\), reject the equal-probability null model at \(\alpha=0.05\).

a) \((10,20,35,25,10)\). b) \(\chi^2=22.50\). c) Reject the null model at \(\alpha=0.05\).

- Check how the sample was selected and how large it is relative to the stated population before computing the statistic. - Verify that the response categories form one complete, nonoverlapping classification of the recorded variable. - Convert the claimed probabilities into expected counts and inspect the smallest expectation before testing. - Compare standardized chi-square contributions rather than assuming the largest raw residual must dominate. - Match the scope of the final conclusion to the population from which the random sample was drawn.

1. The null hypothesis is that the population probabilities are \((0.50,0.25,0.15,0.10)\); the alternative is that the population distribution differs from this model. 2. A simple random sample is stated. The sample size \(200\) is less than \(10\%\) of \(8000\). Each rider is placed in exactly one of the four categories, so the categories are mutually exclusive and exhaustive for the recorded variable. 3. The expected counts are \((100,50,30,20)\), all at least \(5\), so the expected-count condition is met. 4. The category contributions are \(3.24\), \(2\), approximately \(3.333\), and \(0.2\). Thus \(\chi^2\approx8.773\). 5. With four fully specified categories, \(df=4-1=3\). 6. Since \(8.773>7.815\), reject the null hypothesis at \(\alpha=0.05\). The sample provides statistically significant evidence that the current primary fare-payment distribution differs from the claimed distribution. 7. Cash contributes the most, approximately \(3.333\), even though transit card has the largest raw residual magnitude. Cash has a smaller expected count, so its squared residual is divided by a smaller denominator. 8. Because the sample was randomly drawn from the agency's registered riders, the conclusion can reasonably be generalized to that registered-rider population, not automatically to all people who use transit in the city.

a) \(H_0\): The primary fare-payment probabilities are \((0.50,0.25,0.15,0.10)\). \(H_a\): The distribution differs from that model. b) The simple random sample condition is met; \(200<0.10(8000)\); each rider is classified into exactly one of four exhaustive categories; expected counts \((100,50,30,20)\) are all at least \(5\). c) Expected counts: \((100,50,30,20)\); \(\chi^2\approx8.773\); \(df=3\). d) Reject \(H_0\). There is statistically significant evidence at the \(0.05\) level that the registered riders' primary fare-payment distribution differs from the claimed model. e) Cash contributes the most, approximately \(3.333\), even though transit card has the largest raw residual magnitude; cash's smaller expected count makes its standardized contribution larger. f) The conclusion can reasonably be generalized to River City Transit's registered riders.

- Treat the two proposed distributions as two separate null models for the same observed data. - Compare observed counts with each model's own expected counts rather than comparing the two models only to each other. - Keep a failure-to-reject decision distinct from proof that a null model is true. - Ask what quantity changes when the observations stay fixed but the null probabilities change.

1. Under Model A, the expected counts are \((40,30,20,10)\). The contributions are \(0.625\), \(0\), \(1.25\), and \(0\), giving \(\chi^2_A=1.875\). 2. Under Model B, the expected counts are \((55,15,20,10)\). The contributions are approximately \(1.818\), \(15\), \(1.25\), and \(0\), giving \(\chi^2_B\approx18.068\). 3. Since \(1.875<7.815\), fail to reject Model A at \(\alpha=0.05\). Since \(18.068>7.815\), reject Model B at \(\alpha=0.05\). 4. Failing to reject Model A does not prove it is correct; it means the sample does not provide sufficiently strong evidence against it at the chosen significance level. 5. Rejecting Model B provides statistically significant evidence that the population distribution differs from Model B, but the test does not establish a particular alternative model as the unique truth. 6. A goodness-of-fit statistic measures discrepancy relative to a specified null distribution. The same observed counts are close to Model A's expectations but much farther from Model B's expectations, especially in Category 2, so the resulting evidence differs sharply.

a) Model A: expected counts \((40,30,20,10)\), \(\chi^2=1.875\). Model B: expected counts \((55,15,20,10)\), \(\chi^2\approx18.068\). b) Fail to reject Model A; reject Model B at \(\alpha=0.05\). c) Neither statement is justified as phrased. Failure to reject does not prove Model A is correct, and rejection shows evidence against Model B rather than proving a unique alternative model. d) The chi-square statistic measures discrepancy from the particular null model being tested, so changing the expected distribution changes the standardized discrepancies even when the observed data stay fixed.