Free math worksheets

- 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 \(df=5-1=4\).

\(4\)

- A goodness-of-fit null hypothesis should state the entire proposed distribution. - The alternative does not need to predict which category will differ. - Avoid replacing the distributional claim with a statement about a sample count.

1. The null hypothesis specifies the claimed categorical distribution. 2. The alternative says the true distribution differs from the claimed distribution in at least one category.

\(H_0\): The package-size probabilities are \((0.50,0.30,0.20)\). \(H_a\): The package-size distribution is not \((0.50,0.30,0.20)\).

- Expected counts in goodness of fit come directly from the specified null probabilities. - Use the same sample size with each category probability. - Check that your expected counts sum to the total sample size.

1. Multiply the sample size by each null-model probability. 2. The expected counts are \(200\cdot0.50=100\), \(200\cdot0.30=60\), and \(200\cdot0.20=40\).

\((100,60,40)\)

- 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) \(df=5\).

- 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 \(\alpha=0.05\).

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

- Use the null probabilities, not the observed proportions, to form expectations. - Keep the expected count paired with its matching observed category. - Add all category contributions only after calculating them separately.

1. The expected counts are \((40,30,20,10)\). 2. The category contributions are \(\frac{36}{40}=0.9\), \(\frac{36}{30}=1.2\), \(\frac{4}{20}=0.2\), and \(\frac{4}{10}=0.4\). 3. Their sum is \(\chi^2=2.7\). 4. With four fully specified categories, \(df=4-1=3\).

Expected counts: \((40,30,20,10)\); \(\chi^2=2.7\); \(df=3\).

- Compare the p-value directly with the stated significance level. - The null hypothesis is the claimed categorical distribution. - State the conclusion about the population distribution rather than about whether the sample counts are identical to the expected counts.

1. Compare the p-value with the significance level: \(0.030<0.05\). 2. Reject the null hypothesis that the true contact-method distribution matches the company's claimed distribution. 3. The data provide statistically significant evidence that the distribution differs from the claim.

Reject \(H_0\). There is statistically significant evidence at the \(0.05\) level that the customer contact-method distribution differs from the company's claimed distribution.

- 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.

- 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 preexisting 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.

- Use the sample-size constraint before starting the goodness-of-fit calculation. - The null probabilities determine the expected counts independently of the missing observed value. - After finding the statistic, keep the p-value decision separate from the arithmetic that produced it.

1. Observed counts must sum to \(100\), so \(x=100-25-20-15=40\). 2. The expected counts are \((40,30,20,10)\). 3. The contributions are \(0\), \(\frac{25}{30}\approx0.833\), \(0\), and \(\frac{25}{10}=2.5\), so \(\chi^2\approx3.333\). 4. With four categories, \(df=3\). 5. Since \(0.343>0.05\), fail to reject the null hypothesis. The data do not provide statistically significant evidence that the distribution differs from the null model.

a) \(x=40\) b) \(\chi^2\approx3.333\), \(df=3\) c) Fail to reject \(H_0\); there is not statistically significant evidence of a different distribution.

- 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) \((48,30,18,24)\)

- 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.

- Convert the full claimed probability vector into expected counts before comparing categories. - Keep each category's standardized contribution separate; a larger raw residual does not automatically mean a larger chi-square contribution. - Compare the same p-value separately with each significance level. - Distinguish the numerical evidence in the data from the decision rule chosen for interpreting that evidence.

1. \(H_0\): the population category probabilities are \((0.40,0.25,0.20,0.10,0.05)\). \(H_a\): the population distribution differs from that model. 2. The expected counts are \((80,50,40,20,10)\). 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 \(\frac{49}{10}=4.90\), so \(\chi^2=11.98\). 4. With five fully specified categories, \(df=5-1=4\). 5. At \(\alpha=0.05\), \(0.0175<0.05\), so reject \(H_0\). At \(\alpha=0.01\), \(0.0175>0.01\), so fail to reject \(H_0\). 6. Category 5 contributes the most, \(4.90\), even though Category 1 has the largest raw residual magnitude. Category 5 has the smaller expected count, so its squared residual is divided by a much smaller denominator. 7. The p-value summarizes the same sample evidence in both decisions. Changing \(\alpha\) changes the preset rejection threshold, not the data or p-value.

a) \(H_0\): probabilities \((0.40,0.25,0.20,0.10,0.05)\); \(H_a\): the distribution differs. b) Expected counts \((80,50,40,20,10)\); \(\chi^2=11.98\); \(df=4\). c) Reject at \(\alpha=0.05\); fail to reject at \(\alpha=0.01\). d) Category 5 contributes the most, \(4.90\). Its smaller expected count makes its standardized contribution exceed Category 1's despite Category 1 having the larger raw residual magnitude. e) The significance level is the decision threshold; changing it can change the decision without changing the observed evidence or p-value.

- 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.