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) \(df=5\).

- 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.

The student'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.

- 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.

- 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 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.

- The histogram supplies the observed interval counts; the probability model supplies the expected proportions. - Convert each model probability to an expected count using the sample size before computing contributions. - Compare the completed test statistic with the stated critical value.

1. The observed counts are \((6,24,38,25,7)\) for the five intervals. 2. Multiplying the model probabilities by \(100\) gives expected counts \((10,20,40,20,10)\). 3. The contributions are \(1.60\), \(0.80\), \(0.10\), \(1.25\), and \(0.90\), so \(\chi^2=4.65\). 4. Since \(4.65<9.49\), do not reject the proposed delivery-time distribution at \(\alpha=0.05\).

a) Observed: \((6,24,38,25,7)\); expected: \((10,20,40,20,10)\). b) \(\chi^2=4.65\). c) Do not reject the proposed model at \(\alpha=0.05\).

- Use the legend to distinguish the observed and expected frequency polygons. - 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 \(5\).

- Each vertical stack in the dot plot represents the frequency of one possible number of successes. - 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 dot plot 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 \(\alpha=0.05\).

- Read the observed counts from the values printed on the bars before using the null model. - Expected counts come from the sample size and the stated probabilities, not from the bar heights. - Track the contribution from each category separately before making the p-value 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) 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) \((48,30,18,24)\)

- 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. The student 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.

- Read the four observed counts from the labels on the bars. - Equal likelihood makes the expected-count calculation the same for every category. - A critical-value decision compares the test statistic directly with the stated cutoff.

1. From the chart, the observed counts are \((18,27,33,22)\). Equal probabilities give expected counts \((25,25,25,25)\). 2. The contributions are \(\frac{49}{25}=1.96\), \(\frac{4}{25}=0.16\), \(\frac{64}{25}=2.56\), and \(\frac{9}{25}=0.36\). 3. Thus \(\chi^2=5.04\), and Category 3 has the largest contribution. 4. Because \(5.04<7.815\), 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) \(\chi^2=5.04\); Category 3 contributes the most. b) Fail to reject \(H_0\); there is not statistically significant evidence that the categories are not equally likely.

- An ogive gives cumulative counts, so individual interval counts come from successive differences. - Equal probabilities across five intervals imply equal expected counts. - After reconstructing the frequency table, the goodness-of-fit calculation is the usual sum of cell 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 \(\alpha=0.05\).

- Convert the full claimed probability vector into expected counts before comparing categories. - Keep each category's contribution visible so you can diagnose which departure drives the statistic. - Compare the same p-value separately with each significance level. - Distinguish the numerical evidence in the data from the decision rule chosen before 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 \(\frac{400}{80}=5\), \(\frac{100}{50}=2\), \(\frac{100}{40}=2.5\), \(0\), and \(0\), so \(\chi^2=9.5\). 4. With five categories, \(df=4\). 5. At \(\alpha=0.05\), \(0.0497<0.05\), so reject \(H_0\). At \(\alpha=0.01\), \(0.0497>0.01\), so fail to reject \(H_0\). 6. Category 1 contributes the most, with contribution \(5\); its observed count \(100\) is above its expected count \(80\). 7. The p-value summarizes the same sample evidence in both decisions. Changing \(\alpha\) changes the preset threshold required for rejection, not the data or the strength of evidence represented by the p-value.

a) \(H_0\): probabilities \((0.40,0.25,0.20,0.10,0.05)\); \(H_a\): the distribution differs. b) Expected counts \((80,50,40,20,10)\); \(\chi^2=9.5\); \(df=4\). c) Reject at \(\alpha=0.05\); fail to reject at \(\alpha=0.01\). d) Category 1 contributes the most, and it is above expectation. e) The significance level is the decision threshold; changing it can change the decision without changing the observed evidence or p-value.