Free math worksheets

- Identify the type of response variable and the number of populations. - The question asks for a test of a difference, not an interval estimate.

1. The response is binary, and the question compares two independent population proportions. 2. The appropriate procedure is a two-sample z-test for the difference between two population proportions.

A two-sample z-test for the difference between two population proportions.

- Use population proportions, not the observed sample proportions. - Preserve the stated subtraction order.

1. The test compares two population proportions. 2. In the requested order, the parameter is \(p_A-p_B\).

\(p_A-p_B\).

- Decide whether the goal is estimation or evaluation of a claim. - The response is binary and two independent populations are being compared.

1. The research question asks whether a directional claim about the population proportions is supported. 2. A hypothesis test is therefore the primary procedure. 3. The appropriate test is a two-sample z-test for the difference between two population proportions.

Use a two-sample z-test for the difference between two population proportions.

- Translate “equal proportions” into an equation. - Rewrite that equation as a difference between the two proportions.

1. Equality means \(p_1=p_2\). 2. Subtracting gives \(p_1-p_2=0\).

\(0\).

- The word “differ” does not specify a direction. - Write the hypotheses using population parameters rather than sample statistics. - Keep the same parameter expression in both hypotheses.

1. “No difference” is \(p_U-p_R=0\). 2. “Differ” requires a two-sided alternative, \(p_U-p_R\ne0\). 3. Thus \(H_0:p_U-p_R=0\) and \(H_a:p_U-p_R\ne0\).

\(H_0:p_U-p_R=0\); \(H_a:p_U-p_R\ne0\).

- Translate “higher” into the direction of the difference new minus old. - The null hypothesis uses equality. - State hypotheses about the population proportions, not the observed conversions.

1. The null hypothesis states no population difference: \(p_N-p_O=0\). 2. The claim “new is higher” gives the one-sided alternative \(p_N-p_O>0\).

\(H_0:p_N-p_O=0\); \(H_a:p_N-p_O>0\).

- Translate “lower” using the stated order line \(1\) minus line \(2\). - The null hypothesis represents no population difference. - Use population proportions in the hypotheses.

1. No difference is represented by \(p_1-p_2=0\). 2. “Line \(1\) is lower” means \(p_1-p_2<0\). 3. Therefore, \(H_0:p_1-p_2=0\) and \(H_a:p_1-p_2<0\).

\(H_0:p_1-p_2=0\); \(H_a:p_1-p_2<0\).

- Ask whether hypotheses are supposed to describe the samples already observed or the populations being studied. - Distinguish symbols with hats from population-parameter symbols. - Keep the student's two-sided direction while correcting the quantities being tested.

1. Hypotheses describe unknown population parameters, not random sample statistics. 2. The population parameter is \(p_1-p_2\). 3. The correct hypotheses are \(H_0:p_1-p_2=0\) and \(H_a:p_1-p_2\ne0\).

The student used sample statistics instead of population parameters. Correctly: \(H_0:p_1-p_2=0\); \(H_a:p_1-p_2\ne0\).

- Separate sample size from the way the data were collected. - Ask whether each member of the target populations had a random mechanism for selection. - A large sample can still be systematically biased.

1. The respondents were self-selected rather than obtained through independent random samples or randomized assignment. 2. Large sample sizes do not repair the lack of randomization. 3. Therefore, the randomization condition is not satisfied.

No. The samples are voluntary-response samples, not independent random samples or randomized treatment groups.

- Compare each sample size with \(10\%\) of its own population. - Both population-size checks must pass. - Do not use the combined sample size for this condition.

1. For population \(1\), \(0.10(900)=90\), and \(70\le90\). 2. For population \(2\), \(0.10(1200)=120\), but \(140>120\). 3. The condition fails because it must hold for both samples.

No. It fails for sample \(2\) because \(140>120\).

- For this hypothesis test, first combine the successes to estimate the common proportion under \(H_0\). - Use that pooled proportion to find expected successes and failures for each sample. - Check all four expected counts against the threshold.

1. The pooled proportion is \(\hat p_c=\frac{48+52}{80+120}=0.50\). 2. Under the null model, the expected successes and failures are \(80(0.50)=40\) and \(40\) for sample \(1\), and \(120(0.50)=60\) and \(60\) for sample \(2\). 3. All four expected counts are at least \(10\), so the normality condition holds.

Yes. The pooled expected counts are \(40\), \(40\), \(60\), and \(60\), all at least \(10\).

- Pool the observed successes before checking the test's expected counts. - Expected successes and failures are computed under the null model. - The condition fails if any required expected count is below the threshold.

1. The pooled proportion is \(\hat p_c=\frac{4+6}{30+40}=\frac{10}{70}\approx0.1429\). 2. The expected numbers of successes are \(30(0.1429)\approx4.29\) and \(40(0.1429)\approx5.71\). 3. These are below \(10\), so the pooled normality condition fails.

No. The pooled expected success counts are below \(10\) in both samples.

- Separate design conditions from numerical normality conditions. - Use the pooled proportion because the null hypothesis assumes equal success proportions. - Decide whether a population-size condition is relevant to random assignment.

1. Random assignment satisfies the randomization condition. 2. The \(10\%\) condition is unnecessary for this randomized experiment. 3. The pooled proportion is \(\hat p_c=\frac{90+75}{300}=0.55\). 4. Each group has expected successes \(150(0.55)=82.5\) and expected failures \(150(0.45)=67.5\), all at least \(10\). 5. Therefore, the needed conditions are satisfied.

Yes. Randomization and pooled normality are satisfied, and no \(10\%\) check is needed for the randomized experiment.

- Pool successes and total observations across both samples first. - The pooled standard error represents the null model of a common population proportion. - Keep the two sample-size reciprocals separate inside the final factor.

1. The pooled proportion is \(\hat p_c=\frac{78+63}{120+140}=\frac{141}{260}\approx0.5423\). 2. The null standard error is \(\sqrt{\hat p_c(1-\hat p_c)(\frac{1}{120}+\frac{1}{140})}\). 3. Substitution gives \(SE\approx0.0620\).

\(\hat p_c\approx0.5423\) and \(SE\approx0.0620\).

- Find the observed difference in sample proportions first. - Under the null hypothesis, estimate the common proportion by pooling the successes. - Standardize the observed difference using the pooled standard error.

1. \(\hat p_1=0.56\), \(\hat p_2=0.44\), and \(\hat p_1-\hat p_2=0.12\). 2. The pooled proportion is \(\hat p_c=\frac{112+88}{400}=0.50\). 3. The pooled standard error is \(\sqrt{0.50(0.50)(\frac{1}{200}+\frac{1}{200})}=0.05\). 4. Thus \(z=\frac{0.12-0}{0.05}=2.40\).

\(z=2.40\).

- The alternative direction affects the p-value later, not the formula for the z statistic. - Pool the successes because the null hypothesis assumes equal population proportions. - Standardize the observed sample difference relative to the null value \(0\).

1. The sample proportions are \(0.50\) and \(0.40\), so the observed difference is \(0.10\). 2. The pooled proportion is \(\frac{75+60}{300}=0.45\). 3. The pooled standard error is \(\sqrt{0.45(0.55)(\frac{1}{150}+\frac{1}{150})}\approx0.05745\). 4. Therefore, \(z=\frac{0.10}{0.05745\ldots}\approx1.741\).

\(z\approx1.741\).

- Compute both sample proportions before pooling the success counts. - The pooled proportion is used only in the null standard error. - Divide the observed difference by the pooled standard error after centering at the null difference.

1. The sample proportions are \(0.54\) and \(0.48\), so the observed difference is \(0.06\). 2. The pooled proportion is \(\hat p_c=\frac{54+72}{250}=0.5040\). 3. The pooled standard error is \(\sqrt{0.504(0.496)(\frac{1}{100}+\frac{1}{150})}\approx0.0645\). 4. The test statistic is \(z=\frac{0.06}{0.0645477\ldots}\approx0.9295\).

\(\hat p_c=0.5040\), \(SE\approx0.0645\), and \(z\approx0.9295\).

- Identify which part of the z statistic changes when the order of subtraction is reversed. - The pooled standard error uses the same two samples regardless of order. - Check whether the magnitude or only the sign should change.

1. Reversing the subtraction changes the observed difference from \(d\) to \(-d\). 2. The pooled standard error is unchanged because it does not depend on subtraction order. 3. Therefore, the z statistic changes sign from \(2.40\) to \(-2.40\).

\(z=-2.40\).

- The null difference is \(0\). - Use the pooled proportion already supplied to form the null standard error. - Standardize only after the pooled standard error is found.

1. The pooled standard error is \(\sqrt{0.40(0.60)(\frac{1}{100}+\frac{1}{150})}\). 2. This standard error is \(\sqrt{0.004}\approx0.06325\). 3. Therefore, \(z=\frac{0.10-0}{0.06325\ldots}\approx1.581\).

\(z\approx1.581\).

- Work backward from the z-score definition. - The null difference is \(0\), so no additional shift is needed after rescaling. - Multiply rather than divide when undoing the standardization.

1. Under the null hypothesis, the standardized statistic is \(z=\frac{(\hat p_1-\hat p_2)-0}{SE}\). 2. Rearranging gives \(\hat p_1-\hat p_2=z(SE)\). 3. Thus \(\hat p_1-\hat p_2=2.00(0.050)=0.10\).

\(0.10\).

- The direction of the alternative determines which tail is used. - A positive z statistic with a greater-than alternative uses the right tail. - The p-value is an area under the standard normal curve, not the z statistic itself.

1. Because the alternative is \(p_1>p_2\), the p-value is the standard normal area to the right of \(1.74\). 2. \(P(Z\ge1.74)\approx0.0409\).

The p-value is approximately \(0.0409\).

- Match the direction of the shaded tail to the greater-than alternative. - The boundary of the shaded region is the observed z statistic. - Use a standard normal table or calculator to find the area in that tail.

1. The greater-than alternative makes the p-value the area to the right of the observed z statistic. 2. Thus the p-value is \(P(Z\ge1.25)=1-\Phi(1.25)\approx0.1056\).

The p-value is approximately \(0.1056\).

- The not-equal alternative makes this a two-sided test. - Use the magnitude of the observed z statistic to identify equally extreme values in both tails. - Find one tail area and account for its symmetric counterpart.

1. A two-sided alternative requires probability in both tails at least as extreme as \(|z|=2.10\). 2. One tail has area \(P(Z\le-2.10)\approx0.01786\). 3. Doubling gives \(2(0.01786\ldots)\approx0.0357\).

The p-value is approximately \(0.0357\).

- Begin the interpretation by assuming the null hypothesis is true. - Describe a probability about sample results, not a probability that either hypothesis is true. - Use the greater-than alternative to identify what “at least as extreme” means here.

1. A p-value is interpreted under the assumption that the null hypothesis is true. 2. If the two population recommendation proportions are equal, the probability of obtaining a sample difference \(\hat p_1-\hat p_2\) at least as large as the one observed, from random sampling alone, is about \(0.032\).

If the two population recommendation proportions are equal, there is about a \(3.2\%\) chance of obtaining a sample difference in favor of chain \(1\) at least as large as the observed difference from random sampling alone.

- Both shaded tails count because the test is two-sided. - Symmetry means the two shaded areas are equal. - Find one tail area beyond \(1.96\) and double it.

1. For a two-sided test, the p-value includes both standard normal tails beyond \(-1.96\) and \(1.96\). 2. Each tail has area about \(0.0250\). 3. The total p-value is therefore about \(2(0.0250)=0.0500\).

The p-value is approximately \(0.0500\).

- Compare the p-value directly with the stated significance level. - The decision concerns the null hypothesis; the conclusion should address the alternative. - Keep the conclusion about population proportions, not just the sampled students.

1. Compare the p-value with the significance level: \(0.041<0.05\). 2. Reject \(H_0\). 3. There is statistically significant evidence that the population proportion of students who support a later start time is greater in district \(1\) than in district \(2\).

Reject \(H_0\). There is statistically significant evidence at the \(0.05\) level that the support proportion is greater in district \(1\) than in district \(2\).

- First compare the p-value with \(\alpha\). - Distinguish “fail to reject” from proving the null hypothesis. - Phrase the final statement in terms of the evidence for the alternative hypothesis.

1. Since \(0.081>0.05\), fail to reject \(H_0\). 2. Failing to reject a null hypothesis does not establish that it is true, so “accept \(H_0\)” and “the proportions are equal” are too strong. 3. The correct conclusion is that the data do not provide statistically significant evidence at the \(0.05\) level that the two population proportions differ.

Fail to reject \(H_0\). The data do not provide statistically significant evidence that the population proportions differ; they do not prove that the proportions are equal.

- Compare the same p-value separately with each significance level. - A smaller significance level demands stronger evidence before rejection. - Separate the numerical evidence from the rule used to make a decision from that evidence.

1. At \(\alpha=0.05\), \(0.032<0.05\), so reject \(H_0\). 2. At \(\alpha=0.01\), \(0.032>0.01\), so fail to reject \(H_0\). 3. The p-value summarizes the same evidence from the data in both cases. Changing \(\alpha\) changes the threshold required for rejection, so the decision can change without any change in the data.

Reject \(H_0\) at \(\alpha=0.05\), but fail to reject \(H_0\) at \(\alpha=0.01\). The evidence is unchanged; only the rejection threshold changes.

- Use the p-value and significance level to make the test decision first. - Random sampling and random assignment support different kinds of conclusions. - Keep the inferential claim about the two population proportions measured by the survey.

1. Since \(0.006<0.05\), reject \(H_0:p_A=p_B\). 2. There is statistically significant evidence that the population proportions of adults who support the proposal differ between the two cities. 3. Because the samples were random, the inference can be generalized to the adult populations represented by the sampling processes. 4. The study does not impose a treatment, so the result does not establish that living in one city causes a person's opinion.

Reject \(H_0\). There is significant evidence that the two citywide support proportions differ. The random sampling supports generalization to the sampled populations, but the comparison does not establish causation.

- Use the number and location of shaded tails to identify the alternative direction. - Read the z boundary from the graph before finding the standard normal tail area. - Compare the resulting tail area with the stated significance level.

1. Only the right tail is shaded, so the graph represents the greater-than alternative \(H_a:p_1>p_2\). 2. The labeled boundary is \(z=2.05\), so the p-value is \(P(Z\ge2.05)\approx0.0202\). 3. Since \(0.0202<0.05\), reject \(H_0\).

The graph represents \(H_a:p_1>p_2\). The p-value is approximately \(0.0202\), so reject \(H_0\) at \(\alpha=0.05\).

- Ask what assumption about \(p_1\) and \(p_2\) is imposed by the null hypothesis. - A confidence interval estimates a difference without first setting that difference equal to zero. - Connect pooling to the model assumed while calculating the test statistic.

1. A confidence interval estimates the unknown difference without assuming that \(p_1=p_2\), so it estimates each population proportion separately. 2. Its standard error is \(\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}\). 3. The hypothesis test is carried out under \(H_0:p_1=p_2\), so the null model treats the two population proportions as one common value. 4. Therefore the test uses the pooled estimate \(\hat p_c=\frac{x_1+x_2}{n_1+n_2}\) and standard error \(\sqrt{\hat p_c(1-\hat p_c)(\frac{1}{n_1}+\frac{1}{n_2})}\).

The confidence interval uses the unpooled standard error based on \(\hat p_1\) and \(\hat p_2\) separately. The test uses a pooled standard error because \(H_0\) assumes a common population proportion.

- Compare the standard-error formulas used for estimation and for testing equality. - Ask what extra assumption is imposed when the null hypothesis sets the two population proportions equal. - Decide whether two procedures with different standard errors must always give exactly matching boundaries.

1. The interval excludes \(0\), so the confidence interval itself supports a nonzero population difference at the \(95\%\) confidence level. 2. However, the usual two-sample z-interval uses an unpooled standard error based on \(\hat p_1\) and \(\hat p_2\) separately. 3. The standard z-test of \(H_0:p_1=p_2\) uses a pooled standard error based on the common proportion assumed under the null hypothesis. 4. Because those standard errors are not generally identical, the two procedures are not exact mathematical inverses. The pooled test statistic and p-value must be computed from the sample data to guarantee the test decision.

No. Excluding \(0\) supports a difference through the confidence-interval procedure, but it does not by itself guarantee the decision of the standard pooled two-sample z-test because the interval and test use different standard errors.

- Ask what the null hypothesis says about the two population proportions. - Combine the success counts and sample sizes to estimate the common null proportion. - Use that pooled estimate in both terms of the null standard error before standardizing the observed difference.

1. A test of \(H_0:p_1=p_2\) assumes one common population proportion under the null, so its null standard error must use a pooled estimate. 2. The pooled proportion is \(\hat p_c=\frac{62+60}{100+120}=\frac{122}{220}\approx0.5545\). 3. The pooled standard error is \(\sqrt{0.5545(1-0.5545)(\frac{1}{100}+\frac{1}{120})}\approx0.06730\). 4. The observed difference is \(0.62-0.50=0.12\), so \(z=\frac{0.12}{0.067296\ldots}\approx1.783\).

The student's error is using the confidence-interval, unpooled standard error in a null-hypothesis test. The correct pooled test statistic is \(z\approx1.783\).

- The shaded side of the standard normal curve tells you the direction of the alternative. - Read the observed z statistic from the labeled boundary of the shaded tail. - Compare the resulting tail area with \(0.05\) only after identifying the correct direction.

1. The shaded region is in the left tail, so the alternative is \(H_a:p_1<p_2\). 2. The graph labels the observed statistic as \(z=-1.65\). 3. The left-tail area is \(P(Z\le-1.65)\approx0.0495\). 4. Since \(0.0495<0.05\), reject \(H_0\).

The alternative is \(H_a:p_1<p_2\). The p-value is approximately \(0.0495\), so reject \(H_0\) at \(\alpha=0.05\).

- Check whether hypotheses should describe sample statistics or population parameters. - Ask what equality under the null hypothesis implies about estimating the standard error. - Separate the numerical test decision from the strength and direction of the conclusion that the chosen alternative permits. - Recompute the test from the pooled success count before judging the student's numerical work.

1. Hypotheses must be about the population parameters: \(H_0:p_1=p_2\) and \(H_a:p_1\ne p_2\), not about the sample proportions. 2. Under \(H_0\), use a pooled estimate rather than the unpooled confidence-interval standard error. The pooled proportion is \(\hat p_c=\frac{66+45}{100+90}=\frac{111}{190}\approx0.5842\). 3. The pooled standard error is \(\sqrt{0.5842(1-0.5842)(\frac{1}{100}+\frac{1}{90})}\approx0.07161\). 4. The observed difference is \(0.66-0.50=0.16\), so \(z=\frac{0.16}{0.0716105\ldots}\approx2.234\). 5. For a two-sided test, the p-value is \(2P(Z\ge2.234\ldots)\approx0.0255\). 6. Since \(0.0255<0.05\), reject \(H_0\). The data provide statistically significant evidence that the two population proportions differ. The conclusion is not certainty, and a two-sided alternative does not state a pre-specified greater-than direction.

Corrections include using population parameters in the hypotheses, pooling for the null standard error, and replacing “accept” and “definitely” with an evidence-based conclusion. The correct results are \(z\approx2.234\) and p-value \(\approx0.0255\). Reject \(H_0\); there is significant evidence that \(p_1\ne p_2\).

- Define the two population proportions before writing the hypotheses. - Use the pooled proportion to check expected counts and to form the null standard error. - The alternative determines that only the right-tail area contributes to the p-value. - Finish by comparing the p-value with \(\alpha\) and translating the decision back to the two plan populations.

1. Let \(p_A\) and \(p_B\) be the population renewal proportions. Test \(H_0:p_A=p_B\) against \(H_a:p_A>p_B\). 2. The samples are random and independent, and each is less than \(10\%\) of its population. Under the null hypothesis, \(\hat p_c=\frac{117+96}{360}=\frac{213}{360}\approx0.5917\). The expected success counts are \(180(0.5917)=106.5\) in each group and the expected failure counts are \(180(0.4083)=73.5\), so the normal condition is satisfied. 3. The sample proportions are \(\hat p_A=\frac{117}{180}=0.65\) and \(\hat p_B=\frac{96}{180}\approx0.5333\), giving an observed difference of about \(0.1167\). 4. The pooled standard error is \(\sqrt{0.5917(0.4083)(\frac{1}{180}+\frac{1}{180})}\approx0.05181\). 5. The test statistic is \(z=\frac{0.65-0.5333\ldots}{0.0518113\ldots}\approx2.252\). 6. For the greater-than alternative, the p-value is \(P(Z\ge2.252\ldots)\approx0.0122\). 7. Since \(0.0122<0.05\), reject \(H_0\). There is statistically significant evidence that the population renewal proportion is greater for plan A than for plan B.

Conditions are satisfied. With \(H_0:p_A=p_B\) and \(H_a:p_A>p_B\), \(z\approx2.252\) and the p-value is approximately \(0.0122\). Reject \(H_0\); there is significant evidence that plan A has the greater population renewal proportion.