Free math worksheets

- Put the directional research question in the alternative hypothesis. - Ask which hypothesis supplies the fixed population value used when calculating the test statistic. - The null and alternative should be written about the population mean, not the sample mean.

1. The research claim “less than \(40\)” belongs in the alternative hypothesis. 2. The null hypothesis specifies the reference value used to form the test distribution, so it contains equality. 3. The correct hypotheses are \(H_0:\mu=40\) and \(H_a:\mu<40\).

Use \(H_0:\mu=40\) and \(H_a:\mu<40\). The equality belongs in the null hypothesis because the test evaluates sample evidence relative to that fixed reference value.

- Identify the sample quantity used to estimate the standard error in a \(t\)-test. - Ask whether variability can be estimated from a single observed value. - Check the degrees-of-freedom rule when \(n=1\).

1. A one-sample \(t\)-test estimates the standard error using the sample standard deviation \(s\). 2. With only one observation, there is no within-sample variation from which to calculate a sample standard deviation. 3. The degrees of freedom would be \(n-1=0\), so there is no corresponding \(t\)-reference distribution for the standard procedure. 4. Therefore, a standard one-sample \(t\)-test cannot be performed from a single observation.

The test cannot be performed with \(n=1\) because the sample standard deviation is undefined and the procedure would have \(0\) degrees of freedom.

- Read the inequality in the decision rule carefully. - Compare the exact p-value with the significance level, including the equality case. - Distinguish an exact value from a rounded software display.

1. The stated decision rule rejects whenever the p-value is less than or equal to the significance level. 2. Here, \(p=0.050000\) and \(\alpha=0.05\), so \(p=\alpha\). 3. Because equality is included in the rejection rule, reject \(H_0\). 4. This conclusion depends on the p-value being given as exact; a displayed rounded value of \(0.050\) could hide a value slightly above or below \(0.05\).

Reject \(H_0\) under the stated rule because the exact p-value equals \(\alpha\), and the rule uses \(p\le\alpha\).

- Ask whether a hypothesis should describe a sample already observed or a population quantity that is unknown. - Distinguish the symbol for a sample mean from the symbol for a population mean. - Make sure the direction of the alternative matches the manufacturer's question about being below the label value.

1. Hypotheses must be statements about the population parameter, not the sample statistic. 2. Let \(\mu\) be the population mean fill weight for the cereal boxes. 3. The correct hypotheses are \(H_0:\mu=52\) and \(H_a:\mu<52\). 4. The sample mean \(\bar{x}\) is observed from the sample and is used as evidence about the unknown population mean \(\mu\); it is not the parameter being tested.

The error is using \(\bar{x}\), a sample statistic, in the hypotheses. The correct hypotheses are \(H_0:\mu=52\) and \(H_a:\mu<52\), where \(\mu\) is the population mean fill weight.

- Ask what assumption is made before a p-value is calculated. - Separate a probability about possible sample results from a probability about a fixed population parameter. - Base the formal decision on the stated significance level, then translate it back to the original claim.

1. A p-value is calculated assuming the null hypothesis is true; it is not the probability that the null hypothesis is true. 2. The p-value means that if the population mean roast loss were \(15\%\), the probability of obtaining a sample result at least as far from \(15\%\) as the observed result, in either direction, would be about \(0.080\). 3. Since \(0.080>0.05\), fail to reject \(H_0\). 4. The data do not provide convincing evidence that the population mean roast loss differs from \(15\%\).

The p-value is not the probability that \(H_0\) is true. Assuming \(\mu=15\%\), there is about an \(8\%\) chance of getting a sample result at least as extreme as the observed one, in either direction. Because \(0.080>0.05\), fail to reject \(H_0\); there is not convincing evidence that the population mean roast loss differs from \(15\%\).

- A two-sided test counts extreme results on both sides of the reference distribution. - Compare the resulting probability with the significance level only after accounting for both tails. - The distance of a p-value from the cutoff can help communicate the strength of evidence without changing the formal decision.

1. For a two-sided test, the p-value is twice the upper-tail probability beyond \(|t|=2.12\) for \(T_{17}\). 2. The p-value is approximately \(0.0490\). 3. Since \(0.0490<0.05\), reject \(H_0\) at the \(5\%\) significance level. 4. The p-value is only slightly below \(0.05\), so the evidence is just sufficient under this decision rule; it should not be characterized as overwhelmingly strong evidence against \(H_0\).

The two-sided p-value is approximately \(0.0490\). Reject \(H_0\) at \(\alpha=0.05\), but describe the evidence as relatively modest because the p-value is very close to the significance cutoff.

- A test decision describes the strength of evidence in the sample, not certainty about a fixed parameter. - Think about the difference between “not enough evidence against” and “proved true.” - Consider whether a larger or more precise study could potentially lead to a different evidence assessment.

1. A hypothesis test evaluates whether the sample provides sufficiently strong evidence against the null hypothesis under a chosen significance rule. 2. A p-value of \(0.12\) may be too large to reject \(H_0\) at common significance levels such as \(0.05\). 3. Failing to reject \(H_0\) means the data do not provide convincing evidence for the alternative hypothesis at the chosen significance level. 4. It does not prove that the null hypothesis is true or that the population mean equals the null value exactly.

A failure to reject \(H_0\) is not proof that \(H_0\) is true. It means only that the sample does not provide sufficiently strong evidence for the alternative hypothesis at the chosen significance level.

- Decide which measure of population spread is actually known before selecting the reference distribution. - A large sample makes the \(t\)- and standard normal distributions similar, but it does not make an unknown population standard deviation known. - Separate the shape condition from the question of whether population variability is known.

1. The population standard deviation \(\sigma\) is unknown, so sampling uncertainty must be estimated using the sample standard deviation \(s\). 2. The appropriate procedure for a population mean with unknown \(\sigma\) is a one-sample \(t\)-test. 3. A large sample makes the \(t\)-distribution close to the standard normal distribution, but the reference distribution remains \(t\) when \(\sigma\) is estimated by \(s\). 4. Therefore, use a one-sample \(t\)-test with \(df=79\), assuming the other conditions are met.

Use a one-sample \(t\)-test with \(df=79\), not a \(z\)-test. The population standard deviation is unknown, so the sample standard deviation is used and the reference distribution is \(t\).

- Test the sign convention using a simple example in which the after value is larger than the before value. - Translate the verbal claim through the exact definition of the paired difference. - Keep the null centered at the no-change value.

1. An increase in reaction time means \(\text{after}>\text{before}\). 2. Since \(d=\text{before}-\text{after}\), an increase corresponds to negative differences. 3. Therefore, the correct hypotheses are \(H_0:\mu_d=0\) and \(H_a:\mu_d<0\). 4. The student's \(H_a:\mu_d>0\) would instead represent shorter reaction times after training.

The student's direction is incorrect. With \(d=\text{before}-\text{after}\), an increase after training gives negative differences, so use \(H_0:\mu_d=0\) and \(H_a:\mu_d<0\).

- A test statistic compares an observed difference with the amount of sampling variability expected under the null. - Ask what units remain after dividing a difference by its standard error. - The sign gives direction, while the magnitude gives standardized distance from the null value.

1. A \(t\) statistic standardizes the difference between the sample mean and the null hypothesized population mean. 2. The value \(t=-2.4\) means the sample mean is \(2.4\) estimated standard errors below the null value. 3. It does not mean the original measurement difference is \(2.4\) units; the actual difference depends on the standard error.

\(t=-2.4\) means the sample mean is \(2.4\) estimated standard errors below the null hypothesized mean. The number \(2.4\) is on a standardized scale, not necessarily the original measurement scale.

- Identify the parameter value that the null hypothesis claims. - A corresponding confidence interval shows which parameter values remain plausible at that confidence level. - The interval's location also indicates the direction of the observed departure from the null value.

1. A two-sided test at \(\alpha=0.05\) corresponds to checking whether the null value is contained in the \(95\%\) confidence interval for the same population mean. 2. The null value \(15\) is not contained in \((10.2,14.6)\). 3. Therefore, reject \(H_0\). 4. The data provide convincing evidence that the population mean differs from \(15\); the interval suggests it is lower than \(15\).

Reject \(H_0\). Since \(15\) is outside the \(95\%\) confidence interval, there is convincing evidence at \(\alpha=0.05\) that the population mean differs from \(15\), with the observed direction below \(15\).

- Start by identifying what decision corresponds to rejecting the null hypothesis. - For one error, imagine rejecting a true null; for the other, imagine not rejecting when the alternative is true. - Translate each logical error back into the delivery-time claim.

1. A Type I error occurs when the null hypothesis is true but is rejected. 2. In context, that means concluding that the population mean delivery time is greater than \(2.0\) days when the population mean is actually \(2.0\) days. 3. A Type II error occurs when the alternative is true but the null hypothesis is not rejected. 4. In context, that means failing to conclude that the population mean delivery time is greater than \(2.0\) days when it actually is greater than \(2.0\) days.

Type I error: conclude that \(\mu>2.0\) days when \(\mu=2.0\) days. Type II error: fail to conclude that \(\mu>2.0\) days when \(\mu>2.0\) days is actually true.

- Identify the actual observations analyzed in a paired test. - A participant contributes only if a difference can be calculated for that participant. - Degrees of freedom depend on the number of usable differences, not the number of original measurement slots.

1. A paired analysis uses one difference for each participant with both measurements available. 2. The participant with no “after” value has no paired difference, so the lone “before” value cannot be included in the paired \(t\)-test. 3. The analysis therefore uses \(19\) complete differences. 4. A one-sample \(t\)-test on \(19\) differences uses \(df=19-1=18\).

Exclude the incomplete pair from the paired analysis. Use the \(19\) complete differences, giving \(18\) degrees of freedom.

- Interpret the significance level under the assumption that the null hypothesis is true. - Think about repeated use of the same decision rule. - Keep a procedure's error rate separate from a probability assigned to a hypothesis.

1. The significance level is a property of the testing procedure under the assumption that the null hypothesis is true. 2. If the null hypothesis is true and the same testing procedure is repeated many times under the same conditions, about \(1\%\) of those tests would reject the null hypothesis. 3. The significance level is not the probability that the null hypothesis is true.

\(\alpha=0.01\) means the testing procedure has a \(1\%\) Type I error rate: when \(H_0\) is true, it rejects \(H_0\) about \(1\%\) of the time in repeated use. It does not mean there is a \(1\%\) chance that \(H_0\) is true.

- Think of the significance level as controlling how easy it is to enter the rejection region under the null hypothesis. - A stricter rejection rule affects both false rejections and successful detection of a real effect. - Hold the sample size and true alternative value fixed while comparing the two rules.

1. Lowering \(\alpha\) makes the rejection region more difficult to reach when the null hypothesis is true. 2. Therefore, the Type I error rate decreases from \(5\%\) to \(1\%\). 3. With all other features fixed, the more restrictive rejection rule also makes it harder to reject the null when the alternative is true. 4. Therefore, the power decreases and the Type II error probability increases.

Lowering \(\alpha\) from \(0.05\) to \(0.01\) decreases the Type I error rate but also decreases power, so the Type II error probability increases.

- Find the standardized distance produced by Study A first. - Compare the standard errors for the two sample sizes. - The same test statistic requires the raw distance from the null to scale with the standard error.

1. Study A has test statistic \(t=\frac{4}{10/\sqrt{25}}=\frac{4}{2}=2\). 2. In Study B, the standard error is \(\frac{10}{\sqrt{100}}=1\). 3. To produce \(t=2\), the sample mean must be \(2\cdot1=2\) units above the null value. 4. The larger sample needs only half the raw difference because its standard error is half as large.

Study B's sample mean must be \(2\) units above the null value.

- Locate the observed statistic at the center of the reference distribution. - For a one-sided test, consider the area from the observed statistic into the relevant tail. - For a two-sided test, ask what values are at least as far from \(0\) as \(0\) itself.

1. A \(t\)-distribution is symmetric about \(0\), so exactly half of its probability lies above \(0\) and half below \(0\). 2. For a one-sided upper-tail alternative, the p-value is \(P(T\ge0)=0.5\). 3. For a two-sided alternative, outcomes at least as far from \(0\) as an observed statistic of \(0\) include the entire distribution, so the p-value is \(1\).

The two-sided p-value is \(1\). For an upper-tailed one-sided alternative, the p-value is \(0.5\).

- Match a two-sided test at significance level \(\alpha\) with a confidence interval having confidence level \(1-\alpha\). - Compare the same p-value with \(0.05\) and \(0.01\). - Higher-confidence intervals are wider, so inclusion can change as the confidence level rises.

1. Since \(0.030<0.05\), the two-sided test rejects \(H_0:\mu=50\) at \(\alpha=0.05\). Therefore, \(50\) is outside the corresponding \(95\%\) confidence interval. 2. Since \(0.030>0.01\), the test fails to reject \(H_0\) at \(\alpha=0.01\). Therefore, \(50\) is inside the corresponding \(99\%\) confidence interval. 3. The wider \(99\%\) interval can include the null value even when the narrower \(95\%\) interval excludes it.

\(50\) is not in the \(95\%\) confidence interval, but it is in the \(99\%\) confidence interval.

- Match the sample statistic to the population parameter named in the hypotheses. - The null hypothesis is about a population mean. - After choosing the correct point estimate, standardize its distance from the null value.

1. A one-sample \(t\)-test for a population mean uses the sample mean, not the sample median, as the point estimate. 2. The correct test statistic is \(t=\frac{50.8-50}{3.2/\sqrt{16}}=1.00\). 3. With \(df=15\), the two-sided p-value is approximately \(0.333\). 4. The sample median is a different statistic and does not belong in this test statistic for \(\mu\).

The student should use \(\bar{x}\), not the sample median. The correct statistic is \(t=1.00\), with two-sided p-value approximately \(0.333\).

- Hold the rejection rule and sampling variability fixed. - Compare where the sampling distribution is centered under the two true alternative means. - A center farther into the direction of the alternative makes rejection more likely.

1. Power is the probability of rejecting the null hypothesis when the alternative is true. 2. With sample size and variability fixed, sample means are more likely to fall far into the upper rejection region when the true mean is farther above \(\mu_0\). 3. Therefore, Scenario B has greater power because the true effect is larger relative to the sampling variability.

Scenario B has greater power. A true mean farther above the null value is easier for the test to distinguish from \(H_0\) when the other test conditions are unchanged.

- Locate the observed statistic relative to both critical boundaries. - A two-sided rejection region has one part in each tail. - Connect membership in the rejection region with the corresponding p-value comparison to \(\alpha\).

1. The two-sided rejection region consists of test statistics less than \(-2.093\) or greater than \(2.093\). 2. The observed statistic \(-2.20\) lies beyond the lower critical value \(-2.093\). 3. Therefore, reject \(H_0\) at \(\alpha=0.05\). 4. Because the critical-value and p-value approaches give the same decision, the two-sided p-value must be less than \(0.05\).

Reject \(H_0\). Since \(t=-2.20\) lies in the rejection region, the two-sided p-value is less than \(0.05\).

- Separate conditions about the shape of a distribution from conditions about how observations were obtained. - Ask whether knowing one day’s value could give information about the next day’s value. - Increasing the number of observations does not automatically repair a flawed dependence structure.

1. A one-sample \(t\)-test relies on observations that are independent enough for the standard-error calculation and reference distribution to be appropriate. 2. Consecutive daily observations can be associated because operating conditions persist over time. 3. A sample size of \(50\) can help with the shape of the sampling distribution, but it does not remove dependence created by the sampling process. 4. The engineer should address the time dependence or use a sampling design that yields suitably independent observations before applying the usual one-sample \(t\)-test.

The usual one-sample \(t\)-test is not justified merely because \(n=50\). Consecutive daily observations may be dependent, and a large sample does not fix dependence in the data-collection process.

- Compare where each test places its allowed rejection probability. - Think about how far into the upper tail the statistic must fall under each design. - The alternative hypothesis is part of the study plan, not a choice made after inspecting results.

1. An upper-tailed test places the full \(0.05\) rejection probability in the upper tail, while a two-sided test splits the rejection probability between two tails. 2. When the true mean is above the benchmark, the upper-tailed test therefore has a less extreme upper rejection cutoff and greater power to detect that increase. 3. The direction must be specified before examining the data; choosing the tail after seeing the sample would change the error rate and invalidate the planned significance level.

For detecting a true increase, the pre-specified upper-tailed test has greater power than the two-sided test at the same \(\alpha=0.05\). The direction must be chosen in advance, not selected after observing the sample.

- Match the tail calculation to the direction stated in the alternative hypothesis. - A two-sided alternative counts extreme results on both sides of the null value. - Compare the corrected p-value with the significance level only after using the proper tails.

1. The alternative is two-sided, so outcomes at least as far from \(0\) in either direction count as evidence against \(H_0\). 2. The upper-tail probability for \(t=1.50\) with \(35\) degrees of freedom is approximately \(0.0713\). 3. By symmetry, the two-sided p-value is \(2(0.0713)\approx0.1426\). 4. Since \(0.1426>0.05\), fail to reject \(H_0\).

The correct two-sided p-value is approximately \(0.1426\). Fail to reject \(H_0\) at \(\alpha=0.05\).

- Start by identifying the population parameter named in the research question. - A different statistical procedure may target a different parameter even if it seems more robust. - For a large sample, distinguish the shape of the raw population from the shape of the sampling distribution of the mean.

1. The research question concerns the population mean, so replacing the procedure with one that tests a median changes the parameter being studied. 2. The one-sample \(t\)-procedure is designed for inference about the population mean. 3. With a random sample of \(120\), the sampling distribution of the sample mean is typically approximately normal by the central limit theorem unless the population has exceptionally severe features. 4. Thus, strong skewness in the raw incomes does not by itself justify switching to a test of a different parameter.

A test for a median would answer a different question. With a random sample of \(120\), mean-based \(t\) inference can be reasonable despite a strongly skewed population, provided there are no extraordinary features that defeat the large-sample approximation.

- Ask what assumption is made when a p-value is computed. - Distinguish a probability about data from a probability about a hypothesis. - Type II error is a property of a test under a particular alternative, not the complement of an observed p-value.

1. A p-value is calculated under the assumption that the null hypothesis is true; it is not the probability that the null hypothesis itself is true or false. 2. Therefore, \(p=0.20\) does not imply an \(80\%\) probability that \(H_0\) is false. 3. A Type II error probability depends on a specific alternative population mean, the sample size, variability, significance level, and test direction. 4. It cannot be obtained as \(1-p\) from the observed test result.

The p-value is not a probability assigned to \(H_0\), and \(1-p\) is not the Type II error probability. The Type II error rate must be evaluated for a specified alternative and test design.

- Separate a binary decision rule from the continuous information in a p-value. - Compare how far apart the two p-values actually are. - A significance threshold is a decision convention, not a sudden boundary between evidence and no evidence.

1. Under the stated decision rule, Study 1 rejects \(H_0\) because \(0.049<0.05\), while Study 2 fails to reject because \(0.051>0.05\). 2. The two p-values are nevertheless extremely close and represent nearly the same strength of evidence against their respective null hypotheses, assuming comparable designs. 3. Crossing the \(0.05\) threshold changes the formal decision but does not create a large discontinuity in the underlying evidence.

The formal decisions differ at \(\alpha=0.05\), but it is misleading to describe the evidence as strong in one study and absent in the other. \(p=0.049\) and \(p=0.051\) provide very similar evidence.

- Recall what probability a p-value actually describes. - A future study will have a new random sample and a new test statistic. - The probability of rejecting under a specified alternative is a design property of the test, not the complement of one observed p-value.

1. The p-value describes how unusual the observed test statistic would be under the null hypothesis; it is not a probability of replication success. 2. Whether a future study rejects depends on the true population mean, population variability, sample size, significance level, and random sampling variation. 3. The long-run probability of rejection under a particular alternative is the test’s power, which cannot be found as \(1-p\) from one observed study.

The statement is incorrect. \(1-0.03=0.97\) is not the probability that a replication will reject. Replication probability depends on the test’s power under the true population conditions.

- A hypothesis about a mean is evaluated using the sampling variability of the sample mean. - The range describes individual observations, not uncertainty in the population mean estimate. - Compare the sample mean with the null mean in standard-error units.

1. The test concerns the population mean, so the relevant statistic is the sample mean and its standard error, not whether the null value falls between the sample minimum and maximum. 2. The standard error is \(10/\sqrt{100}=1\). 3. The test statistic is \(t=(55-50)/1=5.00\) with \(99\) degrees of freedom. 4. The two-sided p-value is approximately \(0.0000025\). 5. Since the p-value is far below \(0.05\), reject \(H_0\).

\(t=5.00\) and \(p\approx0.0000025\), so reject \(H_0\). The fact that \(50\) lies within the range of individual observations does not determine whether the population mean equals \(50\).

- A test statistic compares quantities measured on the same scale. - Convert the benchmark using the same transformation applied to the observations. - Leaving the numerical null value unchanged after a unit conversion changes the hypothesis itself.

1. Converting the sample data changes the measurement scale, so the hypothesized mean must be converted to the same scale. 2. The Fahrenheit equivalent of \(20^\circ\text{C}\) is \(20\cdot\frac{9}{5}+32=68^\circ\text{F}\). 3. Testing against \(20^\circ\text{F}\) would test a completely different population mean. 4. The correct Fahrenheit-scale null hypothesis is \(H_0:\mu=68^\circ\text{F}\).

The null value must be converted with the data. The correct null hypothesis is \(H_0:\mu=68^\circ\text{F}\), not \(20^\circ\text{F}\).

- Distinguish the people selected from the people whose responses are analyzed. - Ask whether the chance of responding may depend on the variable being studied. - Larger samples reduce random error, not systematic selection bias.

1. The original sample was random, but the analyzed data come only from the respondents. 2. If response probability is related to exercise time, respondents can systematically differ from nonrespondents. 3. The resulting sample mean can therefore be biased for the target population mean. 4. Increasing the original sample size does not remove a systematic nonresponse mechanism, so a standard one-sample \(t\)-test on respondents alone may not support population inference.

Nonresponse can bias the analyzed sample because response is related to the outcome. A large original random sample does not repair systematic missingness among the people who actually provide data.

- Identify the population quantity named in the studio's claim and the direction of the question. - Compare the observed sample mean with the claimed value in units of estimated sampling variability. - The conclusion should reflect both the p-value and the direction stated in the alternative hypothesis.

1. Test \(H_0:\mu=74\) against \(H_a:\mu>74\), where \(\mu\) is the population mean firing time. 2. The random sample and sample-shape conditions support a one-sample \(t\)-test with \(df=24\). 3. The test statistic is \(t=\frac{76.3-74}{5.0/\sqrt{25}}=2.30\). 4. The one-sided p-value is \(P(T_{24}\ge2.30)\approx0.0152\). 5. Because \(0.0152<0.05\), reject \(H_0\). The sample provides convincing evidence that the population mean firing time is greater than \(74\,\text{min}\).

\(H_0:\mu=74\), \(H_a:\mu>74\); \(t=2.30\) with \(df=24\); p-value \(\approx0.0152\). Reject \(H_0\). There is convincing evidence that the population mean firing time exceeds \(74\,\text{min}\).

- Decide what one numerical comparison can be made within each monitor. - The direction of the alternative must match what “decreases” means for the difference you define. - Think about whether observations coming from the same physical unit can reasonably be treated as independent.

1. Let \(d=\text{after time}-\text{before time}\), and let \(\mu_d\) be the population mean paired difference in standby time for monitors like those studied. 2. The hypotheses are \(H_0:\mu_d=0\) and \(H_a:\mu_d<0\). 3. The appropriate procedure is a one-sample \(t\)-test for the population mean difference using the \(18\) paired differences. 4. The before and after values from the same monitor are dependent, so an independent two-sample procedure would ignore the pairing built into the design.

Let \(d=\text{after}-\text{before}\). Test \(H_0:\mu_d=0\) versus \(H_a:\mu_d<0\) with a one-sample \(t\)-test on the paired differences. A two-sample test is inappropriate because each after measurement is linked to the before measurement from the same monitor.

- The alternative should reflect the direction “less than.” - Measure how far the sample mean falls below the claimed mean relative to estimated sampling variability. - Match the p-value tail to the direction of the research question.

1. Test \(H_0:\mu=35\) against \(H_a:\mu<35\). 2. The random sample and sample size \(n=40\) support a one-sample \(t\)-test with \(df=39\). 3. The test statistic is \(t=\frac{33.2-35}{6.4/\sqrt{40}}\approx-1.78\). 4. The lower-tail p-value is approximately \(0.0415\). 5. Because \(0.0415<0.05\), reject \(H_0\). The sample provides convincing evidence that the population mean sound-check time is less than \(35\,\text{min}\).

\(H_0:\mu=35\), \(H_a:\mu<35\); \(t\approx-1.78\), p-value \(\approx0.0415\). Reject \(H_0\). There is convincing evidence that the population mean sound-check time is less than \(35\,\text{min}\).

- Identify the actual observations that enter the one-sample analysis after pairing is taken into account. - The shapes of the original two measurement sets can differ from the shape of their within-device differences. - Check conditions on the data object the test statistic is actually built from.

1. A matched-pairs analysis turns each device's two measurements into one difference. 2. The one-sample \(t\)-test is applied to the sample of paired differences, so the shape condition concerns the distribution of those differences, not the two original measurement distributions separately. 3. The \(22\) differences are roughly symmetric with no outliers, so the stated shape information supports the sample-data condition for the paired \(t\)-test. 4. Other conditions, such as appropriate randomization, would still need to be satisfied.

The relevant shape is the distribution of the \(22\) paired differences. Because those differences are roughly symmetric with no outliers, the stated shape information supports use of the paired one-sample \(t\)-test, assuming the other conditions also hold.

- The sample size and sample standard deviation stay fixed, so the estimated sampling variability is the same in both tests. - Focus on how far the sample mean lies above each proposed null value. - A larger standardized distance in the direction of the alternative produces stronger evidence against the null hypothesis.

1. The standard error for both tests is \(\frac{9}{\sqrt{36}}=1.5\) packages per minute, with \(df=35\). 2. For part a, \(t=\frac{72-70}{1.5}\approx1.33\), giving an upper-tail p-value of approximately \(0.096\). Fail to reject \(H_0\) at \(\alpha=0.05\). 3. For part b, \(t=\frac{72-71.5}{1.5}\approx0.33\), giving an upper-tail p-value of approximately \(0.370\). Fail to reject \(H_0\) at \(\alpha=0.05\). 4. The second null value is closer to the observed sample mean, so its test statistic is closer to \(0\) and its p-value is larger.

a) \(t\approx1.33\), p-value \(\approx0.096\); fail to reject \(H_0\). b) \(t\approx0.33\), p-value \(\approx0.370\); fail to reject \(H_0\). Both tests fail to reject \(H_0\), but the p-values differ because changing the null value changes how far the observed sample mean lies from the value assumed by \(H_0\).

- Consider when the direction of an alternative hypothesis should be decided. - Use the p-value that matches the question the study was designed to test. - Compare that p-value with the stated significance level before writing the conclusion.

1. The direction of a one-sided alternative should be chosen from the research question before examining the sample result, not selected afterward because the observed statistic points in that direction. 2. The planned test was two-sided, so the relevant p-value is \(0.0796\). 3. Because \(0.0796>0.05\), fail to reject \(H_0\). 4. The data do not provide convincing evidence at the \(0.05\) level that the population mean sodium content differs from \(400\,\text{mg}\).

The analyst's switch to a one-sided alternative after seeing the data is not valid for the planned analysis. Using the planned two-sided p-value \(0.0796\), fail to reject \(H_0\); there is not convincing evidence that \(\mu\ne400\,\text{mg}\).

- A \(t\)-statistic alone does not determine a p-value unless the degrees of freedom are also known. - Compare the tail behavior of a \(t\)-distribution with few degrees of freedom to one with many degrees of freedom. - Use the same two-sided tail rule for both tests before comparing with \(0.05\).

1. For Test A, the two-sided p-value is \(2P(T_5\ge2.10)\approx0.0898\). 2. For Test B, the two-sided p-value is \(2P(T_{50}\ge2.10)\approx0.0408\). 3. Test A fails to reject \(H_0\) at \(\alpha=0.05\), while Test B rejects \(H_0\). 4. A \(t\)-distribution with fewer degrees of freedom has heavier tails, so the same standardized statistic corresponds to a larger tail probability.

Test A: p-value \(\approx0.0898\), so fail to reject \(H_0\). Test B: p-value \(\approx0.0408\), so reject \(H_0\). The p-values differ because the two tests use different \(t\)-distributions.

- Work backward from the definition of the one-sample \(t\)-statistic. - The sample size determines the denominator's square-root factor and the degrees of freedom. - For a two-sided test, use both tails beyond the magnitude of the observed statistic.

1. The test statistic satisfies \(-2.00=\frac{18.4-20}{s/\sqrt{25}}\). 2. Since \(\sqrt{25}=5\), solving \(-2.00=\frac{-1.6}{s/5}\) gives \(s=4.0\). 3. The test has \(df=24\). 4. The two-sided p-value is \(2P(T_{24}\ge2.00)\approx0.0569\).

The sample standard deviation is \(4.0\), and the two-sided p-value is approximately \(0.0569\).

- Reduce the matched measurements to one difference per device. - The null value is not zero, so compare the sample mean difference with the specific improvement named in the claim. - Use the direction of the claim to choose the appropriate tail for the p-value.

1. Test \(H_0:\mu_d=2\) against \(H_a:\mu_d>2\). 2. Because the data are paired, use a one-sample \(t\)-test on the \(20\) differences, with \(df=19\). 3. The test statistic is \(t=\frac{3.1-2}{2.4/\sqrt{20}}\approx2.05\). 4. The upper-tail p-value is approximately \(0.0272\). 5. Because \(0.0272<0.05\), reject \(H_0\). There is convincing evidence that the population mean improvement exceeds \(2\) points.

\(t\approx2.05\) with \(df=19\) and p-value \(\approx0.0272\). Reject \(H_0\); there is convincing evidence that \(\mu_d>2\) points.

- With a small sample, inspect the sample shape before relying on a \(t\)-procedure. - Compare the observed sample mean with the null value in standard-error units. - The direction of the alternative determines which tail contributes to the p-value.

1. Because the sample is small, the roughly symmetric shape with no apparent outliers supports using a one-sample \(t\)-procedure. 2. The test statistic is \(t=\frac{104-100}{5/\sqrt{10}}\approx2.530\) with \(9\) degrees of freedom. 3. The upper-tail p-value is approximately \(0.0161\). 4. Since \(0.0161<0.05\), reject \(H_0\). There is statistically significant evidence that the population mean compression strength exceeds \(100\) MPa.

\(t\approx2.530\), \(df=9\), and \(p\approx0.0161\). Reject \(H_0\); the data provide significant evidence that the population mean compression strength is greater than \(100\) MPa. The boxplot supports the small-sample \(t\)-procedure because it shows no strong skewness or outliers.

- Decide whether the response is quantitative before choosing an inference procedure. - For a large random sample, focus on the sampling distribution of the mean rather than requiring the raw data themselves to be normal. - Use the direction of the claim to choose the appropriate tail.

1. The response is quantitative, and the large random sample makes the sampling distribution of the sample mean approximately normal unless the population is extraordinarily skewed. 2. The hypotheses are \(H_0:\mu=10\) and \(H_a:\mu>10\). 3. The test statistic is \(t=\frac{12.6-10}{8/\sqrt{80}}\approx2.907\) with \(79\) degrees of freedom. 4. The upper-tail p-value is approximately \(0.0024\). 5. Since \(0.0024<0.05\), reject \(H_0\). There is significant evidence that the population mean number of absences exceeds \(10\).

A one-sample \(t\)-procedure is reasonable because the variable is quantitative and the random sample is large. \(t\approx2.907\), \(df=79\), and \(p\approx0.0024\), so reject \(H_0\).

- A very large sample can make the standard error quite small. - Separate evidence that a difference exists from the size of the observed difference. - Practical importance requires a contextual benchmark beyond statistical significance.

1. The standard error is \(5/\sqrt{10{,}000}=0.05\) mm. 2. The test statistic is \(t=\frac{50.2-50}{0.05}=4.00\) with \(9999\) degrees of freedom. 3. The two-sided p-value is approximately \(0.000064\), so reject \(H_0\) at \(\alpha=0.05\). 4. The observed difference from the benchmark is only \(0.2\) mm. Whether that size matters in manufacturing depends on tolerances, costs, and consequences, which the p-value does not measure.

\(t=4.00\) and \(p\approx0.000064\), so reject \(H_0\). The result is statistically significant, but the practical importance of a \(0.2\) mm difference must be judged using engineering context rather than the p-value alone.

- For a small sample, check the sample shape as well as the random-sampling condition. - Determine whether the population standard deviation is known or estimated from the sample. - Compare the tail behavior of the normal and \(t\) reference distributions.

1. The random sample and stated shape support a small-sample one-sample \(t\)-procedure. The test statistic is \(t=\frac{105-100}{9/\sqrt{12}}\approx1.925\). 2. A standard normal upper-tail probability at \(1.925\) is approximately \(0.0271\). 3. Because the population standard deviation is unknown and estimated by \(s=9\), the correct reference distribution is \(t\) with \(11\) degrees of freedom. 4. The corresponding upper-tail \(t\) p-value is approximately \(0.0403\), which is larger because the \(t\) distribution has heavier tails.

The student’s normal-tail value is about \(0.0271\), but the correct one-sample \(t\) p-value is about \(0.0403\) with \(11\) degrees of freedom.

- For a small sample, look for information about the population shape. - Use the sample standard deviation to estimate the standard error of the mean. - The direction of the alternative determines the relevant tail.

1. Because the population is stated to be approximately normal, a one-sample \(t\)-procedure can be used even with a very small random sample, provided the observations are independent. 2. The test statistic is \(t=\frac{23-20}{2.5/\sqrt{6}}\approx2.939\) with \(5\) degrees of freedom. 3. The upper-tail p-value is approximately \(0.0161\). 4. Since \(0.0161<0.05\), reject \(H_0\). There is significant evidence that the population mean exceeds \(20\).

\(t\approx2.939\), \(df=5\), and \(p\approx0.0161\). Reject \(H_0\). The small sample is acceptable because the population is stated to be approximately normal.

- Count how many opportunities the researcher may have to declare significance. - The usual significance level assumes a fixed decision rule, not repeated attempts until the threshold is crossed. - Consider what happens under the null hypothesis when many chances to reject are allowed.

1. An ordinary \(\alpha=0.05\) test controls the Type I error rate for a pre-specified analysis of one sample size. 2. Repeatedly testing after additional data creates multiple opportunities to cross the \(0.05\) threshold even when \(H_0\) is true. 3. Stopping specifically when significance appears makes the overall probability of at least one false rejection larger than the nominal \(5\%\) level. 4. A valid sequential design requires a procedure that accounts for the repeated looks at the data rather than applying an unchanged fixed-sample rule each time.

The final ordinary \(t\)-test does not retain a \(5\%\) overall Type I error rate under this optional-stopping rule. Repeated looks at the data increase the chance of a false rejection unless the sequential procedure is adjusted appropriately.

- Turn each before-and-after pair into one change with a consistent subtraction order. - Check the direction of the alternative against what a decrease means for that change. - With only \(10\) paired differences, the shape information matters when deciding whether the test is reasonable.

1. Test \(H_0:\mu_d=0\) versus \(H_a:\mu_d<0\), where \(d=\text{after}-\text{before}\). 2. The sample has \(\bar{d}=-0.102\,\text{s}\) and \(s_d\approx0.0498\,\text{s}\). The random sample and difference distribution support a one-sample paired-difference \(t\)-test with \(df=9\). 3. The test statistic is \(t=\frac{-0.102}{0.0498/\sqrt{10}}\approx-6.47\). 4. The one-sided p-value is approximately \(0.000058\). 5. Since the p-value is less than \(0.01\), reject \(H_0\). There is convincing evidence that the treatment decreases the population mean echo-decay time for halls in the network.

\(H_0:\mu_d=0\), \(H_a:\mu_d<0\); \(t\approx-6.47\), p-value \(\approx0.000058\). Reject \(H_0\). The data provide convincing evidence that the treatment decreases the population mean echo-decay time.

- Summarize the raw sample before comparing it with the hypothesized population mean. - The small sample makes the stated shape information important for using the reference distribution. - The alternative hypothesis tells you which tail of the reference distribution matters.

1. Test \(H_0:\mu=45\%\) versus \(H_a:\mu>45\%\). 2. The sample has \(\bar{x}\approx46.158\%\) and \(s\approx0.810\%\). The random sample and shape support a one-sample \(t\)-test with \(df=11\). 3. The standard error is approximately \(0.810/\sqrt{12}=0.234\%\), giving \(t\approx4.96\). 4. The one-sided p-value is approximately \(0.000216\). 5. Since the p-value is below \(0.01\), reject \(H_0\). There is convincing evidence that the population mean daily relative humidity exceeds \(45\%\).

\(H_0:\mu=45\%\), \(H_a:\mu>45\%\); \(t\approx4.96\), p-value \(\approx0.000216\). Reject \(H_0\). There is convincing evidence that the population mean daily relative humidity is greater than \(45\%\).

- Think about the test statistic value that marks the start of the rejection region for this one-sided test. - Express that boundary in terms of the unknown sample mean rather than starting with a p-value. - The sample standard deviation and sample size determine how far the sample mean must be from the null value.

1. With \(n=16\), the test uses \(df=15\). For an upper-tailed test at \(\alpha=0.05\), the critical value is \(t\approx1.753\). 2. The rejection boundary satisfies \(\frac{\bar{x}-50}{3.2/\sqrt{16}}=1.753\). 3. Since \(\frac{3.2}{\sqrt{16}}=0.8\), the boundary is \(\bar{x}=50+1.753\cdot0.8\approx51.402\) hours. 4. Therefore, the test rejects \(H_0\) when \(\bar{x}\) is greater than approximately \(51.40\) hours.

The sample mean must be greater than approximately \(51.40\) hours for the test to reject \(H_0\) at \(\alpha=0.05\).

- Compute the usual standard error and decision before applying the finite-population factor. - Sampling without replacement from a substantial fraction of the population reduces the standard error. - Recompute the test statistic and p-value using the adjusted standard error, then compare the two decisions.

1. The usual unadjusted standard error is \(9/\sqrt{40}\approx1.423\) minutes. 2. The unadjusted statistic is \(t=\frac{32.3-30}{1.423}\approx1.616\), giving an upper-tail p-value of approximately \(0.0570\). This calculation fails to reject \(H_0\) at \(\alpha=0.05\). 3. The finite-population factor is \(\sqrt{\frac{300-40}{300-1}}\approx0.933\). 4. The adjusted standard error is \(0.933(1.423)\approx1.327\) minutes. 5. The adjusted statistic is \(t=\frac{32.3-30}{1.327}\approx1.733\), giving an upper-tail p-value of approximately \(0.0455\). 6. Using the finite-population adjustment, reject \(H_0\). There is significant evidence at the \(5\%\) level that the company’s mean commute time exceeds \(30\) minutes. 7. The adjustment matters because the sample contains \(40/300\approx13.3\%\) of the finite population, so sampling without replacement reduces the sampling variability.

Unadjusted: \(SE\approx1.423\), \(t\approx1.616\), and \(p\approx0.0570\), so fail to reject \(H_0\). Adjusted: \(SE\approx1.327\), \(t\approx1.733\), and \(p\approx0.0455\), so reject \(H_0\) at \(\alpha=0.05\).