Free math worksheets

- Write the test statistic as a ratio. - Track the units in the numerator and denominator. - Quantities with identical units cancel when divided.

1. For testing \(H_0: \beta=0\), the test statistic is \(t=\frac{1.50-0}{0.60}=2.50\). 2. The numerator and denominator have the same slope units, dollars per hour. 3. Those units cancel in the ratio, so the \(t\) statistic is unitless.

The correct report is \(t=2.50\), with no units. The slope units cancel when the slope difference is divided by its standard error.

- Define the parameter as a population rate of change with units. - Translate “positive” into the direction of the alternative hypothesis. - The null hypothesis for a slope test uses zero.

1. \(\beta\) is the population change in mean accuracy score, in percentage points, associated with one additional hour of practice. 2. “Positive” gives a right-tailed alternative hypothesis, \(H_a:\beta > 0\). 3. The null hypothesis represents no linear association, \(H_0:\beta = 0\).

\(\beta\) is the population change in mean accuracy score, in percentage points, per additional hour of practice. \(H_0:\beta = 0\); \(H_a:\beta > 0\).

- Define the slope as a population rate of change with units. - Translate “negative” into the direction of the alternative hypothesis. - Use zero in the null hypothesis for a slope test.

1. \(\beta\) is the population change in mean temperature, in degrees Fahrenheit, associated with one additional foot from the heater. 2. “Negative” gives a left-tailed alternative hypothesis, \(H_a:\beta < 0\). 3. The null hypothesis represents no linear association, \(H_0:\beta = 0\).

\(\beta\) is the population change in mean temperature, in degrees Fahrenheit, per additional foot from the heater. \(H_0:\beta = 0\); \(H_a:\beta < 0\).

- Define the population slope with the response unit per explanatory-variable unit. - Translate “nonzero” into a two-sided alternative. - Use \(0\) as the null value for testing a regression slope.

1. \(\beta\) is the population change in mean shipping cost, in dollars, associated with one additional pound of package weight. 2. “Nonzero” gives a two-sided alternative hypothesis, \(H_a:\beta \ne 0\). 3. The null hypothesis represents no linear association, \(H_0:\beta = 0\).

\(\beta\) is the population change in mean shipping cost, in dollars, per additional pound of package weight. \(H_0:\beta = 0\); \(H_a:\beta \ne 0\).

- Define the slope with units of feet per inch. - Use the numerical boundary from the claim as the null value. - Translate “greater than” into a right-tailed alternative.

1. \(\beta\) is the population change in mean stream height, in feet, associated with one additional inch of rainfall. 2. The claim “greater than \(2\)” gives the right-tailed alternative \(H_a:\beta > 2\). 3. The null hypothesis uses the boundary value, \(H_0:\beta = 2\).

\(\beta\) is the population change in mean stream height, in feet, per additional inch of rainfall. \(H_0:\beta = 2\); \(H_a:\beta > 2\).

- Define the slope in hours per additional worker. - Use the claimed boundary value in the null hypothesis. - Translate “less than” into a left-tailed alternative.

1. \(\beta\) is the population change in mean completion time, in hours, associated with one additional worker. 2. The claim “less than \(-1\)” gives the left-tailed alternative \(H_a:\beta < -1\). 3. The null hypothesis uses the boundary value, \(H_0:\beta = -1\).

\(\beta\) is the population change in mean completion time, in hours, per additional worker. \(H_0:\beta = -1\); \(H_a:\beta < -1\).

- Define the slope with units of grams per centimeter. - Use the stated comparison value in the null hypothesis. - Translate “differs from” into a two-sided alternative.

1. \(\beta\) is the population change in mean body mass, in grams, associated with one additional centimeter of wing length. 2. “Differs from \(0.5\)” gives the two-sided alternative \(H_a:\beta \ne 0.5\). 3. The null hypothesis uses the specified value, \(H_0:\beta = 0.5\).

\(\beta\) is the population change in mean body mass, in grams, per additional centimeter of wing length. \(H_0:\beta = 0.5\); \(H_a:\beta \ne 0.5\).

- Compare the \(p\)-value directly with \(\alpha\). - Reject \(H_0\) when \(p<\alpha\). - State the conclusion about the population slope and preserve the direction in \(H_a\).

1. Compare the \(p\)-value with the significance level: \(0.0040 < 0.050\). 2. Because \(p < \alpha\), reject \(H_0\). 3. There is convincing evidence that the population slope for predicting bowl diameter from clay mass is positive.

Reject \(H_0\). There is convincing evidence that the population regression slope relating clay mass to bowl diameter is positive.

- Compare \(p\) with \(\alpha\) before making the decision. - When \(p>\alpha\), fail to reject \(H_0\); do not say that \(H_0\) is proven. - State the conclusion using the negative direction from \(H_a\).

1. Compare the \(p\)-value with the significance level: \(0.0810 > 0.050\). 2. Because \(p > \alpha\), fail to reject \(H_0\). 3. There is not convincing evidence that the population slope for predicting sound level from distance is negative.

Fail to reject \(H_0\). There is not convincing evidence that the population regression slope relating distance from a speaker to sound level is negative.

- Compare the \(p\)-value with the stated significance level, not with \(0.05\) automatically. - When \(p>\alpha\), fail to reject \(H_0\). - State the conclusion about whether the population slope is nonzero.

1. Compare the \(p\)-value with the significance level: \(0.0320 > 0.010\). 2. Because \(p > \alpha\), fail to reject \(H_0\). 3. At the \(1\%\) significance level, there is not convincing evidence of a nonzero population slope relating editing passes to errors remaining.

Fail to reject \(H_0\). At \(\alpha = 0.010\), there is not convincing evidence of a nonzero population regression slope relating editing passes to errors remaining.

- Compare the values carefully because \(p\) is close to \(\alpha\). - Reject \(H_0\) whenever \(p<\alpha\). - State the conclusion using the negative direction in \(H_a\).

1. Compare the \(p\)-value with the significance level: \(0.0490 < 0.050\). 2. Because \(p < \alpha\), reject \(H_0\). 3. There is convincing evidence that the population slope for predicting temperature from distance from the heater is negative.

Reject \(H_0\). There is convincing evidence that the population regression slope relating distance from a heater to temperature is negative.

- Compare decimal places carefully when \(p\) and \(\alpha\) are both very small. - Reject \(H_0\) when \(p<\alpha\). - For a two-sided alternative, conclude only that the slope is nonzero unless its direction is also stated from the estimate.

1. Compare the \(p\)-value with the significance level: \(0.0007 < 0.001\). 2. Because \(p < \alpha\), reject \(H_0\). 3. There is convincing evidence that the population slope relating distance from the stage to sound level is nonzero.

Reject \(H_0\). There is convincing evidence of a nonzero population regression slope relating distance from a stage to sound level.

- Use the random-sample and \(10\%\) statements to assess independence. - In the scatterplot, decide whether a straight-line model is reasonable; in the residual plot, look for a systematic shape. - Compare residual spread at low and high volunteer counts, then use the stated residual-distribution information.

1. The random sample and the \(10\%\) condition support independence. 2. The scatterplot shows an approximately linear positive relationship. 3. The residual plot shows no systematic pattern and has roughly similar vertical spread across the explanatory-variable range, supporting linearity and equal variance. 4. The roughly symmetric residual distribution with no strong outliers supports approximate normality. 5. Therefore, the conditions for a \(t\)-test of the population slope are adequately met.

The sampling information and displays adequately support independence, linearity, constant residual variance, and approximate normality, so a \(t\)-test for the population slope is appropriate.

- Separate conditions about the regression model from conditions about how the sample was selected. - Decide whether a convenience sample represents a chance-based selection process. - Appropriate residual plots cannot remove selection bias.

1. The appropriate plots support the model conditions of linearity, constant variance, and approximate normality. 2. However, a convenience sample is not a random sample and may systematically differ from the target population. 3. Because the randomization condition fails, the test does not justify inference about the population slope.

The randomization condition fails because the sample is a convenience sample. Therefore, the \(t\)-test does not justify inference about the population slope, even though the graphical conditions are satisfied.

- Trace the residuals from low to high trail grades and look for a repeated directional pattern. - A suitable linear model should leave points without a curved structure around the zero line. - Identify which regression-inference condition is violated by the pattern.

1. A suitable linear regression model should leave residuals randomly scattered around \(0\). 2. The residual plot shows strong curvature, indicating that a straight line does not adequately model the relationship. 3. Therefore, the linearity condition fails, so the usual \(t\)-test for the slope of this linear model is not appropriate.

The linearity condition fails because the residual plot shows strong curvature. Therefore, the usual \(t\)-test for the linear-model slope is not appropriate.

- Compare the vertical range of residuals for small and large numbers of exhibits. - Decide whether the error spread is approximately stable across the plot. - Connect a widening residual spread to the standard-error assumption used by the slope test.

1. The usual slope \(t\)-test assumes that residual variance is approximately constant across values of the explanatory variable. 2. The residual plot shows a sharply widening vertical spread as the number of exhibits increases. 3. Therefore, the constant-variance condition is not met, so the usual standard error and \(t\)-test are not reliable without a different model or inference method.

The constant-variance condition fails because the residual spread increases sharply. Therefore, the usual standard error and slope \(t\)-test are not reliable.

- Match a two-sided \(\alpha=0.05\) test with a \(95\%\) confidence interval. - Locate the null value \(0\) relative to the interval. - Excluding the null value corresponds to rejecting \(H_0\).

1. A two-sided test at \(\alpha = 0.05\) rejects \(H_0\) exactly when the null value is outside the matching \(95\%\) confidence interval. 2. The null value \(0\) is not in \((0.12, 0.88)\). 3. Therefore, reject \(H_0\).

Reject \(H_0\) because the null value \(0\) is outside the matching \(95\%\) confidence interval.

- Match a two-sided \(\alpha=0.05\) test with a \(95\%\) confidence interval. - Check whether the null value \(0\) lies between the endpoints. - Including the null value corresponds to failing to reject \(H_0\).

1. A two-sided test at \(\alpha = 0.05\) rejects \(H_0\) exactly when the null value is outside the matching \(95\%\) confidence interval. 2. The null value \(0\) lies in \((-0.35, 0.46)\). 3. Therefore, fail to reject \(H_0\).

Fail to reject \(H_0\) because the null value \(0\) is inside the matching \(95\%\) confidence interval.

- Use the specified null value \(2.0\), not \(0\). - Check whether \(2.0\) lies between the interval endpoints. - A null value inside the matching confidence interval leads to failing to reject \(H_0\).

1. A two-sided test at \(\alpha = 0.05\) rejects \(H_0\) exactly when the null value is outside the matching \(95\%\) confidence interval. 2. The null value \(2.0\) lies in \((1.10, 2.90)\). 3. Therefore, fail to reject \(H_0\).

Fail to reject \(H_0\) because the null value \(2.0\) is inside the matching \(95\%\) confidence interval.

- Compare \(p=0.12\) with the stated significance level. - Distinguish “fail to reject” from “accept as true.” - Consider that weak evidence can result from limited sample information, not only from a true null hypothesis.

1. Since \(0.12 > 0.05\), the formal decision is to fail to reject \(H_0\). 2. Failing to reject \(H_0\) means that the data do not provide convincing evidence against the null hypothesis. 3. It does not prove that the population slope equals the null value. The study may simply lack enough evidence to detect a real effect.

The reasoning is incorrect. The correct decision is to fail to reject \(H_0\), which means the data do not provide convincing evidence against it; this does not prove that \(H_0\) is true.

- Identify what rejecting a zero-slope null hypothesis establishes. - Distinguish statistical association from cause and effect. - Check whether the study randomly assigned the explanatory variable.

1. Rejecting \(H_0:\beta = 0\) provides evidence of a nonzero linear association between kayak length and top speed in the population. 2. Because the study is observational, other variables may explain some or all of the association. 3. Therefore, the test does not establish that changing kayak length causes a change in top speed.

The conclusion is not justified. The test supports a nonzero linear association, but an observational study cannot establish that kayak length causes top speed to change.

- Ask when the direction of \(H_a\) was chosen. - A valid one-sided test commits to one tail before seeing the sample slope. - Consider what happens if either a large positive or a large negative slope can trigger a favorable one-sided test.

1. The direction of a one-sided alternative must be specified before examining the sample result. 2. Choosing the direction that matches the observed slope makes unusually large results in either direction capable of producing a small reported one-sided \(p\)-value. 3. This data-dependent choice understates the true evidence and inflates the Type I error rate. A two-sided test should be used unless a directional alternative was justified in advance.

The procedure is invalid. The one-sided direction should be chosen before examining the data; choosing it afterward makes the \(p\)-value misleading. Use a two-sided test unless the direction was specified in advance.

- Ask whether a hypothesis should describe the sample or the population. - Identify the parameter that represents the long-run relationship of interest. - Match the direction of the alternative to the research question.

1. Hypotheses must be stated about the population parameter, not the sample statistic. 2. Let \(\beta\) be the population slope relating weekly rehearsal time to mean performance accuracy. 3. The correct hypotheses are \(H_0: \beta=0\) and \(H_a: \beta>0\).

The student used the sample slope \(b\) instead of the population slope \(\beta\). The correct hypotheses are \(H_0: \beta=0\) and \(H_a: \beta>0\).

- Begin with the definition of a Type I error. - Translate the null hypothesis into the context of the two variables. - State the mistaken conclusion and the true condition together.

1. A Type I error occurs when the null hypothesis is true but is rejected. 2. Here, \(H_0\) states that the population slope is \(0\). 3. A Type I error would be concluding that greater weekly training time is associated with a positive change in mean improvement score when the true population slope is actually \(0\).

A Type I error is concluding that the population slope is positive when the true population slope is \(0\).

- Begin with the definition of a Type II error. - Translate the alternative hypothesis into the context of the variables. - State what the analysis would fail to detect.

1. A Type II error occurs when the alternative is true but the null hypothesis is not rejected. 2. Here, the alternative says that the population slope is negative. 3. A Type II error would be failing to find evidence of a negative population slope when machine age truly is associated with a decrease in mean production speed.

A Type II error is failing to reject \(H_0\) when the true population slope is negative.

- Think about when a significance threshold should be selected relative to seeing the data. - Compare the reported probability with the threshold that was specified in advance. - Consider why changing a decision rule after seeing the result can make false positives easier to obtain.

1. The significance level should be chosen before examining the test result because it sets the tolerated probability of a Type I error under the null hypothesis. 2. Changing \(\alpha\) after seeing the \(p\)-value makes the decision rule data-dependent and inflates the chance of declaring significance opportunistically. 3. Using the prespecified level, \(p=0.032>0.01\), so the researcher fails to reject the null hypothesis.

The researcher should not raise \(\alpha\) after seeing the \(p\)-value. Using the prespecified \(\alpha=0.01\), the decision is to fail to reject the null hypothesis because \(0.032>0.01\).

- Identify what assumption is made when a \(p\)-value is calculated. - Distinguish the probability of data under a hypothesis from the probability that the hypothesis is true. - State the result in terms of outcomes at least as extreme as the one observed.

1. The null hypothesis treats \(\beta=0\) as the condition under which the sampling distribution is evaluated. 2. The \(p\)-value is the probability, assuming \(H_0\) is true, of obtaining a test statistic at least as extreme as the observed one in the direction or directions specified by the alternative. 3. It is not the probability that \(H_0\) itself is true.

The statement is incorrect. A \(p\)-value of \(0.030\) means that, if \(\beta=0\), results at least as extreme as the observed result would occur with probability \(0.030\) under the test model. It does not mean there is a \(3\%\) probability that \(\beta=0\).

- Identify which side of the \(t\) distribution corresponds to the alternative. - Compare the observed statistic with the given cutoff for rejection. - State the conclusion as evidence about how mean improvement score changes with weekly practice time.

1. The one-sided rejection region is \(t>1.725\). 2. The observed statistic \(t=1.70\) does not enter the rejection region. 3. Equivalently, the one-sided \(p\)-value is about \(0.0523\), which is greater than \(0.05\). 4. Therefore, fail to reject \(H_0\). The sample does not provide sufficient evidence at the \(5\%\) level that mean improvement score increases as weekly practice time increases.

Fail to reject \(H_0\). Since \(1.70<1.725\), there is insufficient evidence at \(\alpha=0.05\) that mean improvement score increases with weekly practice time.

- Use the connection between a two-sided test and a confidence interval with the matching confidence level. - Check whether the null value lies inside or outside the interval. - The reject-or-fail-to-reject decision gives a range for the \(p\)-value, not necessarily its exact value.

1. A two-sided test at \(\alpha=0.05\) rejects a null value that lies outside the matching \(95\%\) confidence interval. 2. Since \(0\) is outside \((1.2, 3.4)\), the corresponding two-sided test has \(p<0.05\). 3. The confidence interval alone does not provide the exact tail area, so it does not determine the exact \(p\)-value.

The two-sided test rejects \(H_0: \beta=0\) at \(\alpha=0.05\), and its \(p\)-value is less than \(0.05\). The exact \(p\)-value cannot be determined from the interval alone.

- Match the uncertainty measure to the parameter being tested. - Subtract the null slope from the estimated slope. - Check that the numerator and denominator have the same units.

1. A slope test standardizes the estimated slope using the standard error of the slope. 2. The test statistic is \(t=\frac{0.60-0}{0.20}=3.0\). 3. The residual standard deviation \(s=4.0\) measures typical vertical prediction error, while \(SE_b\) measures uncertainty in the estimated slope. 4. The numerator has slope units, so the denominator must also have slope units.

The test statistic is \(t=3.0\). The denominator must be the slope standard error \(SE_b=0.20\), not the residual standard deviation \(s=4.0\).

- Think about where a statistic close to zero lies in a null \(t\) distribution. - A very small two-sided \(p\)-value requires a statistic far from zero in magnitude. - Use the stated degrees of freedom to check the numerical tail area.

1. A test statistic of \(t=0.20\) lies very close to \(0\), so it is not far into either tail of the \(t\) distribution. 2. With \(30\) degrees of freedom, the two-sided \(p\)-value for \(|t|=0.20\) is approximately \(0.843\). 3. Therefore, \(p<0.001\) is incompatible with the reported test statistic and degrees of freedom.

The report is inconsistent. For \(t=0.20\) with \(30\) degrees of freedom, the two-sided \(p\)-value is approximately \(0.843\), not less than \(0.001\).

- Ask whether a two-sided tail area depends on the sign or the distance from zero. - The two statistics are equally far from zero. - Keep evidence strength separate from the direction of the estimated effect.

1. A two-sided \(p\)-value depends on the magnitude \(|t|\), because evidence in both tails is counted. 2. Tests A and B have the same \(|t|=2.50\) and the same degrees of freedom, so their two-sided \(p\)-values are equal. 3. The sign of \(t\) indicates the direction of the estimated slope relative to the null value: positive for Test A and negative for Test B.

The two tests have equal two-sided \(p\)-values. The sign of \(t\) indicates the direction of the estimated slope relative to the null, while the two-sided \(p\)-value measures evidence against the null in either direction.

- Use \(t=\frac{b-\beta_0}{SE_b}\) with \(\beta_0=0\). - Regression slope tests use \(df=n-2\). - Because the alternative is two-sided, double the one-tail probability beyond \(|t|\).

1. Compute the test statistic: \(t = \frac{b-\beta_0}{SE_b} = \frac{0.700-0}{0.190} \approx 3.684\). 2. The degrees of freedom are \(df = n - 2 = 14 - 2 = 12\). 3. For a two-sided test, \(p = 2P(T_{12} \ge |3.684|) \approx 0.003\).

\(t \approx 3.684\), \(df = 12\), and \(p \approx 0.003\).

- Compute \(t=\frac{b-\beta_0}{SE_b}\) and preserve its sign. - Use \(df=n-2\). - For \(H_a:\beta>0\), the \(p\)-value is the area to the right of the observed \(t\), even when \(t\) is negative.

1. Compute the test statistic: \(t = \frac{b-\beta_0}{SE_b} = \frac{-0.920-0}{0.210} \approx -4.381\). 2. The degrees of freedom are \(df = n - 2 = 16 - 2 = 14\). 3. The alternative is right-tailed, so \(p = P(T_{14} \ge -4.381) \approx 0.999686 \approx 1.000\). 4. The \(p\)-value is very large because the observed slope is strongly negative, opposite the positive direction in \(H_a\).

\(t \approx -4.381\), \(df = 14\), and \(p \approx 1.000\).

- Use \(t=\frac{b-\beta_0}{SE_b}\) with \(\beta_0=0\). - Use \(df=n-2\) for the slope test. - For a two-sided alternative, double the upper-tail probability beyond \(|t|\); report a very small \(p\)-value as an inequality rather than \(0\).

1. Compute the test statistic: \(t = \frac{b-\beta_0}{SE_b} = \frac{1.360-0}{0.250} = 5.440\). 2. The degrees of freedom are \(df = n - 2 = 20 - 2 = 18\). 3. For a two-sided test, \(p = 2P(T_{18} \ge 5.440) \approx 0.000036\), so \(p < 0.001\).

\(t = 5.440\), \(df = 18\), and \(p < 0.001\).

- Subtract the null slope \(1.0\), not \(0\), when computing the test statistic. - Use \(df=n-2\). - Double the one-tail probability because the alternative is two-sided.

1. Compute the test statistic: \(t = \frac{b-\beta_0}{SE_b} = \frac{2.020-1.0}{0.310} \approx 3.290\). 2. The degrees of freedom are \(df = n - 2 = 26 - 2 = 24\). 3. For a two-sided test, \(p = 2P(T_{24} \ge |3.290|) \approx 0.003\).

\(t \approx 3.290\), \(df = 24\), and \(p \approx 0.003\).

- Carefully subtract the negative null value when computing \(b-\beta_0\). - Use \(df=n-2\). - For \(H_a:\beta>-0.5\), use the area to the right of the observed \(t\).

1. Compute the test statistic using the null value \(-0.5\): \(t = \frac{b-\beta_0}{SE_b} = \frac{-2.240-(-0.5)}{0.330} \approx -5.273\). 2. The degrees of freedom are \(df = n - 2 = 28 - 2 = 26\). 3. The alternative is right-tailed, so \(p = P(T_{26} \ge -5.273) \approx 0.999992 \approx 1.000\). 4. The observed slope is far below the null value, opposite the direction in \(H_a\), so the \(p\)-value is very large.

\(t \approx -5.273\), \(df = 26\), and \(p \approx 1.000\).

- For \(H_a:\beta>0\), use the area to the right of \(t=2.10\). - A two-sided test counts equally extreme results in both tails. - The doubling relationship applies because the \(t\)-distribution is symmetric and the statistic is in the one-sided alternative’s direction.

1. For the positive one-sided alternative, the \(p\)-value is the upper-tail probability: \(P(T_{20} \ge 2.10) \approx 0.0243\). 2. For the two-sided alternative, include equally extreme values in both tails: \(2P(T_{20} \ge 2.10) \approx 2(0.0243) \approx 0.0486\). 3. Because the observed \(t\) statistic is in the direction of the one-sided alternative, the two-sided \(p\)-value is twice the one-sided \(p\)-value.

One-sided \(p \approx 0.0243\); two-sided \(p \approx 0.0486\). Here, the two-sided \(p\)-value is twice the one-sided value.

- Use the explanatory-variable row rather than the constant row. - Compare the slope row’s \(p\)-value with \(\alpha\). - State the conclusion about the population slope in the context of musicians and setup time.

1. Use the musicians row because it contains the estimated slope. The output gives \(t = 4.40\) and \(p = 0.0003\). 2. The degrees of freedom are \(df = n - 2 = 22 - 2 = 20\). 3. Since \(0.0003 < 0.05\), reject \(H_0\). 4. There is convincing evidence that the population slope relating the number of musicians to setup time is nonzero.

Reject \(H_0\). There is convincing evidence of a nonzero population regression slope relating the number of musicians to setup time.

- Match the tail of the probability calculation to the direction of the alternative hypothesis. - Use the stated degrees of freedom with the observed test statistic. - Compare the resulting probability with the significance level.

1. For a left-tailed test, the \(p\)-value is the probability of a \(t\)-statistic at or below \(-2.34\) with \(17\) degrees of freedom. 2. This probability is \(p\approx0.016\). 3. Since \(0.016<0.025\), reject \(H_0\).

\(p\approx0.016\). Reject \(H_0\) at \(\alpha=0.025\); the data provide evidence that the population slope is negative.

- Think about how additional independent observations affect uncertainty in an estimate. - Consider what makes a true effect easier to distinguish from the null value. - Focus on the general direction of the change rather than an exact numerical amount.

1. With comparable data quality and a larger sample, the slope estimate generally has a smaller standard error. 2. A smaller standard error makes a fixed nonzero true slope easier to distinguish from the null value. 3. Therefore, the test generally has greater power to detect the positive slope.

The standard error generally decreases, and the power of the test generally increases.

- Check whether the observed statistic points in the same direction as the prespecified alternative. - Use the symmetry of the reference distribution to relate one-sided and two-sided probabilities. - Compare the appropriate probability with the significance level.

1. The observed test statistic is in the direction specified by \(H_a: \beta<0\). 2. For a symmetric \(t\)-distribution, the corresponding one-sided \(p\)-value is half the two-sided value: \(0.084\div2=0.042\). 3. Since \(0.042<0.05\), reject \(H_0\).

The one-sided \(p\)-value is \(0.042\). Reject \(H_0\) at \(\alpha=0.05\).

- Separate evidence against a null hypothesis from the size of an estimated effect. - Ask what information a \(p\)-value provides and what it does not provide. - Compare the estimated slope with the practical benchmark given in the problem.

1. The small \(p\)-value provides strong evidence against \(H_0: \beta=0\), so the data support a nonzero population slope under the test conditions. 2. Statistical significance does not measure the practical size or strength of the relationship. 3. The estimated slope is only \(0.018\) response units per one-unit increase in the explanatory variable, which is below the stated practical threshold of \(0.10\). 4. Therefore, the result can be statistically significant while the estimated effect is practically small.

The report is incorrect. The tiny \(p\)-value is evidence that the population slope is nonzero, but it does not show that the relationship is strong or practically important. The estimated slope \(0.018\) is below the stated practical threshold of \(0.10\).

- Check which null value the software output is actually testing. - Recenter the slope estimate around the null value in the research question. - Use the sample size to identify the appropriate degrees of freedom before finding the tail probability.

1. The software’s reported \(p\)-value corresponds to the null value \(0\), so it cannot be reused for a test whose null slope is \(1.000\). 2. For the requested test, \(t=\frac{1.800-1.000}{0.400}=2.000\) with \(df=22-2=20\). 3. The two-sided \(p\)-value for \(|t|=2.000\) with \(20\) degrees of freedom is approximately \(0.059\).

The student is incorrect. The reported result tests \(H_0: \beta=0\), not \(H_0: \beta=1.000\). For the requested test, \(t=2.000\) with \(df=20\), and the two-sided \(p\)-value is approximately \(0.059\).

- Use the sign of the test statistic together with the direction of the one-sided alternative. - Split the two-sided tail probability between the two symmetric tails. - Decide whether the observed statistic lies in the tail specified by the alternative or in the opposite tail.

1. A two-sided \(p\)-value of \(0.012\) places \(0.006\) probability in each tail beyond the observed magnitude of the test statistic. 2. Because the observed test statistic is negative but the alternative is \(\beta>0\), the relevant upper-tail probability includes almost the entire distribution except the lower tail of probability \(0.006\). 3. The one-sided \(p\)-value is therefore \(1-0.006=0.994\). 4. The data point strongly in the direction opposite the prespecified alternative, so they provide no evidence for a positive population slope.

The appropriate one-sided \(p\)-value is \(0.994\). The observed slope is in the direction opposite \(H_a: \beta>0\), so the relevant tail probability is very large.

- Think of \(\alpha\) as controlling how much of the null distribution is placed in the rejection region. - Ask what happens when that rejection region becomes larger. - Consider the tradeoff between false positives and the ability to detect a real slope.

1. Using \(\alpha=0.05\) makes the rejection region larger, so less extreme sample evidence is required to reject the null hypothesis. 2. Under the null hypothesis, the probability of a Type I error rises from \(0.01\) to \(0.05\). 3. When the alternative is true, the larger rejection region makes rejection more likely, so the test has greater power, all else being equal.

Using \(\alpha=0.05\) makes rejection easier, increases the Type I error probability from \(0.01\) to \(0.05\), and increases the test’s power when the alternative is true.

- Track how reversing the response scale affects the fitted rate of change. - Distinguish the sign of an estimate from the size of its uncertainty. - A two-sided test treats equally extreme positive and negative test statistics symmetrically.

1. Multiplying every response by \(-1\) changes the slope estimate from \(0.60\) to \(-0.60\). 2. The standard error measures the magnitude of sampling uncertainty, so multiplying the response by \(-1\) leaves its value at \(0.20\). 3. The new test statistic is \(t=\frac{-0.60-0}{0.20}=-3.00\). 4. A two-sided test depends on \(|t|\), so the two-sided \(p\)-value is unchanged.

The new slope is \(-0.60\), the standard error is \(0.20\), and the test statistic is \(-3.00\). The two-sided \(p\)-value is unchanged.

- Standardize each slope estimate using its own standard error. - Compare how many standard errors each estimate lies from the null value. - A test reflects both effect size and uncertainty.

1. For Study A, \(t=\frac{0.50}{0.10}=5.00\). With \(df=30\), the two-sided \(p\)-value is less than \(0.001\). 2. For Study B, \(t=\frac{0.50}{0.40}=1.25\). With \(df=30\), the two-sided \(p\)-value is approximately \(0.221\). 3. The studies have the same estimated slope but different uncertainty. The smaller standard error in Study A makes the estimate much farther from \(0\) in standard-error units. 4. Statistical significance depends on the estimated effect relative to its uncertainty, not on the slope estimate alone.

Study A: \(t=5.00\), \(p<0.001\). Study B: \(t=1.25\), \(p\approx0.221\). The conclusions differ because the same estimated slope is much more precise in Study A.

- Convert the null value using the same unit change that converts the fitted slope. - Compare the standardized distance from the null before and after rescaling. - A numerical slope value has meaning only together with its explanatory-variable units.

1. Multiplying the explanatory variable by \(10\) divides every response-per-unit slope by \(10\). 2. Therefore, the original null value \(\beta=2.00\) corresponds to \(\beta^*=0.20\). 3. The transformed test statistic is \(t=\frac{0.24-0.20}{0.06}\approx0.667\), the same value as \(\frac{2.40-2.00}{0.60}\). 4. Testing \(\beta^*=2.00\) would correspond to an original-scale slope of \(20.00\), so it is not the same scientific hypothesis.

Test \(H_0: \beta^*=0.20\). The test statistic is \(t\approx0.667\). Leaving the null value at \(2.00\) after changing units would test a different slope.

- Distinguish the exact \(p\)-value from a value rounded for display. - Think about which exact values can round to \(0.050\). - A decision at a boundary requires enough numerical precision to know which side of the threshold the exact value lies on.

1. The usual decision rule rejects when the exact \(p\)-value is at most the chosen significance level. 2. A displayed value of \(0.050\) rounded to three decimals could represent an exact value slightly below \(0.050\), exactly \(0.050\), or slightly above \(0.050\). 3. Therefore, the rounded display alone is not enough to make a definite boundary decision at \(\alpha=0.050\). 4. More numerical precision from the software is needed before applying the decision rule.

The student’s conclusion is not justified from the rounded display alone. Because \(0.050\) is rounded, the exact \(p\)-value could lie on either side of \(0.050\). More precision is needed to decide whether to reject at \(\alpha=0.050\).

- Identify the reference distribution used by each test. - The numerical test statistic alone does not specify the relevant tail area. - Compare how \(t\) distributions change as the degrees of freedom increase.

1. A \(t\)-test \(p\)-value depends on both the test statistic and the degrees of freedom because different \(t\) distributions have different tail thicknesses. 2. For \(t=2.00\) with \(df=5\), the two-sided \(p\)-value is approximately \(0.102\). 3. For \(t=2.00\) with \(df=50\), the two-sided \(p\)-value is approximately \(0.051\). 4. The larger degrees of freedom produce thinner tails, so the same \(|t|\) corresponds to a smaller tail probability.

Test A: \(p\approx0.102\). Test B: \(p\approx0.051\). The claim is incorrect because a \(t\)-test \(p\)-value depends on the degrees of freedom as well as on the test statistic.

- Check the conditions behind the inference procedure, not just the size of the reported \(p\)-value. - Think about what long runs of residuals with the same sign say about observations close together in time. - Decide whether the standard regression test is trustworthy when one of its sampling conditions is doubtful.

1. The usual slope test requires the observations, and therefore the regression errors, to be independent. 2. Long runs of residuals with the same sign in time order are evidence of serial dependence rather than independent errors. 3. With this condition in doubt, the standard error and resulting \(t\)-test may not have the stated sampling behavior. 4. Therefore, the reported \(p\)-value of \(0.004\) should not be treated as valid evidence from the usual slope test until the dependence is addressed.

No. The time-order residual pattern suggests that the independence condition is violated, so the usual slope-test \(p\)-value is not reliable as reported.

- Consider which regression-inference conditions matter especially when the sample is small. - Ask what an extreme residual can do to both the error distribution and the fitted slope. - A small \(p\)-value does not repair a failed model condition.

1. With a small sample, the slope \(t\)-procedure is sensitive to strong departures from the nearly normal error condition and to extreme observations. 2. The isolated large residual is evidence that this condition is doubtful and may also represent an influential observation. 3. Therefore, the reference \(t\) distribution used to obtain \(p=0.03\) may not be appropriate for these data. 4. The reported significance should not be treated as dependable without investigating the unusual observation and the model conditions.

No. With only \(10\) observations, the extreme residual makes the usual slope \(t\)-test questionable, so the reported \(p=0.03\) is not enough by itself for a dependable conclusion.

- Separate what the slope test says from what the study design contributes. - Ask how random assignment affects potential confounding variables. - Keep any causal conclusion within the population and treatment range actually studied.

1. The slope test provides evidence of a positive linear relationship between assigned fertilizer amount and mean growth over the studied range. 2. Random assignment tends to balance other explanatory factors across fertilizer levels, reducing confounding. 3. Because the explanatory variable was deliberately randomized and the model conditions are supported, the positive relationship can be interpreted as evidence that fertilizer amount causes a change in mean growth for the experimental population and range studied. 4. The conclusion should not automatically be generalized beyond the experimental conditions or fertilizer range.

The randomized experiment can support a causal conclusion: within the studied range and population, increasing the assigned fertilizer amount affects mean growth. An observational slope association alone would not justify that causal claim.

- Substitute the sample correlation and sample size into the provided test-statistic formula. - A simple linear regression slope test uses \(n-2\) degrees of freedom. - Use both tails because the alternative allows slopes in either direction.

1. Substitute \(r=0.54\) and \(n=30\): \(t=0.54\sqrt{\frac{28}{1-0.54^2}}\approx3.395\). 2. The residual degrees of freedom are \(30-2=28\). 3. For \(t\approx3.395\) with \(28\) degrees of freedom, the two-sided \(p\)-value is approximately \(0.00207\). 4. Since \(0.00207<0.01\), reject \(H_0\). The sample provides evidence of a nonzero population slope relating weekly tutoring time to improvement in assessment score.

\(t\approx3.395\), \(df=28\), and \(p\approx0.00207\). Reject \(H_0\) at \(\alpha=0.01\); there is evidence of a nonzero population regression slope.