Justifying claims: confidence interval practice

The rule for every part of this problem: a value inside the interval is plausible, so the data give no convincing evidence against it; a value outside the interval is not plausible at this confidence level, so the data give convincing evidence against it.

(a) "More than half" is the claim $p > 0.50$, so the value to check is 0.50. Compare it to the endpoints: $0.50 < 0.507$, so 0.50 sits below the entire interval.

(a) Every plausible value for $p$ is therefore greater than 0.50, and that is what justifies the claim. There is convincing evidence that more than half of all registered voters in this town support the bond measure.

(b) 0.54 is the sample proportion, $\hat{p} = \frac{486}{900} = 0.54$, and it is the center of the interval. The interval says 0.54 is plausible. It says exactly the same about 0.51, 0.55, and 0.57.

(b) So the claim is not supported as stated. Failing to rule a value out is not evidence that it is the truth. The supportable version names the range: the data are consistent with true support anywhere from about 50.7% to 57.3%.

(c) The error is attaching a probability to a finished interval. The endpoints 0.507 and 0.573 are fixed numbers and $p$ is a fixed number, so this interval either contains $p$ or it does not. The 95% belongs to the method: in repeated random sampling with $n = 900$, about 95% of the intervals built this way capture $p$.

$\hat{p} = \frac{92}{400} = 0.23$.

Conditions. Random: the 400 students were randomly sampled. 10%: $400 \leq 0.10(6200) = 620$, so the sample is under 10% of the district. Large Counts: $n\hat{p} = 92 \geq 10$ and $n(1 - \hat{p}) = 308 \geq 10$. All three hold, so a one-sample z interval is appropriate.

$SE = \sqrt{\frac{0.23(0.77)}{400}} = \sqrt{\frac{0.1771}{400}} = \sqrt{0.00044275} = 0.021042$.

At 95% confidence the critical value (z-star, written $z^*$) is $z^* = 1.96$, so $ME = 1.96(0.021042) = 0.041242$.

(a) Interval: $0.23 \pm 0.041242 = (0.188758,\ 0.271242)$, about $(0.189,\ 0.271)$.

(b) Compare 0.30 to the endpoints. $0.30 > 0.271$, so 0.30 lies above the entire interval and is not a plausible value at this confidence level. There is convincing evidence against the assumption, and the evidence points lower: every plausible value sits below 27.2%.

(c) Compare 0.25. $0.189 < 0.25 < 0.271$, so 0.25 is inside the interval. The parent group's figure stays plausible and the interval gives no convincing evidence against it. Note what you may not add: the interval does not show the figure is 25%.

(d) The justification names the confidence level, both endpoints, the parameter in context, and the verdict in the language of plausible values.

$\hat{p} = \frac{220}{500} = 0.44$. Conditions: the sample is random, $500 \leq 0.10(21000) = 2100$, and $n\hat{p} = 220 \geq 10$ with $n(1 - \hat{p}) = 280 \geq 10$.

$SE = \sqrt{\frac{0.44(0.56)}{500}} = \sqrt{\frac{0.2464}{500}} = \sqrt{0.0004928} = 0.022199$. The sample does not change between the two parts, so this one standard error serves both intervals.

(a) At 90% confidence $z^* = 1.645$, so $ME = 1.645(0.022199) = 0.036518$ and the interval is $0.44 \pm 0.036518 = (0.403482,\ 0.476518)$, about $(0.403,\ 0.477)$.

(a) Judge 0.48: $0.48 > 0.476518$, so it sits just above the interval and is not plausible at this level. There is convincing evidence against the 48% claim. Judge 0.50: it is above the interval as well, so there is convincing evidence against that claim too.

(b) At 99% confidence $z^* = 2.576$, so $ME = 2.576(0.022199) = 0.057185$ and the interval is $0.44 \pm 0.057185 = (0.382815,\ 0.497185)$, about $(0.383,\ 0.497)$.

(b) Judge 0.48 again: $0.383 < 0.48 < 0.497$, so 0.48 is now inside and stays plausible. There is no convincing evidence against the 48% claim at this level. Judge 0.50: $0.50 > 0.497185$, still above the interval, so the evidence against the half-of-households claim survives the change.

(c) The 99% interval is wider from the same data because $z^*$ grew from 1.645 to 2.576, and a wider interval rules out fewer values. Demanding a higher long-run capture rate means accepting a longer list of plausible values, and 0.48 is on the longer list.

(c) What it does not show: nothing about the county changed between the two parts, and neither interval is the wrong one. This is why a justification has to state its confidence level. A value ruled out at 90% and left plausible at 99% is a borderline value, and naming the level is what keeps the two answers from contradicting each other.

Get the numbers straight first. $\hat{p}_1 = \frac{95}{125} = 0.76$ and $\hat{p}_2 = \frac{88}{125} = 0.704$, so the point estimate is $0.76 - 0.704 = 0.056$. Large Counts holds on the observed counts: 95 and 30 in the new group, 88 and 37 in the standard group.

(a) Zero is the value to check for a difference. Because $-0.054 < 0 < 0.166$, a difference of 0 is a plausible value, so there is not convincing evidence that the two inhalers differ in the true proportion of patients reporting no symptoms.

(a) The interval also runs from negative to positive, so it settles no direction either. A standard-inhaler advantage of about 5 percentage points and a new-inhaler advantage of about 17 points are both plausible on this evidence.

(b) First error: you never accept a null hypothesis, you only fail to reject it. Failing to rule out equality is not evidence for equality, and an interval containing 0 is a fail-to-reject verdict in another form. If the inhalers really do differ and this study missed it, that is a Type II error, which is exactly the possibility the word "accept" pretends away.

(b) Second error: "work equally well" pulls 0 out of the interval and reports it as the truth, when 0 has no better claim on being right than 0.10 does. Write that 0 is plausible, not that the difference is 0.

(c) Not supported. "No better" is the claim that the new inhaler's true proportion is at most the standard one's, and the interval reaches up to 0.166, so a real advantage for the new formulation of up to about 17 percentage points is among the plausible values. The honest conclusion is that 125 patients per group did not settle this comparison in either direction.

$\hat{p}_N = \frac{168}{400} = 0.42$ and $\hat{p}_S = \frac{252}{450} = 0.56$, so the point estimate is $\hat{p}_N - \hat{p}_S = -0.14$. Conditions hold: two independent random samples, $400 \leq 0.10(9000) = 900$ and $450 \leq 0.10(12000) = 1200$, and the four observed counts 168, 232, 252, and 198 all clear 10.

(a) Both endpoints are negative, so 0 is not in the interval and every plausible value of $p_N - p_S$ is negative. There is convincing evidence that the true proportion of tutoring users is higher on the South campus.

(a) Say how much, because the interval carries that too. The South campus lead is plausibly between about 7.3 and 20.7 percentage points.

(b) No. Close to 0 is not 0. Compare endpoints as signed numbers: the interval runs from $-0.207$ up to $-0.073$, and 0 is above $-0.073$, so 0 lies outside. If you want the verdict to reflect how borderline the evidence is, raise the confidence level and see whether 0 comes back inside, rather than waving the endpoint in.

(c) Reversing the order of subtraction swaps the endpoints and negates them, giving $(0.073,\ 0.207)$. Both endpoints are positive now, and 0 is still outside, so the same evidence supports the same conclusion: the South campus rate is higher.

(c) The verdict never depends on the order, but the sentence does. State your order in the first line and hold it to the last, because a conclusion naming the wrong campus loses the point even when every number is right.

Set the scene. $\hat{p}_A = \frac{570}{1000} = 0.57$ and $\hat{p}_B = \frac{520}{1000} = 0.52$, so the point estimate is 0.05, and the four observed counts 570, 430, 520, and 480 all clear 10.

Statement 1 is wrong. The interval is computed, so its endpoints are fixed numbers, and the true difference is a fixed number. This interval either contains it or it does not. The 95% belongs to the method across repeated samples, which is why the required wording is "95% confident" and not "95% chance".

Statement 2 is correct. A 95% interval for a difference lines up with a two-sided test at $\alpha = 1 - 0.95 = 0.05$, and 0 outside the interval matches rejecting $H_0$. One caution the course expects you to know: the interval uses each sample's own proportion in its standard error while the test pools the samples, so on a borderline problem the two can disagree. If the question asks for a test, run the test.

Statement 3 is wrong. 0.05 is the point estimate, the center of the interval, not something the interval establishes. Every difference from 0.006 to 0.094 is supported, and 0.05 has no special standing among them.

Statement 4 is wrong, and worth naming precisely. A p-value is not the probability that $H_0$ is true. It is computed by assuming $H_0$ is true and asking how often a difference at least this extreme would appear, so it cannot also be a probability about $H_0$. Here $H_0$ is a statement about two fixed population proportions: it is either right or wrong, and no probability attaches to it. The 0.05 is the long-run rate at which this procedure rejects a true null.

Statement 5 is correct. It names the confidence level, gives both endpoints, identifies the parameter and both populations with the order of subtraction attached, and takes the direction from the fact that both endpoints are positive.

$\hat{p}_1 = \frac{195}{300} = 0.65$ and $\hat{p}_2 = \frac{200}{400} = 0.50$, so the point estimate is $0.65 - 0.50 = 0.15$.

Conditions. Two independent random samples. 10%: $300 \leq 0.10(6000) = 600$ enrolled and $400 \leq 0.10(14000) = 1400$ not enrolled. Large Counts on the observed counts: 195, 105, 200, and 200, all at least 10.

Standard error, unpooled because this is an interval rather than a test: $\frac{0.65(0.35)}{300} = \frac{0.2275}{300} = 0.00075833$ and $\frac{0.50(0.50)}{400} = \frac{0.25}{400} = 0.000625$, so $SE = \sqrt{0.00075833 + 0.000625} = \sqrt{0.00138333} = 0.037193$.

(a) At 99% confidence $z^* = 2.576$, so $ME = 2.576(0.037193) = 0.09581$ and the interval is $0.15 \pm 0.09581 = (0.05419,\ 0.24581)$, about $(0.054,\ 0.246)$.

(b) Both endpoints are positive, so 0 is outside the interval and every plausible difference favors the enrolled group. The justification: we are 99% confident that the interval from 0.054 to 0.246 contains the true difference in the proportions meeting the step goal, enrolled minus not enrolled, and because 0 is not in that interval there is convincing evidence that enrolled employees meet the goal at a higher rate.

(c) Not as stated. 15 percentage points is the point estimate, and the interval runs from about 5.4 to 24.6 points. The supportable claim names that range. A gap of 5 points and a gap of 24 points are very different stories for a company deciding what the program is worth, and this study does not choose between them.

(d) No, for a reason that has nothing to do with the arithmetic. Employees chose whether to join, so these are two random samples from two self-selected groups, not a randomized experiment. Employees who join a wellness program may already walk more, and no confidence interval touches that confounding. The interval supports a claim about a difference in rates, never a claim about what produced the difference.

$\hat{p}_1 = \frac{415}{5000} = 0.083$ and $\hat{p}_2 = \frac{350}{5000} = 0.07$, so the point estimate is $0.013$. Random assignment gives two independent groups, no 10% condition applies to a randomized experiment, and the observed counts 415, 4585, 350, and 4650 all clear 10.

(a) Yes. Both endpoints are positive, so 0 is not a plausible difference and every plausible difference favors the new page. Because visitors were randomly assigned to a page, this experiment supports a cause and effect conclusion for the visitors in the study: the new page increases the completion rate.

(b) No, and the comparison is the same one you used for 0. The claimed value 0.05 sits above the entire interval, since $0.05 > 0.0234$, so it is not plausible. There is convincing evidence against the manager's claim.

(b) Notice that justification runs in both directions. An interval rules out values above it just as firmly as values below it, so it can refute a claim that the effect is large as readily as a claim that there is no effect. Here the plausible gains run from about 0.26 to 2.34 percentage points.

(c) Statistical significance and practical importance are different questions. With 5,000 visitors per group, $SE = \sqrt{\frac{0.083(0.917)}{5000} + \frac{0.07(0.93)}{5000}} = \sqrt{0.0000152222 + 0.00001302} = \sqrt{0.0000282422} = 0.005314$, so the observed gap of 1.3 percentage points is $\frac{0.013}{0.005314} = 2.45$ standard errors, enough to push 0 out of a 95% interval. Large samples detect small real differences.

(c) Whether an increase of roughly 0.3 to 2.3 percentage points is worth rebuilding the checkout is a business judgment, not a statistical one. What the interval contributes is a bound on the size of the effect, and here it caps the gain below 2.4 percentage points.