Unit 4 · 10–20% of the exam Beta

Unit 4 practice test: Inference for Quantitative Data: Means

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value, an interval or the setup; sets share one data set, plot or table.

Covers: Sampling Distributions for Sample Means; Constructing a Confidence Interval for a Population Mean or Population Mean Difference; Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference; Setting Up a Test for a Population Mean or Population Mean Difference; Carrying Out a Test for a Population Mean or Population Mean Difference; Sampling Distributions for the Difference Between Two Sample Means; Constructing a Confidence Interval for the Difference Between Two Population Means; Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means; Setting Up a Test for the Difference Between Two Population Means; Carrying Out a Test for the Difference Between Two Population Means.

Table

A tutoring program's claim

A tutoring company claims its 6-week program raises math test scores by more than 10 points on average. A district took a random sample of 40 students who completed the program and compared each student's score before and after (invented data). The differences (after minus before) show no strong skew.

Summary of the 40 differences, after minus before, in points
Sample size	Mean difference	Standard deviation of differences	95% confidence interval for the mean difference
40	11.2	11.9	from 7.4 to 15.0 points

1. Does the interval give convincing evidence that students improve at all, on average?

No, because 10 is in the interval
Yes, because every value in the interval is greater than 0
No, because some students' scores went down
Yes, because the sample mean of 11.2 points is well above 0 points

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The claim "some improvement" means $\mu_d > 0$ . Since 0 is outside the interval and every plausible value is positive, there is convincing evidence of improvement.

No, because 10 is in the interval: Whether 10 is inside matters for the company's claim. For "any improvement" the value to check is 0.
Correct: Yes, because every value in the interval is greater than 0: A mean difference of 0 (no change) is not plausible, since the interval runs from 7.4 to 15.0.
No, because some students' scores went down: Individual students can decline while the mean difference is clearly positive.
Yes, because the sample mean of 11.2 points is well above 0 points: A positive sample mean alone could be chance. The interval, entirely above 0, is the evidence.

2. The district wants a narrower 95% interval in its next study. What should it change?

Use a larger random sample of students
Use a 99% confidence level
Compare two separate groups of students instead of before and after
Leave out the students whose scores dropped

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Margin of error $= t^*\frac{s_d}{\sqrt{n}}$ : a larger sample shrinks it. Four times the students roughly halves it.

Correct: Use a larger random sample of students: A larger $n$ makes $\frac{s_d}{\sqrt{n}}$ smaller (and $t^*$ slightly smaller), so the interval narrows at the same confidence.
Use a 99% confidence level: A higher level needs a larger $t^*$ , which widens the interval.
Compare two separate groups of students instead of before and after: Two separate groups add student-to-student variation, which usually widens the interval.
Leave out the students whose scores dropped: Dropping data that do not fit is not honest and makes the sample unrepresentative.

3. A parent reads the interval and says, "So 95% of students gain between 7.4 and 15.0 points." What is wrong?

Nothing; that is the meaning of 95% confidence
It should say 95% of the 40 sampled students
The interval estimates the mean gain, not individual gains
It should say that 5% of students gain less than 7.4 points

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A confidence interval is about a parameter, here $\mu_d$ . With $s_d = 11.9$ , single students' gains spread far wider than 7.4 to 15.0.

Nothing; that is the meaning of 95% confidence: 95% confidence describes how often the method captures the true mean, not where 95% of individual values fall.
It should say 95% of the 40 sampled students: Even for the sample, the interval does not describe individual gains.
Correct: The interval estimates the mean gain, not individual gains: Individual differences have standard deviation 11.9 points, so many students gain far more or less than this range.
It should say that 5% of students gain less than 7.4 points: The interval makes no statement about the share of students above or below a value.

Table

Apartment rents in two cities

A housing group took independent random samples of one-bedroom apartment listings in cities A and B and compared monthly rents (invented data). Both samples are large and small shares of each city's listings. The 99% two-sample t interval for $\mu_A - \mu_B$ , the difference in true mean monthly rent, is shown.

The 99% interval for the difference in mean monthly rent
Parameter	Confidence level	Interval
$\mu_A - \mu_B$ (city A minus city B)	99%	from $-45$ to 130 dollars

4. A reporter writes, "The study proves that rents are the same in the two cities." What is wrong?

Nothing is wrong, because 0 is inside the interval from $-45$ to 130
The interval shows city A is more expensive
A 99% interval is too wide to say anything
0 is plausible, but so are differences up to 130 dollars

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Failing to find convincing evidence of a difference is not evidence of no difference.

Nothing is wrong, because 0 is inside the interval from $-45$ to 130: Being in the interval makes 0 plausible, not proven.
The interval shows city A is more expensive: Negative values are in the interval too, so no direction is established.
A 99% interval is too wide to say anything: It says something useful: no convincing evidence of a difference, and how big a difference is plausible.
Correct: 0 is plausible, but so are differences up to 130 dollars: A plausible value is not a proven one. The interval also allows city A to cost up to 130 dollars more, or city B up to 45 dollars more.

5. If the group had built a 95% interval from the same large samples instead, which statement is true? (For large samples, $t^*$ is close to 2.58 for 99% and 1.96 for 95%.)

It would be narrower, about $-24$ to 109 dollars, and still contain 0
It would be wider, so it would contain 0 with more room to spare
It would be the same interval, since the data are the same
It would be narrower and would exclude 0, since 95% is less strict

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Center 42.5 dollars; 95% margin about 66 dollars; interval about $-24$ to 109 dollars, which contains 0. The conclusion does not change.

Correct: It would be narrower, about $-24$ to 109 dollars, and still contain 0: Same center, $\frac{-45 + 130}{2} = 42.5$ ; the margin shrinks from 87.5 to about $87.5 \times \frac{1.96}{2.58} = 66$ , so the interval still reaches below 0.
It would be wider, so it would contain 0 with more room to spare: A lower confidence level gives a narrower interval, not a wider one.
It would be the same interval, since the data are the same: The critical value changes with the confidence level, so the margin changes.
It would be narrower and would exclude 0, since 95% is less strict: It is narrower, but its lower end, about $-24$ , is still below 0.

6. What range of differences in mean rent does the interval treat as plausible?

From city A costing 45 dollars more to city A costing 130 dollars more
From city B costing 45 dollars more to city A costing 130 dollars more
Differences near the center of 42.5 dollars, and not the ends
Differences in individual rents from $-45$ to 130 dollars

Show the answer

$\mu_A - \mu_B$ from $-45$ to 130: B up to 45 dollars more expensive, or A up to 130 dollars more expensive, on average.

From city A costing 45 dollars more to city A costing 130 dollars more: A negative difference, $\mu_A - \mu_B = -45$ , means city B costs more.
Correct: From city B costing 45 dollars more to city A costing 130 dollars more: Negative values mean B is more expensive on average, positive values mean A is, so the interval covers both directions.
Differences near the center of 42.5 dollars, and not the ends: Every value in the interval is plausible, not just the center.
Differences in individual rents from $-45$ to 130 dollars: The interval is about the difference in mean rents, not individual apartments.

7. Incomes in a town are strongly skewed right. A researcher takes a random sample of 8 households. What shape does the sampling distribution of $\bar{x}$ have?

Approximately normal, because sample means are normal
Skewed left, the opposite of the population
Skewed right, though less skewed than the incomes
Exactly as skewed as the incomes

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For a skewed population and a small sample, the sampling distribution of $\bar{x}$ is skewed in the same direction, less strongly. A normal model would need $n \ge 30$ .

Approximately normal, because sample means are normal: Sample means become approximately normal only for a normal population or a large sample.
Skewed left, the opposite of the population: Averaging pulls the shape toward normal; it does not flip the direction of skew.
Correct: Skewed right, though less skewed than the incomes: With $n = 8$ the central limit theorem cannot be counted on, so some of the population's right skew remains.
Exactly as skewed as the incomes: Averaging 8 values already reduces the skew, though not enough to call the shape normal.

8. As the degrees of freedom increase, what happens to $t^*$ for 95% confidence?

It increases toward 3
It decreases toward $z^* = 1.96$
It stays the same, at 1.96, whatever the sample size
It decreases toward 0

Show the answer

Table B: the 95% column falls from 12.71 at df 1 toward 1.96 in the infinite row.

It increases toward 3: The heaviest tails, and the largest $t^*$ , come with the fewest degrees of freedom.
Correct: It decreases toward $z^* = 1.96$ : More data make $s$ a steadier estimate of $\sigma$ , so the t distribution gets closer to the standard normal: 2.776 at df 4, 2.064 at df 24, 1.984 at df 100.
It stays the same, at 1.96, whatever the sample size: $t^*$ is larger than 1.96 at each finite df, and the gap shrinks as df grows.
It decreases toward 0: $t^*$ approaches $z^* = 1.96$ , not 0: the middle 95% of a normal curve still has width.

9. A student sees a sample mean of 27 minutes for battery charging time and then writes $H_a: \mu < 30$ because 27 is below 30. The original question was whether the mean differs from 30. What is the error?

The direction of $H_a$ was chosen from the data instead of the question
The hypothesized mean should be 27
There is no error, because the data showed which direction to test
The alternative should be $\mu > 30$

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Hypotheses are set from the research question, before looking at the data.

Correct: The direction of $H_a$ was chosen from the data instead of the question: The question asked about a difference in either direction, so $H_a: \mu \ne 30$ . Letting the sample pick the side inflates the chance of a false rejection.
The hypothesized mean should be 27: The hypothesized mean comes from the question (30); 27 is the sample result.
There is no error, because the data showed which direction to test: Choosing the side after seeing the data makes the test look stronger than it is.
The alternative should be $\mu > 30$ : The question is two-sided, so neither one-sided alternative fits.

10. A random sample of 20 students at a large university slept a mean of 7.55 hours a night with standard deviation 1.09 hours. For $H_0: \mu = 8$ against $H_a: \mu \ne 8$ , what is the p-value?

Type a number. A table answer and a calculator answer are both accepted.

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$t$ $= \frac{7.55 - 8}{1.09/\sqrt{20}}$ $= -1.85$ , $\text{df} = 19$ . Two-sided: $p$ $= 2P(t < -1.85)$ $= 0.080$ .

Answer: 0.0805

11. Independent random samples of 60 nurses are taken at each of two hospitals, one with 450 nurses and one with 2,000. Which condition check is needed for the standard deviation formula?

The total of 120 has to be less than 10% of the 2,450 nurses
Each sample has to be less than 10% of its own hospital's nurses
The two hospitals need the same number of nurses
Each sample needs at least 10 nurses

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Independence within each sample: $n_1 < 0.10N_1$ and $n_2 < 0.10N_2$ . Here $60 > 45$ at the smaller hospital.

The total of 120 has to be less than 10% of the 2,450 nurses: Each sample is drawn from its own population, so each is checked separately.
Correct: Each sample has to be less than 10% of its own hospital's nurses: The 10% condition is checked for each sample. 60 is less than 10% of 2,000 (200) but not of 450 (45), so it fails for the first hospital.
The two hospitals need the same number of nurses: Population sizes can differ; each sample just needs to be a small share of its own population.
Each sample needs at least 10 nurses: That is not a condition for means; the large counts condition is for proportions.

12. A 95% interval for $\mu_A - \mu_B$ is from 2.1 to 6.5. What is the 95% interval for $\mu_B - \mu_A$ from the same data?

From 2.1 to 6.5
From $-2.1$ to $-6.5$
From $-6.5$ to $-2.1$
From 6.5 to 2.1

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If $\mu_A - \mu_B$ is between 2.1 and 6.5, then $\mu_B - \mu_A$ is between $-6.5$ and $-2.1$ .

From 2.1 to 6.5: The new parameter is the negative of the old one, so the interval changes sign.
From $-2.1$ to $-6.5$ : Write the lower end first: $-6.5$ is less than $-2.1$ .
Correct: From $-6.5$ to $-2.1$ : Reversing the order of subtraction flips the sign of every value: the interval keeps its width and changes sides of 0.
From 6.5 to 2.1: That reverses the ends without changing signs, and the lower end must come first.

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Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5.