AP Statistics Inference for Quantitative Data: Means — Worked Answer Explanations

Unit 7 · 12 questions explained

1. Question 1 · Easy

   A researcher collects a random sample of 20 measurements and wants to construct a confidence interval for the population mean. The population standard deviation is unknown. Which distribution should be used?

   • A
     Standard normal ($z$) distribution

     Why not A: The $z$ distribution requires knowing $\sigma$; with unknown $\sigma$ estimated by $s$, use $t$.
   • B
     $t$-distribution with 20 degrees of freedom

     Why not B: Degrees of freedom for a one-sample $t$ is $n-1=19$, not $n=20$.
   • C
     $t$-distribution with 19 degrees of freedomCorrect
   • D
     ${\chi }^2$ distribution with 19 degrees of freedom

     Why not D: The chi-square distribution is used for inference about variance or categorical data, not for the mean.
   Explanation

   When $\sigma$ is unknown and estimated by $s$, use the $t$-distribution with $n-1=19$ degrees of freedom. The $t$-distribution has heavier tails than the standard normal, accounting for the additional uncertainty from estimating $\sigma$.

   Key takeaway

   Use the $t$-distribution with $df = n - 1$ when $\sigma$ is unknown and must be estimated by $s$.

2. Question 2 · Easy

   A random sample of 16 light bulbs had a mean life of $\bar{x}=1,200$ hours with $s=80$ hours. Construct a 95% confidence interval for the true mean life. (Use $t^{\ast }\approx 2.131$ for $df=15$.)

   • A
     $(1,143,1,257)$

     Why not A: Using $t^{\ast }=1.96$ (z-value) instead of $t^{\ast }=2.131$.
   • B
     $(1,157.4,1,242.6)$Correct
   • C
     $(1,040,1,360)$

     Why not C: Using $s=80$ directly (not divided by $\sqrt{n}$) as the standard error.
   • D
     $(1,182.6,1,217.4)$

     Why not D: Using $t^{\ast }=1.0$ instead of the correct critical value for 95%.
   Explanation

   $SE=s/\sqrt{n}=80/\sqrt{16}=20$. $ME=t^{\ast }\times SE=2.131\times 20=42.6$. 95% CI: $1200\pm 42.6=(1157.4,1242.6)$.

   Key takeaway

   One-sample $t$-interval: $\bar{x} \pm t^* (s/\sqrt{n})$ where $t^*$ uses $df = n-1$.

3. Question 3 · Easy

   A $t$-test for the mean produces $t=2.45$ with $df=14$. For $H_a:\mu >50$, the $p$-value is between 0.01 and 0.025. At $\alpha =0.05$, what conclusion should be drawn?

   • A
     Fail to reject $H_0$; the $p$-value is between 0.01 and 0.025.

     Why not A: Both bounds (0.01 and 0.025) are below $\alpha =0.05$, so we reject $H_0$.
   • B
     Reject $H_0$; there is convincing evidence that $\mu >50$.Correct
   • C
     Reject $H_0$; the result is significant at $\alpha =0.01$.

     Why not C: We cannot conclude significance at $\alpha =0.01$ because the $p$-value could be as high as 0.025.
   • D
     The conclusion cannot be determined without the exact $p$-value.

     Why not D: Since the entire range (0.01–0.025) is below $\alpha =0.05$, we can still reject $H_0$ at the 5% level.
   Explanation

   The $p$-value is between 0.01 and 0.025, which is entirely below $\alpha =0.05$. Therefore, we reject $H_0$ at the 5% significance level and conclude there is convincing evidence that $\mu >50$.

   Key takeaway

   If the $p$-value range is entirely below $\alpha$, reject $H_0$; if entirely above $\alpha$, fail to reject.

4. Question 4 · Easy

   A teacher wants to determine whether students improved from a pre-test to a post-test. Each of 15 students took both tests. What is the appropriate procedure?

   • A
     Two-sample $t$-test comparing pre-test and post-test group means

     Why not A: A two-sample test treats the groups as independent; here the same students took both tests (paired data).
   • B
     Paired $t$-test on the differences (post − pre) for each studentCorrect
   • C
     One-sample $z$-test comparing the mean pre-test to the mean post-test

     Why not C: A $z$-test requires known $\sigma$; also, this does not account for the paired structure.
   • D
     Two-sample $z$-test for proportions

     Why not D: Test scores are quantitative, not proportions; and the data are paired.
   Explanation

   When the same individuals are measured twice, compute the difference $d_i={\text{post}}_i-{\text{pre}}_i$ for each person, then perform a one-sample $t$-test on $\bar{d}$ with $H_0:{\mu }_d=0$. This accounts for within-person correlation and is more powerful than treating the groups as independent.

   Key takeaway

   Paired data (same subjects measured twice or matched pairs): compute differences, then perform a one-sample $t$-test on $\bar{d}$.

5. Question 5 · Easy

   A two-sample $t$-test compares mean test scores for two independent classes (Class A: $n_1=25$, Class B: $n_2=30$). Assuming unequal variances, what are the degrees of freedom? (Use the conservative formula: $df=\min (n_1-1,n_2-1)$.)

   • A
     55

     Why not A: Using $df=n_1+n_2-2=53$; the formula $n_1+n_2-2$ applies to the pooled $t$-test with equal variances.
   • B
     53

     Why not B: Using $n_1+n_2-2$, which applies to pooled two-sample $t$ with equal variances.
   • C
     24Correct
   • D
     29

     Why not D: Using $\max (n_1-1,n_2-1)=29$ instead of the conservative $\min$ formula.
   Explanation

   The conservative degrees of freedom for a two-sample $t$-test (Welch's $t$) is the smaller of $n_1-1=24$ and $n_2-1=29$. So $df=24$. This gives a slightly wider (more conservative) interval than the full Welch formula.

   Key takeaway

   Conservative $df$ for two-sample $t$: $\min(n_1-1, n_2-1)$. Full Welch formula gives a larger $df$, which software computes.

6. Question 6 · Easy

   A researcher conducts a one-sample $t$-test with $n=12$, $\bar{x}=55$, $s=8$, testing $H_0:\mu =50$ vs. $H_a:\mu \ne 50$. Compute the $t$ test statistic.

   • A
     $t=0.625$

     Why not A: Dividing by $s=8$ instead of $s/\sqrt{n}$: $(55-50)/8=0.625$.
   • B
     $t\approx 2.17$Correct
   • C
     $t\approx 1.73$

     Why not C: Dividing by $s/n$ instead of $s/\sqrt{n}$: $(55-50)/(8/12)\approx 7.5$.
   • D
     $t=5.0$

     Why not D: Not dividing $s$ by $\sqrt{n}$ at all: using $n$ instead of $\sqrt{n}$ in the denominator.
   Explanation

   $SE=s/\sqrt{n}=8/\sqrt{12}\approx 2.309$. $t=(\bar{x}-{\mu }_0)/SE=(55-50)/2.309\approx 2.17$ with $df=11$.

   Key takeaway

   One-sample $t$ statistic: $t = (\bar{x} - \mu_0)/(s/\sqrt{n})$ with $df = n - 1$.

7. Question 7 · Easy

   Two independent random samples are taken from two populations. Group 1: $n_1=20$, ${\bar{x}}_1=45$, $s_1=6$. Group 2: $n_2=20$, ${\bar{x}}_2=41$, $s_2=7$. What is the test statistic for testing $H_0:{\mu }_1={\mu }_2$?

   • A
     $t\approx 0.615$

     Why not A: Forgetting to take the square root of the sum of variances/n.
   • B
     $t\approx 1.96$Correct
   • C
     $t\approx 4.0$

     Why not C: Using $s_1+s_2$ instead of $\sqrt{s_1^2/n_1+s_2^2/n_2}$ in the denominator.
   • D
     $t\approx 0.44$

     Why not D: Dividing by $n$ without squaring $s$ first in the standard error formula.
   Explanation

   $SE=\sqrt{s_1^2/n_1+s_2^2/n_2}=\sqrt{36/20+49/20}=\sqrt{1.8+2.45}=\sqrt{4.25}\approx 2.06$. $t=(45-41)/2.06\approx 1.94\approx 1.96$.

   Key takeaway

   Two-sample $t$ statistic: $t = (\bar{x}_1 - \bar{x}_2)/\sqrt{s_1^2/n_1 + s_2^2/n_2}$.

8. Question 8 · Easy

   A study finds a statistically significant difference between two teaching methods at $\alpha =0.05$. The mean score difference is 2 points on a 100-point exam. What is the most important follow-up consideration?

   • A
     The study should be replicated because $n$ must have been very small.

     Why not A: Very small $n$ makes it harder, not easier, to detect differences; a 2-point difference with significance suggests a large $n$.
   • B
     The 2-point difference may be statistically significant but not practically meaningful for educational decisions.Correct
   • C
     A one-sided test should have been used instead of a two-sided test.

     Why not C: Changing the test direction post-hoc is not valid; the result is already significant.
   • D
     The confidence interval should be computed to replace the hypothesis test.

     Why not D: Confidence intervals and hypothesis tests are complementary; neither replaces the other, and practical significance concerns both.
   Explanation

   A 2-point difference on a 100-point exam is tiny and unlikely to change how teachers teach. With large enough samples, even trivially small differences become statistically significant. Always consider effect size and practical significance alongside $p$-values.

   Key takeaway

   Statistical significance does not equal practical significance; always evaluate the effect size and its real-world meaning.

9. Question 9 · Medium

   Conditions for a one-sample $t$-test include randomness, independence (10% condition), and normality. A sample of $n=8$ shows a slightly skewed dotplot with no outliers. Is the normality condition met?

   • A
     No; $n=8<30$, so normality cannot be assumed by the CLT.

     Why not A: For small samples, we check the data plot, not just the sample size. CLT is for large $n$, but small samples from roughly normal populations are acceptable.
   • B
     Yes; slight skew with no outliers in a small sample is considered acceptable for the $t$-procedure.Correct
   • C
     No; the $t$-procedure always requires $n\ge 30$.

     Why not C: The $t$-distribution was designed for small samples; the CLT's $n\ge 30$ rule is for the normal approximation with unknown shape, not for the $t$-test.
   • D
     Yes; the $t$-procedure is valid for any sample distribution.

     Why not D: The $t$-procedure is not valid for small samples with strong skew or outliers, which would indicate the population is far from normal.
   Explanation

   For the $t$-procedure, normality can be assumed if: the population is known to be normal, $n\ge 30$ (CLT), or for small $n$ the sample shows no outliers and no strong skewness. Slight skewness with $n=8$ and no outliers is considered acceptable.

   Key takeaway

   Normality for $t$-procedures: OK if $n \geq 30$, or small $n$ with no outliers/strong skew; always check the data plot.

10. Question 10 · Medium

    A paired $t$-test produces $\bar{d}=5.2$, $s_d=7.1$, $n=18$. Compute the $t$ statistic for testing $H_0:{\mu }_d=0$ vs. $H_a:{\mu }_d>0$.

    • A
      $t\approx 0.73$

      Why not A: Using $s_d=7.1$ in the denominator without dividing by $\sqrt{n}$: $5.2/7.1\approx 0.73$.
    • B
      $t\approx 3.11$Correct
    • C
      $t\approx 3.56$

      Why not C: Using $df=n=18$ to find $SE$; should use $SE=s_d/\sqrt{n}=7.1/\sqrt{18}$.
    • D
      $t\approx 1.67$

      Why not D: Dividing by $s_d/n$ instead of $s_d/\sqrt{n}$.
    Explanation

    $S{E}_d=s_d/\sqrt{n}=7.1/\sqrt{18}\approx 7.1/4.243\approx 1.673$. $t=\bar{d}/S{E}_d=5.2/1.673\approx 3.11$ with $df=17$.

    Key takeaway

    Paired $t$ statistic: $t = \bar{d}/(s_d/\sqrt{n})$ with $df = n - 1$ where $n$ = number of pairs.

11. Question 11 · Hard

    A 95% confidence interval for the difference in mean scores ${\mu }_1-{\mu }_2$ between two schools is $(-3,11)$. What can you conclude at $\alpha =0.05$?

    • A
      There is convincing evidence that School 1 scores higher than School 2.

      Why not A: The interval includes 0, meaning the two-sided test fails to reject $H_0:{\mu }_1={\mu }_2$.
    • B
      We fail to reject $H_0:{\mu }_1={\mu }_2$ because 0 is contained in the confidence interval.Correct
    • C
      The mean scores are definitely the same for both schools.

      Why not C: Failing to reject $H_0$ does not prove $H_0$ is true.
    • D
      School 1 scores are higher on average because the midpoint (4) is positive.

      Why not D: The midpoint does not determine statistical significance; the interval contains 0, so the difference is not significant.
    Explanation

    Because 0 is contained in the 95% CI for ${\mu }_1-{\mu }_2$, the two-sided test at $\alpha =0.05$ fails to reject $H_0:{\mu }_1={\mu }_2$. We cannot conclude there is a significant difference between the schools.

    Key takeaway

    If 0 is in the confidence interval for $\mu_1 - \mu_2$, fail to reject $H_0: \mu_1 = \mu_2$ at the corresponding $\alpha$.

12. Question 12 · Hard

    A one-sample $t$-test for the mean yields $t=3.8$ with $df=24$ and a two-sided $p$-value of 0.0009. A student concludes that because $p=0.0009$, there is only a 0.09% chance the null hypothesis is true. Identify the error.

    • A
      The student should use a one-sided $p$-value, which would be 0.00045.

      Why not A: The problem states a two-sided test; the error is the interpretation, not the direction.
    • B
      The $p$-value is the probability of obtaining data this extreme if $H_0$ is true, not the probability that $H_0$ is true.Correct
    • C
      With $p=0.0009$, the researcher should use $\alpha =0.001$ instead of 0.05.

      Why not C: $\alpha$ is set in advance; the $p$-value is not used to choose $\alpha$ retroactively.
    • D
      A $t$-value of 3.8 with $df=24$ does not correspond to $p=0.0009$.

      Why not D: For $df=24$, $t=3.8$ does indeed correspond to a very small two-sided $p$-value, approximately 0.001.
    Explanation

    The $p$-value is $P(|T|\ge 3.8\mid H_0\text{ true})$ — the probability of observing data this extreme assuming $H_0$ is true. It is not the probability that $H_0$ is true. $H_0$ is a fixed statement (true or false); probabilities are not assigned to hypotheses in frequentist statistics.

    Key takeaway

    $p$-value is the probability of the observed data (or more extreme) given $H_0$ is true — not the probability that $H_0$ is true.